<![CDATA[The Knot]]>https://alex1222.com/RSS for NodeFri, 27 Mar 2026 14:31:26 GMT<![CDATA[Fix: Aliyun OSS Internal IP Overlaps with Tailscale CGNAT and Breaks Access (Debian)]]>

0.1 Contents

1 Fix: Aliyun OSS Internal IP Overlaps with Tailscale CGNAT and Breaks Access (Debian)

1.1 Problem

Tailscale assigns node IPv4 addresses from the CGNAT range 100.64.0.0/10 (RFC 6598).
Aliyun OSS internal endpoints can also resolve to IPs within this same range.

For example, according to Aliyun official documentation, the Hong Kong OSS internal endpoint oss-cn-hongkong-internal.aliyuncs.com may resolve to addresses in:

  • 100.115.61.0/24
  • 100.99.103.0/24
  • 100.99.104.0/24
  • 100.99.106.0/24

I first noticed the issue when OSS access via the internal endpoint failed on an Aliyun ECS instance.
I also saw similar failures when accessing the Aliyun internal Debian package mirror.

Switching Debian mirror traffic to a public mirror is a workable workaround.
However, switching OSS traffic to the public endpoint is not ideal because it can increase costs: internal OSS traffic is free, while public endpoint traffic is billed.

Packet-level testing showed that traffic to these /24 ranges left via the Aliyun internal interface, and replies arrived at the interface, but applications still timed out (for example, ping).

1.2 Root cause

From Tailscale documentation: when netfilter-mode is enabled, Tailscale installs iptables rules to prevent CGNAT spoofing:

Traffic from 100.64.0.0/10 that does not arrive through tailscale0 is dropped.

In practice, the ts-input chain includes a rule like:

  • -A ts-input -s 100.64.0.0/10 ! -i tailscale0 -j DROP

So replies from OSS hosts (for example, 100.115.61.1) that arrive on ens5 (or another non-tailscale0 interface) are dropped before they reach user-space applications—even though packet capture confirms the replies are arriving.

1.3 Solution

Add a narrow allowlist exception for the specific OSS /24 ranges before Tailscale’s CGNAT anti-spoof DROP rule.

The simplest approach is to insert RETURN rules into ts-input:

  • match packets with source IPs in the specific OSS /24 ranges
  • only when they arrive on interfaces other than tailscale0

This keeps Tailscale’s anti-spoof protection for the rest of 100.64.0.0/10, while allowing known legitimate ranges on your LAN/VPC path.

Disabling netfilter integration entirely also fixes the problem, but weakens protection more broadly.
A targeted allowlist is the safer compromise.

1.4 One-time manual fix (immediate)

Insert exceptions into ts-input before the CGNAT DROP rule:

sudo iptables -I ts-input 3 -s 100.115.61.0/24 ! -i tailscale0 -j RETURN
sudo iptables -I ts-input 3 -s 100.99.103.0/24 ! -i tailscale0 -j RETURN
sudo iptables -I ts-input 3 -s 100.99.104.0/24 ! -i tailscale0 -j RETURN
sudo iptables -I ts-input 3 -s 100.99.106.0/24 ! -i tailscale0 -j RETURN

Verify rule order:

sudo iptables -S ts-input

Verify connectivity:

ping -c 3 100.115.61.1

Tailscale may recreate or rewrite ts-* iptables chains during restart or upgrade.
For reliability, re-apply the allowlist after tailscaled starts.

1.5.1 Script

Create:

/usr/local/sbin/tailscale-cgnat-allow-corp.sh

#!/bin/sh
set -eu

# OSS/internal CGNAT subnets that must be allowed on non-tailscale interfaces
SUBNETS="
100.115.61.0/24
100.99.103.0/24
100.99.104.0/24
100.99.106.0/24
"

# Wait briefly for tailscaled to install ts-input chain (avoid boot race)
for i in 1 2 3 4 5; do
  if iptables -S ts-input >/dev/null 2>&1; then
    break
  fi
  sleep 1
done

# Insert RETURN rules near the top of ts-input, before the CGNAT DROP.
# Use -C first to keep it idempotent (safe to run repeatedly).
for net in $SUBNETS; do
  if ! iptables -C ts-input -s "$net" \! -i tailscale0 -j RETURN 2>/dev/null; then
    iptables -I ts-input 3 -s "$net" \! -i tailscale0 -j RETURN
  fi
done

Make it executable:

sudo chmod 0755 /usr/local/sbin/tailscale-cgnat-allow-corp.sh

1.5.2 systemd unit

Create:

/etc/systemd/system/tailscale-cgnat-allow-corp.service

[Unit]
Description=Allow selected CGNAT subnets outside tailscale0 (Tailscale anti-spoof override)
After=tailscaled.service network-online.target
Wants=tailscaled.service network-online.target

[Service]
Type=oneshot
ExecStart=/usr/local/sbin/tailscale-cgnat-allow-corp.sh

[Install]
WantedBy=multi-user.target

Enable and run:

sudo systemctl daemon-reload
sudo systemctl enable --now tailscale-cgnat-allow-corp.service

1.5.3 Verify persistence across Tailscale restart

sudo systemctl restart tailscaled
sudo iptables -S ts-input | sed -n '1,30p'
ping -c 3 100.115.61.1
]]>https://alex1222.com/posts/2026-02-18-Fix-Aliyun-OSS-Internal-Overlaps-With-Tailscale.mdhttps://alex1222.com/posts/2026-02-18-Fix-Aliyun-OSS-Internal-Overlaps-With-Tailscale.mdWed, 18 Feb 2026 00:00:00 GMT<![CDATA[How to Fix 'Steam Encountered an Unexpected Error During Startup']]>

0.1 Contents

1 How to Fix 'Steam Encountered an Unexpected Error During Startup'

1.1 Introduction

If you've been using Steam without issues and suddenly encounter the error message "Steam encountered an unexpected error during startup," you're not alone. This frustrating error typically appears after a Steam update that didn't go smoothly. What's particularly annoying is that even a fresh installation might not resolve the issue, as Steam automatically updates to the latest version upon launch—which could be the problematic version.

In this blog post, we'll walk through a solution that has worked for many users by forcing Steam to downgrade to a previous, stable version.

1.2 Why This Error Occurs

The "Steam encountered an unexpected error during startup" message usually appears when:

  1. A recent Steam client update has compatibility issues with your system
  2. Steam files became corrupted during an update process
  3. There's a conflict with another application or system component

1.3 Solution: Force Steam to Downgrade

The most effective fix is to force Steam to downgrade to a previous, working version. Here's how to do it using Whisky, a popular compatibility layer for running Windows applications:

1.3.1 Step 1: Set Up the Downgrade Process

  1. Open Whisky and locate your Steam installation
  2. Click on the "Config" dropdown

The Config Dropdown

  1. Find the "Arguments" section

The Arguments Section

  1. Copy and paste the following command:
-forcesteamupdate -forcepackagedownload -overridepackageurl http://web.archive.org/web/20240520if_/media.steampowered.com/client -exitsteam
  1. Click "Run" in the bottom right corner

Steam will now begin downloading a downgraded version and will close automatically once the process completes.

1.3.2 Step 2: Configure Steam to Avoid Future Updates

After Steam has closed:

  1. Go back to your programs list in Whisky
  2. Click the gear icon on Steam again (as in Step 1)
  3. Delete the previous command line from the Arguments section
  4. Replace it with:
-noverifyfiles -nobootstrapupdate -skipinitialbootstrap -norepairfiles -overridepackageurl
  1. Press "Run" again

The second set of arguments prevents Steam from automatically updating to the problematic version, keeping your Steam client functioning correctly.

1.4 Why This Works

This solution works by:

  1. First forcing Steam to download a specific archived version from the Web Archive
  2. Then preventing Steam from verifying, repairing, or updating its files in subsequent launches

1.5 Conclusion

The "Steam encountered an unexpected error during startup" issue can be particularly frustrating since traditional reinstallation methods often won't work. By forcing a downgrade and preventing automatic updates, you can get back to your gaming library without the hassle.

If you continue to experience issues, you might want to check if there are any new stable updates available before trying other more invasive troubleshooting methods.

Have you encountered other Steam errors that were difficult to resolve? Share your experiences in the comments below!

]]>https://alex1222.com/posts/2025-03-12-Fix-Steam-Start-Up-Error.mdhttps://alex1222.com/posts/2025-03-12-Fix-Steam-Start-Up-Error.mdWed, 12 Mar 2025 00:00:00 GMT<![CDATA[Group Theory Note]]>

0.1 Contents

1 Group Theory Note

1.1 Introduction

This is my personal note on group theory.

1.2 Group Actions

1.2.1 Orbit and Stabilizer

1.2.2 Burnside's Lemma

1.2.2.1 Group With 2 Conjugacy Classes

Group G with 2 conjugacy classes, are particularly interesting. As only finite groups have 2 conjugacy classes, well be C_2. And any group with 2 conjugacy classes is simple. Thus, it may serve as a good example to infinite simple groups.

Theorem 1. If G is a group with 2 conjugacy classes, then G is simple.

Proof: As 1 itself is a conjugacy class, the other conjugacy class must be all the other elements.

Let N be a non-trivial normal subgroup of G. Take x \in N such that x\neq 1. Then for any y\in G, y\neq 1, there exists g\in G such that gxg^{-1} = y. Thus, N=G.■

Theorem 2. If G is a finite group with 2 conjugacy classes, then G is isomorphic to C_2.

Proof: As 1 itself is a conjugacy class, the other conjugacy class must be all the other elements.

By Burnside's lemma, we have the following equation:

2 = \frac{1}{|G|}\sum_{g\in G}|C(g)| = 1 + \frac{1}{|G|}\sum_{g\neq 1, g \in G}|C(g)|.

Where C(g) = \{x\in G \mid gxg^{-1} = x\}, which is the elements that is fixed by g. As g must fix 1 and itself, thus, for any g\neq 1, |C(g)| \ge 2.

Therefore, we have the following inequality:

2= 1 + \frac{1}{|G|}\sum_{g\neq 1, g \in G}|C(g)| \ge 1 + \frac{2}{|G|}(|G|-1) = 2.

Which then gives us the equality:

|G| \le 2.

■

1.3 Abelian Group

In this section, we will only focus on Abelian groups. And we will use the additive notation for the group operation.

1.3.1 Free Abelian Group

Definition 1. An Abelian group F is free Abelian if it is a direct sum of infinite cyclic groups. More precisely, there is a subset X \subseteq F of elements of infinite order, called a basis of F, with F = \sum_{x\in X}; i.e., F = \sum\mathbb{Z}.

Theorem 3. There is an infinite p-primary group \mathbb{Z}(p^{\infty}) each of whose proper subgroups is finite and cyclic.

Proof: We define a group G have generators:

X = \{x_0,x_1,\dots,x_i,\dots\}, i\in\mathbb{N}

with relations:

\{ px_0, x_0 - px_1, x_1 - px_2, \dots, x_i - px_{i+1}, \dots \}

We let F be the free Abelian group generated by X. And let R \le F be the subgroup generated by the relations. Let a_i = x_i + R \in F/R = G, as \{x_0,x_1,\dots\} form a basis of F, \{a_0,a_1,\dots\} form a basis of G. And as pa_0 = 0, pa_1 = a_0, pa_2 = a_1, \dots, for every a_i, we have p^{i+1}a_i = 0. So G is a p-primary group.

To prove G is infinite, we prove that a_i are distinct. Assume a_i = a_j for some i< j. Then by previous reasoning, we have a_i = p^{j-i} a_j, which implies (p^{j-i}-1) a_j= 0. And previous reasoning also shows that p^{j+1}a_j = 0. Thus, the order of a_j must divides \gcd(p^{j+1} , p^{j-i}-1) = 1. Which implies a_j = 0, which is a contradiction.

Take H to be a proper subgroup of G. By previous argument, for all j< i, a_j is in the cyclic group generated by a_i. Thus, we have

G = \bigcup_{i=0}^{\infty} \langle a_i \rangle.

If H is finite, then there must be n such that H \subseteq \langle a_n \rangle. Then H is a subgroup of finite and cyclic group, thus H is finite and cyclic.

If H is infinite, then there must be a sequence of elements in H, say \{h_1, h_2, \dots\}, such that h_i \in \langle a_{n_i} \rangle, and h_i \notin \langle a_{n_i-1} \rangle. Assume h_i = ka_{n_i}, then \gcd(k,p) must be 1, as otherwise h_i will be in \langle a_{n_i-1} \rangle. Then h_i also generate \langle a_{n_i} \rangle, which implies a_{n_i} \in H. And thus, a_i \in H, \forall i, which implies H = G.■

1.3.2 Divisible and Reduced Groups

Definition 2. A group G is called divisible if for every a \in G and every positive integer n, there exists an element b \in G such that nb = a.

Example 1. The group \mathbb{Z}(p^\infty) is divisible.

We define a group \mathbb{Z}(p^\infty) have generators:

X = \{x_0,x_1,\dots,x_i,\dots\}, i\in\mathbb{N}

with relations:

\{ px_0, x_0 - px_1, x_1 - px_2, \dots, x_i - px_{i+1}, \dots \}

We let F be the free Abelian group generated by X. And let R \le F be the subgroup generated by the relations. Let a_i = x_i + R \in F/R = G, as \{x_0,x_1,\dots\} form a basis of F, \{a_0,a_1,\dots\} form a basis of G. Also, pa_0 = 0, pa_1 = a_0, pa_2 = a_1, \dots, therefore we have, for all j< i, a_j is in the cyclic group generated by a_i. Thus, we have

G = \bigcup_{i=0}^{\infty} \langle a_i \rangle.

We take n=p^r m such that \gcd(p,m)=1. Then for any element b \in \mathbb{Z}(p^\infty), we have n such that b \in \langle a_n \rangle. Assume b = ka_n, we also take c = ka_{n+r}. Then, by the relation of generators, we have p^r c = b. As m is co-prime with p, by Bézout's identity, there exists x,y such that mx + p^{n+r+1}y = 1. We take d = xc. Then

\begin{aligned} nd &= p^r m xc \\ & = p^r(1-p^{n+r+1}y)c \\ & = p^r c - p^{n+2r+1}y c \\ & = b - 0 \\ & = b. \end{aligned}

Which finish the proof that \mathbb{Z}(p^\infty) is divisible.

Theorem 4 (Injective Property, Baer, 1940). Let D be a divisible group and let A be a subgroup of a group B. If f: A \rightarrow D is a homomorphism, then f can be extended to a homomorphism \phi: B \rightarrow D; that is, the following diagram commutes:

#align(center, commutative-diagram( node((1, 0), [$0$]), node((1, 1), [$A$]), node((1, 2), [$B$]), node((0, 1), [$D$]), arr((1, 0), (1, 1), []), arr((1, 1), (1, 2), [$i$],

Proof: This theorem can be proved by Zorn's lemma, or an equivalent proof of well-ordering principle.

Consider \mathcal{F} to be the set of all pairs (C, g), where C is a subgroup of B containing A and g: C \rightarrow D is a homomorphism extending f. Note that \mathcal{F} is non-empty as (A,f) \in \mathcal{F}.

We define a partial order on \mathcal{F} by (C_1, g_1) \le (C_2, g_2) if C_1 \subseteq C_2 and g_2|_{C_1} = g_1. Then for any chain \{(C_i, g_i)\}_{i\in I} in \mathcal{F}, we can define C = \bigcup_{i\in I} C_i and g: C \rightarrow D by g(c) = g_i(c) for c\in C_i. It is easy to check g is well-defined and g is a homomorphism. Thus, (C,g) is an upper bound of the chain.

By Zorn's lemma, there exists a maximal element (B', g') in \mathcal{F}. If B' \neq B, then there exists b \in B \setminus B'. And we discuss whether B'\cap \langle b \rangle=0.

If B'\cap \langle b \rangle=0, then we can extend g' to B' + \langle b \rangle by defining g'(b) = 0. Which contradicts the maximality of B'.

If B'\cap \langle b \rangle \neq 0, then we can take k to be the smallest positive integer such that kb \in B'. Then as D is divisible, we can find d \in D such that kd = kb. Then we can extend g' to B' + \langle b \rangle by defining g'(b) = d. And it is easy to check that g' is a homomorphism. Which contradicts the maximality of B'.■

Follows from the injective property, we have the following corollary:

Corollary 1. If a divisible group D is a subgroup of a group B, then D is a direct summand of B.

Proof: We consider the following diagram:

#align(center, commutative-diagram( node((1, 0), [$0$]), node((1, 1), [$D$]), node((1, 2), [$B$]), node((0, 1), [$D$]), arr((1, 0), (1, 1), []), arr((1, 1), (1, 2), [$i$],

As D is divisible, we can extend 1_D to \varphi: B \rightarrow D. By theorem in group extension, we have B = D \oplus \ker \varphi.■

Definition 3. If G is a group, then dG is the subgroup of G generated by all divisible subgroups of G.

Note that dG is invariant under all automorphism of G. As image of any divisible subgroup is also divisible.

Lemma 1. For any group F, dG is the unique maximal divisible subgroup of G.

Proof: We first prove that dG is divisible. For any a \in dG and n \in \mathbb{N}, there exists H which is a divisible subgroup of G such that a \in H. As H is divisible, there exists b \in H such that nb = a. And as H is a subgroup of dG, b \in dG.

The maximal is obvious, as if there exists a divisible subgroup H such that dG \subset H \subset G, then H \subset dG, by the definition of dG.

It is unique as if there exists another maximal divisible subgroup H, then \langle H, dG \rangle is also a divisible subgroup, then \langle H, dG \rangle \subset dG.■

Definition 4. A group G is called reduced if dG = 0.

Theorem 5. For every group G, there exists a decomposition G = dG \oplus R, where dG is divisible and R is reduced.

Proof: Since dG is divisible, by corollary of injective property, dG is a direct summand of G. Thus, there exists R\le G such that G = dG \oplus R.

Next, we prove that R is reduced. If R is not reduced, then there exists a non-zero element r \in R such that for every n \in \mathbb{N}, r is divisible by n. Then take an element a \in dG, then a+r \in dG \oplus R is also divisible by n for every n \in \mathbb{N}. Which contradicts the maximally of dG.■

Every abelian group is an extension of the torsion group tG by a torsion free subgroup, but tG need not to be a direct summand of G. However, dG is a direct summand of G.

Definition 5. Given a group G define G[n] = \{x\in G \mid nx = 0\}.

Lemma 2. If G and H are divisible p-primary group, then G \cong H if and only if G[p] \cong H[p].

Proof: The "only if" part is obvious. For the "if" part, we define \phi: G[p] \rightarrow H[p], to be the isomorphism between G[p] and H[p]. Then \phi is also a homomorphism from G[p] to H. By previous lemma, we can extend \phi to a homomorphism \Phi: G \rightarrow H. And we prove that \Phi is injective first.

For injectivity, we prove this by induction on n \ge 1, that if p^n x = 0, and \Phi(x) = 0, then x = 0. The base case n=1 is obvious, as now x \in G[p], which means 0=\Phi(x) = \phi(x), and as \phi is isomorphic, x = 0. Now we assume the statement is true for n-1, and we prove it for n. If p^n x = 0, then p^{n-1} (px) = 0, also \Phi(x) = 0 must implies \Phi(px) = 0, thus by the induction hypothesis, px = 0, which implies x = 0 by the base case.

As \Phi is injective, \Phi: G \rightarrow \text{Im}(\Phi) is an isomorphism. And we define \phi': \text{Im}(\Phi) \rightarrow G as the inverse of \Phi. Also, note that the inverse of \phi' extends the inverse of \phi, as it is easy to check \phi'\mid H[p] = \phi^{-1}. As \phi' is a homomorphism from a subgroup of H to G, by previous lemma, we can extend \phi' to a homomorphism \Phi': H \rightarrow G. As \phi' is a surjection by definition, \Phi' is also a surjection. And also as \Phi' extends \phi^{-1}, apply a similar reasoning as above, we have \Phi' is injective. Thus, \Phi' is an isomorphism.■

Theorem 6. Every divisible group D is a direct sum of copies of \mathbb{Q} and of copies of \mathbb{Z}(p^\infty) for various primes p.

Proof: It is easy to check that tD, the torsion group of D is divisible, so tD is a direct summand of D. Thus, there exists V \subset D such that D= tD \oplus V. As now V is torsion free, and divisible, it is a vector space over \mathbb{Q}, and thus, V is a direct sum of copies of \mathbb{Q}.

Now we prove that tD is a direct sum of copies of \mathbb{Z}(p^\infty), for various primes p. For prime p, let B_p = \{ x \in tD \mid p^nx = 0 \text{ for some } n \in \mathbb{N} \}. Then B_p is a divisible p-primary group. It is easy to check that tD = \bigoplus_{p} B_p. So, we only focus on B_p.

It is easy to check that the group B_p[p] is now a vector space over \mathbb{F}_p, let r_p be the dimension of B_p[p]. By previous lemma, B_p is isomorphic to direct sum of r_p copies of \mathbb{Z}(p^\infty).■

There is an analogy between theorems about free Abelian groups and divisible groups, that may be formalized as follows. Given a commutative diagram containing exact sequences, then its dual diagram is the diagram obtained by reversing all the arrows. For example, the dual diagram of 0 \rightarrow A \rightarrow B is B^* \rightarrow A^* \rightarrow 0, and this leads on the say that "subgroup" and "quotient" are dual concepts.

Theorem 7. Every group G can be embedded in a divisible group D.

Proof: As every Abelian group can be written as F/R, where F is a free Abelian group and R is a subgroup of F. Now F = \sum \mathbb{Z}, so that F \le \sum \mathbb{Q}. Hence, G = F/R = (\sum\mathbb{Z})/R \le (\sum\mathbb{Q})/R. As (\sum\mathbb{Q})/R is a quotient of a divisible group, it is divisible.■

Corollary 2. A group G is divisible if and only if it is a direct summand of any group containing it.

Proof: By previous theorem, we can embed G in a divisible group D. Then G is a direct summand of D. And any direct summand of divisible group is divisible.■

1.3.2.1 Exercises

Exercise 1. If 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 is an exact sequence and if A and C are reduced, then B is reduced.

Proof: We name the maps in the exact sequence as follows:

#align(center, commutative-diagram( node((0, 0), [$0$]), node((0, 1), [$A$]), node((0, 2), [$B$]), node((0, 3), [$C$]), node((0, 4), [$0$]), arr((0, 0), (0, 1), []), arr((0, 1), (0, 2), [$i$]), arr((0, 2), (0, 3), [$j$]), arr((0, 3), (0, 4), []), ))

As i is injective, we now assume A \le B. Assume B is not reduced, then dB \neq 0. Then, j(dB) is divisible, as it is a quotient of divisible group. Because C is reduced, j(dB) = 0, which implies, dB \le \ker j = \text{Im} i = A. As A is reduced, dB = 0.■

Exercise 2.

  • Prove that \mathbb{Q}/\mathbb{Z} = \sum_p \mathbb{Z}(p^\infty).
  • Prove that \mathbb{Q}/\mathbb{Z}[n] \cong \mathbb{Z}_n.

Proof: As \mathbb{Q}/\mathbb{Z} is a quotient of divisible group, it is divisible. Also given \frac{a}{b} \in \mathbb{Q}/\mathbb{Z}, we have b \frac{a}{b} = 0. Thus, t\mathbb{Q}/\mathbb{Z} = \mathbb{Q}/\mathbb{Z}, and \mathbb{Q}/\mathbb{Z} contains no summand of \mathbb{Q}.

It is easy to check that:

\mathbb{Q}/\mathbb{Z}[p] = \{ \frac{a}{b} \in \mathbb{Q}/\mathbb{Z} \mid \gcd(a,b) = 1, b = p \}.

Then it is a cyclic group of order p, which is isomorphic to \mathbb{Z}(p^\infty)[p]. Then by previous theorem and lemma, we have, \mathbb{Q}/\mathbb{Z} = \sum_p \mathbb{Z}(p^\infty).

For the second part, we factorize n = p_1^{r_1} \dots p_k^{r_k}. Then we have:

\begin{aligned} \mathbb{Q}/\mathbb{Z}[n] &= (\sum_p \mathbb{Z}(p^\infty))[n] \\ &= \sum_p (\mathbb{Z}(p^\infty)[n]) \\ &= \sum_{i = 1}^{k} (\mathbb{Z}(p_i^\infty)[p_i^{r_i}]) \\ &= \sum_{i = 1}^{k} \mathbb{Z}_{p_i^{r_i}} \\ &= \mathbb{Z}_n. \end{aligned}

■

Exercise 3. Prove that a group G is divisible if and only if G has the injective property.

Proof: The "only if" part is given by previous theorem. For the "if" part, we prove that if a group D contains G, G is a direct summand of D. And then by previous corollary, G is divisible.

By injective property, we have the following diagram, where \varphi is an extension of 1_G.

#align(center, commutative-diagram( node((1, 0), [$0$]), node((1, 1), [$G$]), node((1, 2), [$D$]), node((0, 1), [$G$]), arr((1, 0), (1, 1), []), arr((1, 1), (1, 2), [$i$],

Then by previous theorem about group extension, G is a direct summand of D.■

Exercise 4. A group G is divisible if and only if G = pG for all prime p.

Proof: If G is divisible, then for any element g \in G, as G is divisible, there exists h \in G such that ph = g. Thus, we have G \le pG. And as pG \le G obviously, we have G = pG.

Then, if G = pG for all prime p, then for any n\in \mathbb{N}, we do induction on n. The base case n=1 is obvious, any element g \in G is divisible by 1. Now we assume that any element g \in G is divisible by all natural numbers less than n, and we prove that g is divisible by n. If n is a prime, then as G = nG, there exists h \in G such that nh = g. If n is not a prime, then there exists n_1 \neq 1 and n_2 \neq 1 such that n = n_1 n_2. As g is now divisible by n_1 by induction hypothesis, there exists h_1 \in G such that n_1 h_1 = g. And as h_1 is now divisible by n_2 by induction hypothesis, there exists h_2 \in G such that n_2 h_2 = h_1. Thus, n h_2 = g.■

Exercise 5. A p-primary group G is divisible if and only if G=pG.

Proof: If G is divisible, then for any element g \in G, as G is divisible, there exists h \in G such that ph = g. Thus, we have G \le pG. And as pG \le G obviously, we have G = pG.

Then, if G = pG, then for any n\in \mathbb{N}, we do induction on n. The base case n=1 is obvious, any element g \in G is divisible by 1. Now we assume that any element g \in G is divisible by all natural numbers less than n, and we prove that g is divisible by n. If n = p, then as G = pG, there exists h \in G such that ph = g. If n is a prime that is not p, then we assume the order of g is p^r, and by Bézout identity, there exists x,y such that p^rx + ny = 1. Then we have n (yg) = ny g = (1-p^rx)g = g, thus g is divisible by n. If n is not a prime, then there exists n_1 \neq 1 and n_2 \neq 1 such that n = n_1 n_2. As g is now divisible by n_1 by induction hypothesis, there exists h_1 \in G such that n_1 h_1 = g. And as h_1 is now divisible by n_2 by induction hypothesis, there exists h_2 \in G such that n_2 h_2 = h_1.■

Exercise 6. The following are equivalent for a group G:

  1. G is divisible.
  2. Every non-zero quotient of G is infinite.
  3. G has no maximal proper subgroup.

Proof: From 1 to 2: If G is divisible, then for any non-zero quotient H of G, H is divisible. And by previous theorem, H is a direct summand of \mathbb{Q}, and \mathbb{Z}(p^\infty), thus H is infinite.

From 2 to 3: We assume G has H as a maximal proper subgroup, and derive a contradiction. As H is maximal, G/H is simple. And by 2, G/H is infinite. Take x \in G/H, if x has finite order, then \langle x \rangle is a proper subgroup of G/H, contradicts the simplicity of G/H. If x has infinite order, then \langle 2x \rangle is a proper subgroup of G/H, contradicts the simplicity of G/H.

From 3 to 1: We assume that G has no maximal proper subgroup, and G is now not divisible, and tries to derive a contradiction. As G is not divisible, then there exists p such that G \neq pG. And we assume \varphi: G \rightarrow G/pG is the canonical map. Then G/pG[p] = G/pG, obviously. Thus, G/pG is a vector space over \mathbb{F}_p, we select a basis of G/pG, namely \{x_1, x_2, \dots\}. Then, we define \phi: G/pG \rightarrow \mathbb{F}_p by \phi(m_1 x_2 + m_2 x_2 + \dots) = m_1. Then \phi is a surjection. And \phi\varphi: G \rightarrow \mathbb{F}_p is a surjection. Thus, \ker{\phi\varphi} is a maximal proper subgroup of G, which contradicts the assumption.■

Exercise 7. If G and H are divisible groups each of which is isomorphic to a subgroup of the other, then G \cong H. Is this true if we drop the adjective "divisible"?

Proof: Let \varphi: G \rightarrow H be the embedding of G to the subgroup of H that is isomorphic to G. And let \psi: H \rightarrow G be the embedding of H to the subgroup of G that is isomorphic to H. Then for any prime p, we can restrict \varphi and \psi to G[p] and H[p], and we have \varphi[p]: G[p] \rightarrow H[p] and \psi[p]: H[p] \rightarrow G[p]. As G[p] and H[p] are now vector spaces over \mathbb{F}_p, and as we have the embedding as above, the dimension of G[p] is less than or equal to the dimension of H[p], and vice versa. Thus, the dimension of G[p] is equal to the dimension of H[p]. And thus, G[p] is isomorphic to H[p]. By previous lemma, tG is isomorphic to tH.

For the torsion free part, we define G' = G/tG and H' = H/tH. And \varphi' : G' \rightarrow H' by \phi'(g + tG) = \varphi(g) + tH, and \psi' : H' \rightarrow G' by \psi'(h + tH) = \psi(h) + tG. To prove \varphi' is well-defined, we assume g + tG = g' + tG, then there exists t \in tG such that g = g' + t, thus we have:

\begin{aligned} \varphi'(g+tG) - \varphi'(g'+ tG) &= \varphi(g) - \varphi(g') + tH \\ &= \varphi(g') + \varphi(t) - \varphi(g') + tH \\ &= \varphi(t) + tH \\ \end{aligned}

As t\in tG, \varphi(t) \in tH, thus \varphi'(g+tG) = \varphi'(g'+ tG). In the same way, we can prove \psi' is well-defined.

We then check that \varphi' is injective. Assume \varphi'(g + tG) = 0 + tH, then \varphi(g) \in tH. As \varphi is an isomorphism between G and \text{Img}(\varphi), the order of g is same as the order of \varphi(g) in \text{Img}(\varphi), thus g \in tG. In the same way, we can prove \psi' is injective.

Also, G/tG and H/tH are divisible, thus are vector spaces over \mathbb{Q}, and as both \varphi' and \psi' are injective, the dimension of G/tG is equal to the dimension of H/tH.

Thus, G is isomorphic to H.

If we drop the adjective "divisible", then the statement is not true. For instance, we take G = \mathbb{Q}^{\omega} = \prod_{i\in \mathbb{N}} \mathbb{Q}. And H = Z \times G. Then G is obviously divisible, but H is not.

We define \varphi: G \rightarrow H by \varphi((q_i)_{i\in \mathbb{N}}) = (0, (q_i)_{i\in \mathbb{N}}). And \psi: H \rightarrow G by \psi((z, (q_i)_{i\in \mathbb{N}})) = (z, (q_{i-1})_{i\in \mathbb{N}\setminus\{0\}}). Then both \varphi and \psi are injective, which are both embedding. But G is not isomorphic to H, as one is divisible and the other is not.■

Exercise 8.

  1. Prove that the following groups are all isomorphic:
    1. \mathbb{R}/\mathbb{Z}.
    2. The circle group \mathbf{T}.
    3. \prod_p\mathbb{Z}(p^{\infty}).
    4. \mathbb{R}\oplus(\mathbb{Q}/\mathbb{Z}).
    5. \mathbb{C}^{\times}.
  2. Prove that t(\mathbb{C}^{\times})\cong \mathbb{Q}/\mathbb{Z}.

Proof: The isomorphism between \mathbb{R}/\mathbb{Z} and the circle group is obvious.

By consider a real number as decimal representation, the cardinality of \mathbb{R}/\mathbb{Z} is the same as \prod_p\mathbb{Z}(p^{\infty}). And it is also easy to verify that the torsion group of \mathbb{R}/\mathbb{Z} and \prod_p\mathbb{Z}(p^{\infty}) is isomorphic to \sum_p\mathbb{Z}(p^{\infty}). Thus, (\mathbb{R}/\mathbb{Z})/t(\mathbb{R}/\mathbb{Z}) and (\prod_p\mathbb{Z}(p^{\infty}))/t(\prod_p\mathbb{Z}(p^{\infty})) must have the same cardinality, by looking at the cardinal number division. And we could say (\mathbb{R}/\mathbb{Z})/t(\mathbb{R}/\mathbb{Z}) and (\prod_p\mathbb{Z}(p^{\infty}))/t(\prod_p\mathbb{Z}(p^{\infty})) have the say dimension, again by cardinal number arithmetic. Thus, they are isomorphic.

As \mathbb{Q}/\mathbb{Z} \cong \sum_p\mathbb{Z}(p^{\infty}), (\mathbb{R}\oplus(\mathbb{Q}/\mathbb{Z})) / t(\mathbb{R}\oplus(\mathbb{Q}/\mathbb{Z})) = \mathbb{R}. Thus, we only need to verify that \mathbb{R} is isomorphic to (\mathbb{R}/\mathbb{Z})/(\mathbb{Q}/\mathbb{Z}), which is equal to \mathbb{R}/\mathbb{Q}. As \mathbb{R} is an infinite dimensional vector space over \mathbb{Q}, quotient by \mathbb{Q} will minus one dimension, thus, the dimension of \mathbb{R}/\mathbb{Q} is equal the dimension of \mathbb{R}, which verify that \mathbb{R} is isomorphic to \mathbb{R}/\mathbb{Q}.

As \mathbb{C}^{\times} is isomorphic to \mathbb{R}_{+}^{\times}\oplus\mathbf{T}, which \mathbb{R}_{+}^{\times} is the multiplicative group of positive real numbers. And \mathbb{R}_{+}^{\times} is isomorphic to \mathbb{R}, by the natural logarithm map. Thus, \mathbb{C}^{\times}\cong \mathbb{R} \oplus \mathbf{T}, and t(\mathbb{C}^{\times}) = \mathbb{Q}/\mathbb{Z}. Also (\mathbb{C}^{\times})/t(\mathbb{C}^{\times}) \cong \mathbb{R} \oplus ((\mathbb{R}/\mathbb{Z})/(\mathbb{R}/\mathbb{Z})). By above discussion, we already now that (\mathbb{R}/\mathbb{Z})/(\mathbb{R}/\mathbb{Z}) \cong \mathbb{R}, thus, (\mathbb{C}^{\times})/t(\mathbb{C}^{\times}) \cong \mathbb{R} \oplus \mathbb{R}. It is obvious that \mathbb{R} \oplus \mathbb{R} is then a vector space over \mathbb{Q}, and the dimension of it is equal to square of the dimension of \mathbb{R}. By the cardinal number arithmetic, square of infinite cardinal number is still the same as the original infinite cardinal number. Thus, we have \mathbb{R} \oplus \mathbb{R} \cong \mathbb{R}. Then by previous result, we have \mathbb{R}/\mathbb{Z} \cong \mathbb{C}^{\times}.■

]]>https://alex1222.com/posts/2024-11-26-Group-Theory-Note.mdhttps://alex1222.com/posts/2024-11-26-Group-Theory-Note.mdTue, 26 Nov 2024 00:00:00 GMT<![CDATA[Install K3S on Machine do not have Internet Access]]>

0.1 Contents

1 Install K3S on Machine do not have Internet Access

1.1 Introduction

K3S is a lightweight Kubernetes distribution. It is a great choice for edge computing and IoT devices. I currently have a machine with only ipv6 access, and I want to install K3S on it. Sadly, k3s install script need to download binary from GitHub, which is known to be not ipv6 friendly currently.

1.2 Prerequisites

1.2.1 Firewall

Make sure the machine have open several ports to allow it to communicate with other nodes.

ufw allow 6443/tcp #apiserver
ufw allow from 10.42.0.0/16 to any #pods
ufw allow from 10.43.0.0/16 to any #services

1.2.2 Kernel Modules

Through trying, I found that the machine need to have br_netfilter and overlay kernel modules loaded. Use the following command to check, if it gives a fatal message, it probabily means k3s may not supported on this machine. Also, see https://github.com/k3s-io/k3s/issues/1236.

modprobe br_netfilter
modprobe overlay

1.3 Solution

First, from the k3s release page download the airgap image, and transfer it to the machine you want to install k3s on. Then, download the install script from https://get.k3s.io, and put it in the same directory as the airgap image, and make sure the script is executable.

chmod +x install.sh

Then, download the k3s binary from the release page, and put it into where ever you install executable, normally it is /usr/local/bin.

mv k3s /usr/local/bin
chmod +x /usr/local/bin/k3s

Now, you can run the install script:

INSTALL_K3S_SKIP_DOWNLOAD=true INSTALL_K3S_EXEC='server --token=SECRET' ./install.sh

To join another node to the cluster, you can run:

INSTALL_K3S_SKIP_DOWNLOAD=true INSTALL_K3S_EXEC='server --server https://server-ip:6443 --token=SECRET' ./install.sh

1.4 Uninstall K3S

1.4.1 Uninstall K3S on Server

/usr/local/bin/k3s-uninstall.sh

1.4.2 Uninstall K3S on Agent

/usr/local/bin/k3s-agent-uninstall.sh
]]>https://alex1222.com/posts/2024-08-16-Install-K3S.mdhttps://alex1222.com/posts/2024-08-16-Install-K3S.mdFri, 16 Aug 2024 00:00:00 GMT<![CDATA[Markov Chain Note]]>

0.1 Contents

0.2 Stochastic Process

A stochastic process is a collection of random variables X_t indexed by time t.

0.3 Markov Property

A stochastic process \{X_t\} has the Markov property if for any n \geq 0, any t_0 < t_1 < \cdots < t_n, and any B \in \mathcal{B}, we have

P(X_{t_{n+1}} \in B | X_{t_0}, X_{t_1}, \ldots, X_{t_n}) = P(X_{t_{n+1}} \in B | X_{t_n}).

0.4 Markov Chain

A Markov chain is a stochastic process with the Markov property.

0.4.1 Transition Probability

The transition probability at time n of a Markov chain is defined as

P_{ij} = P(X_{n+1} = j | X_n = i).

0.4.2 Transition Probability Matrix

The transition probability matrix at time n of a Markov chain is a matrix P = (P_{ij}) where P_{ij} is the transition probability from state i to state j.

0.4.3 n-Step Transition Probability

The n-step transition probability from time m to n+m of a Markov chain is defined as

P_{ij}^{(n)} = P(X_{n+m} = j | X_m = i).

0.4.4 Time-Homogeneous Markov Chain

A Markov chain is time-homogeneous if the transition probabilities are independent of time.

In this note, we focus on time-homogeneous Markov chains.

0.5 Chapman-Kolmogorov Equation

The Chapman-Kolmogorov equation states that for a time-homogeneous Markov chain, the (n+m)\text{-step} transition probability is given by

P_{ij}^{(n+m)} = \sum_{k} P_{ik}^{(n)} P_{kj}^{(m)}.

0.6 Hitting Probability

The hitting probability of a state j starting from state i is defined as

f_{ij} = P(\text{state } j \text{ is hit starting from state } i)

By intuition, the hitting probability could be calculated by

f_{ij} = \sum_{n=1}^{\infty} P(\text{state } j \text{ is first hit at time } n | X_0 = i).

Thus, we have the following definition.

0.6.1 First-Visit(Hitting) Probability

The first-visit probability of a state j starting from state i is defined as

f_{ij}^{(n)} = P(\text{state } j \text{ is first hit at time } n | X_0 = i).

0.6.1.1 Convergence of First-Visit Probability

The first-visit probability will always converge to 0 as n goes to infinity.

Proof:

If this is not the case, then the sum of hitting probabilities will be greater than 1, which is a contradiction.

0.6.2 Hitting Time for a Given Path

The hitting time of a state j to state i for a given path i_0, i_1, \ldots is defined as

T_{j}^i = \min\{n \geq 1: X_n = i | X_0 = j\}

Sometimes we use T^i if it is clear that the starting state is j.

0.6.3 Mean Hitting Time (Mean First-Visit Time)

The average hitting time of a state j to state i is defined as

\beta_{j}^i = E[T_{j}^i | X_0 = j].

0.7 Communication

We say that the state i communicates with state j if

P_{ij}^{(n)} > 0 \text{ for some } n \geq 0

and is denoted by i \rightarrow j.

0.7.1 Inter-Communication

Two states i and j are said to inter-communicate if i \rightarrow j and j \rightarrow i.

0.7.2 Irreducible Class of States

A set of states C is said to be an irreducible class of states if any state in C communicates with any other state in C.

0.7.3 Closed Class of States

A set of states C is said to be a closed class of states if any state in C communicates only with other states in C.

0.7.4 Absorbing State

A state i is said to be an absorbing state if itself forms a closed class of states.

0.8 Recurrence

0.8.1 Recurrent and Transient State

A state i is said to be recurrent if the hitting probability

f_{ii} = 1.

A state i is said to be transient if it is not recurrent.

0.8.2 Recurrent and Transient Class

A class of states C is said to be recurrent if any state in C is recurrent.

A class of states C is said to be transient if any state in C is transient.

0.8.3 Two Important Theorems

  1. A state i is recurrent if and only if
\sum_{n=1}^{\infty} P_{ii}^{(n)} = \infty.

Proof:

We define the following generating function:

f(s) = \sum_{n=1}^{\infty} f_{ii}^{(n)} s^n

Then, the probability of hitting state i at time n for the k-th time is the coefficient of s^n in the expansion of f(s)^k.

Thus, we the transition probability:

P_{ii}^{(n)} = \sum_{k=1}^{n} P(\text{hitting }i\text{ at time }n\text{ for the }k\text{-th time})

which should equal to the coefficient of s^n in the expansion of \sum_{k=1}^{n}f^k(s).

As the coefficient of s^i for i<k in f(s)^k is 0, we have

\begin{align} P_{ii}^{(n)} &= \text{coefficient of }s^n\text{ in }\sum_{k=1}^{n} f^k(s) \\ &= \text{coefficient of }s^n\text{ in }f^{n+1}(s) + \text{coefficient of }s^n\text{ in }\sum_{k=1}^{n} f^k(s) \\ &= \text{coefficient of }s^n\text{ in }\sum_{k=1}^{n+1} f^k(s) \text{(by repeating similar steps)} \\ &= \text{coefficient of }s^n\text{ in }\sum_{k=1}^{\infty} f^k(s) \end{align}

Thus, we have

\sum_{n=1}^{\infty} P_{ii}^{(n)} = \sum_{n=1}^{\infty} f^k(1)

If i is recurrent, then f(1) = 1 the previous sum diverges.

If i is transient, then f(1) < 1 and the previous sum is a sum of a geometric series, which converges.

  1. If i is transient, then for all j
P_{ji}^{(n)} \rightarrow 0 \text{ as } n \rightarrow \infty.

Proof:

If the condition does not satisfy, we assume that a=\lim_{n\rightarrow\infty}P_{ji}^{(n)}, and as i is transient, let \alpha = f_{ii}<1. We choose \epsilon, such that \frac{a(1-\alpha)}{3+\alpha} > \epsilon >0. Then there exist N such that for all n\ge N, a-\epsilon<P_{ji}^{(n)}<a+\epsilon, and f_{ji}^{(n)} < \epsilon and \sum_{k=1}^{n}f_{ii}^{(n)} > \alpha-\epsilon

For n>2N+1, we have

\begin{align} P_{ji}^{(n)} &= f_{ji}^{(n)} + \sum_{k=1}^{n-1}P_{ji}^{k}f_{ii}^{n-k} \\ &\le \epsilon+\sum_{k=1}^{n-1}P_{ji}^{k}f_{ii}^{n-k} \\ &\le \epsilon+\epsilon+\sum_{k=N}^{n-1}P_{ji}^{k}f_{ii}^{n-k} \\ &\le 2\epsilon+\sum_{k=N}^{n-1}(a+\epsilon)f_{ii}^{n-k} \\ &= 2\epsilon+(a+\epsilon)\sum_{k=N}^{n-1}f_{ii}^{n-k} \\ &\le 2\epsilon + (a+\epsilon)\alpha \\ &< a-\epsilon \end{align}

which contradicts the assumption.

0.8.4 Finite State Space Must have Recurrent State

If the state space is finite, then there must be at least one recurrent state.

Proof:

If all state are transient, then by the second theorem in previous section, we have P_{ji}^{(n)} \rightarrow 0 as n \rightarrow \infty for all i and j, thus \sum_{i\in S} P_{ji}^{(n)} tends to 0 as n goes to infinity.

However, \sum_{i\in S} P_{ji}^{(n)} is constantly 1 as it is the sum of the transition probability of state j to all other states, which is a contradiction.

0.8.5 Intercommunication and Recurrence

If two states intercommunicate, then they are either both recurrent or both transient.

Proof:

Given two states i and j that intercommunicate.

Suppose i is recurrent, then

\sum_{n=1}^{\infty} P_{ii}^{(n)} = \infty

As i and j intercommunicate, there either exists n such that P_{ij}^{(n)} > 0, and m such that P_{ji}^{(m)} > 0,

Then we have

P_{jj}^{(n+k+m)} \ge P_{ij}^{(n)}P_{jj}^{(k)}P_{ji}^{(m)}

Thus,

\begin{align} \sum_{\alpha=1}^{\infty} P_{jj}^{(\alpha)} &\ge \sum_{\alpha=n+m+1}^{\infty} P_{jj}^{(\alpha)} \\ &\ge \sum_{\alpha=1}^{\infty} P_{ij}^{(n)}P_{jj}^{(\alpha)}P_{ji}^{(m)} \\ &= P_{ij}^{(n)}P_{ji}^{(m)}\sum_{\alpha=1}^{\infty} P_{jj}^{(\alpha)} \end{align}

which is infinite as the sum of P_{jj}^{(\alpha)} is infinite.

Then, j is recurrent.

0.8.6 Number of Visits to a State for a Given Path

The number of visits to a state j for a given path i_0, i_1, \ldots is defined as

N_j = \sum_{n=1}^{\infty} I(X_n = j).

0.8.6.1 Number of Visits at Infinity

\begin{align} P(N_j = \infty) = 1 &\text{ if } j \text{ is recurrent} \\ P(N_j = \infty) = 0 &\text{ if } j \text{ is transient} \end{align}

0.8.7 Decomposition of State Space

The state space can be decomposed into a set of Irreducible Closed Recurrent classes and a set of Transient classes.

Proof:

Given state space I, as the InterCommunicate relation is an equivalence relation, we have a partition of I into Inter Communicate classes. Let the classes be I_{\alpha}, \alpha \in A.

As all states in an InnerCommunicate class have the same Recurrence, let

\begin{equation} B = \{\alpha \in A | I_{\alpha} \text{ is Transient}\} \end{equation}

Let C = A - B. Then, for all \alpha \in C, I_{\alpha} is Recurrent.

We then prove that I_{\alpha} is Closed for all \alpha \in C., which finished the proof of this theorem.

If I_{\alpha} is not closed, then there exists i \in I_{\alpha} and j \notin I_{\alpha}, and n \in \mathbb{N} such that P_{ij}^{(n)} > 0.

Also, as i \in I_{\alpha}, i is Recurrent, and thus

\begin{equation} f_{ii} = 1 \end{equation}

Also, in this case, j can not Communicate with i, as if it does, j will be in I_{\alpha}, which means

\begin{equation} f_{ji} = 0 \end{equation}

Then we have

\begin{align*} 1 &= f_{ii} \\ &= \sum_{k\in I} P_{ik}^{(n)} f_{ki} \\ &= P_{ij}^{(n)}*f_{ji} + \sum_{k\in I, k\neq j} P_{ik}^{(n)} f_{ki} \\ &= \sum_{k\in I, k\neq j} P_{ik}^{(n)} f_{ki} \\ &\le \sum_{k\in I, k\neq j} P_{ik}^{(n)} \\ &= 1- P_{ij}^{(n)} \\ &< 1 \end{align*}

This is a contradiction. Thus, I_{\alpha} is closed.

0.8.8 Mean Recurrence Time

The mean recurrence time of a state i is defined as

\mu_i = E[T_i^i | X_0 = i] = \sum_{n=1}^{\infty} n f_{ii}^{(n)}.

0.8.8.1 Mean Recurrence Time for a Transient State

The mean recurrence time for a transient state is infinite.

0.8.8.2 Positive and Null Recurrent State

A state i is said to be positive recurrent if the mean recurrence time:

\mu_i < \infty.

A state i is said to be null recurrent if the mean recurrence time:

\mu_i = \infty.

0.8.8.3 Identification of Positive and Null Recurrent State

A null recurrent state has similar properties to a transient state:

A recurrent state is null recurrent if and only if

\lim_{n\rightarrow \infty}P_{ii}^{(n)} = 0

0.8.8.4 State in the Same Irreducible Class have the Same Recurrence (Positive or Null)

If two states are in the same irreducible class, then they are either both positive recurrent or both null recurrent.

0.8.8.5 Finite State Space Must have Positive Recurrent State

If the state space is finite, then there must be at least one positive recurrent state.

0.9 Periodicity

0.9.1 Period of a State

The period of a state i is defined as

d_i = \gcd\{n \geq 1: P_{ii}^{(n)} > 0\}

If d_{i} is 1, then the state is said to be aperiodic.

where \gcd is the greatest common divisor.

0.9.2 Period of a Class

All states in the same irreducible class have the same period.

0.10 Stationary Distribution

0.10.1 Stationary Distribution

A distribution \pi = (\pi_1, \pi_2, \ldots) is said to be a stationary distribution of a Markov chain if

\pi = \pi P

where P is the transition probability matrix.

0.10.2 Existence and Uniqueness of Stationary Distribution

If a Markov chain is irreducible and positive recurrent, then it has a unique stationary distribution.

And the stationary distribution is given by

\pi_i = \frac{1}{\mu_i}

where \mu_i is the mean recurrence time of state i.

0.10.3 Ergodic Markov Chain

A Markov chain is said to be ergodic if it is irreducible, positive recurrent and aperiodic.

0.10.3.1 Limiting Distribution

Given an ergodic Markov chain, and any vector \pi^{(0)},

The limit:

\lim_{n\rightarrow \infty} \pi^{(0)} P^n = \pi

where \pi is the stationary distribution.

0.11 Discrete Renewal Process

Define a sequence of independent distributed random variables \{X_n\} with positive integer state space.

Define a stochastic process \{T_n\} as

T_n = T_{n-1} + X_n

where T_0 = 0.

Then, \{T_n\} is a discrete renewal process.

0.11.1 Other Ways to Define Discrete Renewal Process

This first definition is defined by the inter-arrival time. We can directly define the process by the renewal time T_n, which are not independent.

We can also define the process by assign each time n a renewal probability u_n, which is the probability that the process is renewed at time n.

Or we can define the process by the number of renewal N_n, which is the number of renewals in the time interval [0, n].

0.11.2 Mean of Number of Renewals

If we are given the renewal probability u_n, then the mean of the number of renewals in the time interval [0, n] is

E[N_n] = \sum_{k=1}^{n} u_k.

0.11.3 Renewal Process and Markov Chain

If we are given a Markov chain with state space S = \{1, 2, \ldots\}, and we set Y_0 = 1, and we said the process is renewed whenever it hits state 1, then the process is a discrete renewal process.

And the renewal probability is given by

u_n = P(\text{the process is renewed at time } n) = P(Y_n = 1|Y_0=1) = P_{11}^{(n)}

0.11.3.1 Inter-renewal Distribution of Discrete Renewal Process Defined by Markov Chain

The inter-renewal distribution of the discrete renewal process defined by a Markov chain is given by

f_k = P(\text{the precess first reach }1\text{ at } k | Y_0 = 1) = f_{11}^{(k)}

which is the first renewal probability.

0.11.3.2 The Limit Renewal Theorem of Ergodic Markov Chain

Given an ergodic Markov chain with state space S = \{1, 2, \ldots\}, and we set Y_0 = 1, and we said the process is renewed whenever it hits state 1, and we set the number of renewals in the time interval [0, n] as N_n, then

\lim_{n\rightarrow \infty} \frac{E[N_n]}{n} = \frac{1}{E[T_1]}

where E[T_1] is the mean recurrence time of state 1.

0.12 Branching Process

0.12.1 Offspring Distribution

An offspring distribution is a probability distribution of the number of offspring of each individual.

0.12.2 Discrete Branching Process

A discrete branching process with offspring distribution Z is a stochastic process \{X_n\} defined as

X_n = \sum_{i=1}^{X_{n-1}} Z

And X_n is called the size of the nth generation.

0.12.3 The Probability Generating Function of Discrete Branching Process

The probability generating function of a discrete branching process is defined as

F(s) = E[s^{Z}]

where Z is the offspring distribution.

0.12.4 Mean Generation Size

If we are starting from X_{0} = 1, then the mean generation size of the nth generation is given by

E[X_n] = E[Z]^n

where E[Z] is the mean of the offspring distribution.

0.12.5 Total Progeny

The total progeny of the branching process is defined as the total number of individuals that ever exist in the process.

T = \sum_{n=0}^{\infty} X_n

0.12.5.1 Probability Generating Function of Total Progeny

Let G(s) be the probability generating function of the total progeny T, then

G(s) = sF(G(s))

0.12.6 Extinction

We say that the branching process is extinct if X_n = 0 for some n.

0.12.6.1 Probability of Extinction

The probability of extinction is given by the ever visit probability of state 0 starting from state 1.

\gamma = f_{10} = P(\text{extinction}) = \sum_{n=1}^{\infty} f_{10}^{(n)}
0.12.6.1.1 Calculation of Probability of Extinction

The probability of extinction is the least non-negative solution of the equation

s = F(s)

0.13 Random Walk

0.13.1 Simple Random Walk

A simple random walk is a stochastic process \{X_n\} defined as

X_n = X_{n-1} + Y_n

where Y_n is a sequence of independent and identically distributed random variables.

In this note, we only focusing on Y to be a Bernoulli random variable with

P(Y = 1) = p, P(Y = -1) = 1 - p

0.13.1.1 Symmetric Random Walk

If p = 0.5, then the random walk is said to be symmetric.

0.13.1.2 Number of Possible Path

The number of possible paths of a simple random walk from (0,i) to (n,j) is denoted as N_n(i,j).

0.13.1.3 Crossing Condition

Let the number N_{n}^{m}(i,j) denote the number of possible paths of a simple random walk from (0,i) to (n,j) that is at (k,m) for some k.

]]>https://alex1222.com/posts/2024-05-29-Markov-Chain-Note.mdhttps://alex1222.com/posts/2024-05-29-Markov-Chain-Note.mdWed, 29 May 2024 00:00:00 GMT<![CDATA[Quantum Mechanics Note]]>

0.1 Contents

0.2 Probability Basics

0.2.1 Random Variable

A random variable X is a function that maps the sample space \Omega to the real number line \mathbb{R}.

0.2.2 Probability Density Function

The probability density function f(x) of a random variable X is a function that describes the likelihood of the random variable to take on a specific value.

0.2.3 Cumulative Distribution Function

The cumulative distribution function F(x) of a random variable X is a function that describes the probability that the random variable takes on a value less than or equal to x.

F(x) = P(X \leq x) = \int_{-\infty}^x f(x) dx.

0.2.4 Expectation

The expectation of a random variable X is the average value of the random variable.

<X> = \int_{-\infty}^{\infty} x f(x) dx.

0.2.5 Variance

The variance of a random variable X is a measure of how much the values of the random variable vary.

\Delta X^2 = <(X - <X>)^2> = <X^2> - <X>^2.

0.2.6 Standard Deviation

The standard deviation of a random variable X is the square root of the variance.

\Delta X = \sqrt{\Delta X^2}.

0.2.7 Probability Amplitude

The probability amplitude \psi(x) of a random variable X is a complex value function that the likelihood of the random variable to take on value x is given by |\psi(x)|^2.

In other words, the probability density function f(x) is given by

f(x) = |\psi(x)|^2.

0.3 Calculus Basics

0.3.1 Gaussian Integral

The Gaussian integral is given by

\int_{-\infty}^{\infty} e^{-\frac{x^2}{\sigma^2}} dx = \sqrt{\pi}\sigma.

0.3.1.1 Odd Powers of x in Gaussian Integral

The Gaussian integral with odd powers of x is given by

\int_{-\infty}^{\infty} x^{2n+1} e^{-\frac{x^2}{\sigma^2}} dx = 0.

As this is an odd function, the integral is zero.

0.3.1.2 Even Powers of x in Gaussian Integral

The even powers of x in Gaussian integral can be calculated using Feynman's trick.

\begin{align} \int_{-\infty}^{\infty} x^{2n} e^{-ax^2} dx &= \int_{-\infty}^{\infty} \left(-\frac{\partial}{\partial a}\right)^n e^{-ax^2} \\ &= \left(-\frac{\partial}{\partial a}\right)^n \int_{-\infty}^{\infty} e^{-ax^2} dx \\ &= \left(-\frac{\partial}{\partial a}\right)^n \sqrt{\frac{\pi}{a}} \end{align}

0.4 Wave Function

We can postulate that the state of a particle is described by a complex value wave function \Psi(x, t), where x is the position and t is the time.

0.4.1 Born Rule

The probability density to find the particle at position x is given by |\Psi(x, t)|^2. And thus, the probability to find the particle in an measurable area A is given by

P(A) = \int_A |\Psi(x, t)|^2 dx.

0.4.2 Continuity Requirement of Wave Function

The wave function \Psi(x, t) must be continuous and differentiable.

0.4.3 Normalizing Wave Function

The wave function \Psi(x, t) defined on space X is normalized if

\int_{X} |\Psi(x, t)|^2 dx = 1.

If the wave function is not normalized, and if A=\int_{X} |\Psi(x, t)|^2 dx is finite and greater than 0, then the normalized wave function is given by

\Psi_{\text{normalized}}(x, t) = \frac{\Psi(x, t)}{\sqrt{A}}.

0.4.4 Superposition Principle of Wave Functions

If \Psi_1(x, t) and \Psi_2(x, t) are two wave functions, then the superposition of the wave functions is given by

\Psi(x, t) = c_1 \Psi_1(x, t) + c_2 \Psi_2(x, t),

where c_1 and c_2 are complex numbers. And \Psi is also a wave function.

0.4.5 Interference of Wave Functions

The interference of wave functions is a phenomenon where two wave functions \Psi_1(x, t) and \Psi_2(x, t) interfere with each other to form a new wave function \Psi(x, t).

Let \Psi(x, t) = \Psi_1(x, t) + \Psi_2(x, t). Then the probability density of the new wave function is given by

|\Psi(x, t)|^2 = |\Psi_1(x, t)|^2 + |\Psi_2(x, t)|^2 + \Psi_1(x, t)^* \Psi_2(x, t) + \Psi_1(x, t) \Psi_2(x, t)^*

And the \Psi_1(x, t)^* \Psi_2(x, t) + \Psi_1(x, t) \Psi_2(x, t)^* is usually called the interference term.

0.4.6 Classical Plane Wave

The classical plane wave is given by

\Psi(x, t) = A e^{ikx - i\omega t}

where A is the amplitude, k is the wave number, and \omega is the angular frequency.

Note: The classical plane wave is not normalizable, as the integral of the probability density is infinite. In this case, we can use \Psi(x, t) = A e^{ikx - i\omega t -iax^2} where a is a relatively small positive constant. The new wave function is normalizable and have similar behaviour to the plane wave near the origin.

0.5 Schrödinger Equation

0.5.1 Reduced Planck Constant

The reduced Planck constant is given by

\hbar = \frac{h}{2\pi} \approx 1.0545718 \times 10^{-34} \text{ J s}.

where h is the Planck constant.

0.5.2 Quantization Rule

The quantization rule is given by

\begin{align} \hat{p} &= -i \hbar \nabla \\ \hat{E} &= i \hbar \frac{\partial}{\partial t} \end{align}

0.5.3 Schrödinger Equation for Free Particle

In classical mechanics, the energy of a free particle is given by

E = T + V = \frac{p^2}{2m} + V(x).

By applying the quantization rule, the Schrödinger equation (SE) for a free particle with wave equation \psi is given by

\begin{align} \hat{E}\psi &= \left[\frac{\hat{p}^2}{2m} + V\right]\psi \\ i \hbar \frac{\partial \psi}{\partial t} &= \left[-\frac{\hbar^2}{2m} \nabla^2 + V\right]\psi = \left[-\frac{\hbar^2}{2m} \Delta + V\right]\psi \end{align}

0.5.4 Hamiltonian Operator

The Hamiltonian operator \hat{H} is given by

\hat{H} = \frac{\hat{p}^2}{2m} + V(x)

Using the Hamiltonian operator, the Schrödinger equation for a free particle is given by

\hat{E} \frac{\partial \psi}{\partial t} = \hat{H}\psi.

0.5.5 Using Separation of Variables to Solve Schrödinger Equation

The Schrödinger equation can be solved using separation of variables. Let \psi(x, t) = \phi(x) \tau(t). Then the Schrödinger equation becomes

\begin{align} \hat{E}\psi &= \hat{H}\psi \\ \phi \hat{E} \tau &= \tau \hat{H} \phi \\ \frac{\hat{E} \tau}{\tau} &= \frac{\hat{H} \phi}{\phi} = E \\ i\hbar \frac{\partial}{\partial t} \tau &= E \tau \\ \tau &= e^{-iEt/\hbar} \\ \hat{H} \phi &= E \phi \end{align}

Assume, \phi_1,\phi_2,\ldots solve \hat{H} \phi = E \phi with E_1,E_2,\ldots. Then the general solution is given by

\phi(x) = \sum_{n=1}^{\infty} c_n \phi_n(x)e^{-iE_nt/\hbar}.

0.5.6 Time-Independent Schrödinger Equation

By the previous discussion, the task of solving the Schrödinger equation is reduced to solving the equation

\hat{H} \phi_n = E_n \phi_n.

This equation is called the time-independent Schrödinger equation (TISE).

We always call E_n eigenvalues of H. And it is also known as:

  1. Energy lLevels
  2. Eigenenergies
  3. Energy Eigenvalues

And \phi_n are called energy eigenstates of H. And it is also known as:

  1. Stationary States
  2. Energy Eigenfunctions

0.5.6.1 Degeneracy of TISE Solutions

If two or more energy eigenstates have the same energy eigenvalue, then the energy eigenvalue is said to be degenerate.

If two eigenstates have the same eigenvalue, we call it degeneracy of order 2 or twice degenerate.

0.6 Double Slit Experiment

Consider the following double slit experiment setup.

Double Slit Experiment

We can assume the wave function that go through the slits A and B are the same, and is given by the analogue of plane wave \psi(r, t) = \frac{e^{ikx-iwt}}{r}.

Then, for any point P on the screen, the wave function is the superposition of the wave functions from the slits A and B.

\begin{align} \Psi(x, t) &= \psi_A(r_1, t) + \psi_B(r_2, t) \\ &= e^{ikr_1-iwt} + e^{ikr_2-iwt} \\ &= e^{-iwt}\left[\frac{e^{ikr_1}}{r_1} + \frac{e^{ikr_2}}{r_2}\right] \end{align}

where r_1 and r_2 are the distances from the slits A and B to the point P. And is given by

\begin{align} r_1 &= \sqrt{L^2 + (x-d)^2} \\ r_2 &= \sqrt{L^2 + (x+d)^2} \end{align}

Without normalization, the probability density of the wave function is proportional to

\begin{align} |\Psi(x, t)|^2 &= \left|\frac{e^{ikr_1}}{r_1} + \frac{e^{ikr_2}}{r_2}\right|^2 \\ &= \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_1r_2}\left[e^{ik(r_1-r_2)} + e^{-ik(r_1-r_2)}\right] \\ &\approx \frac{1}{L^2}\left[2 + 2\cos(k(r_1-r_2))\right] \\ &= \frac{1}{L^2}\left[2 + 2\cos\left( k \left( \sqrt{L^2 + (x+d)^2} - \sqrt{L^2 + (x-d)^2} \right) \right) \right] \\ &= \frac{1}{L^2}\left[2 + 2\cos\left( k \frac{ 4dx }{ \sqrt{L^2 + (x+d)^2} + \sqrt{L^2 + (x-d)^2} } \right) \right] \\ &\approx \frac{1}{L^2}\left[2 + 2\cos\left( k \frac{ 4dx }{ 2L } \right)\right] \quad\text{As } L \gg x,d \\ &= \frac{1}{L^2}\left[2 + 2\cos\left( k \frac{ 2dx }{ L } \right)\right] \end{align}

We can see that the probability density of the wave function is proportional to a transformed cosine function, which explains the interference pattern on the screen.

0.7 Parity of TISE Solutions

0.7.1 Even potential and Parity

Given a potential V(x): \mathbb{R}\rightarrow \mathbb{C}, the potential is said to be even if

V(x) = V(-x)

In such cases, we can assume the TISE solution has definite parity, that is:

\phi(x) = \pm \phi(-x)

0.7.2 Real Potential and Parity

Given a real potential V(x), we can assume the TISE solution to have definite parity regarding conjugation, that is:

\phi(x) = \phi^*(x)

0.8 States of TISE Solutions

In general there are three kinds of states of TISE solutions:

  1. Bound States
  2. Scattering States
  3. Non-physical States

0.8.1 Bound States

Bound states are states where the energy eigenvalue E_n is less than the potential energy at infinity and greater than the minimum potential energy. In such cases, the TISE solution is normalizable, and the energy eigenvalue is quantized.

Bound States

0.8.2 Scattering States

Scattering states are states where the energy eigenvalue E_n is greater than the potential energy at infinity. In such cases, the TISE solution is not normalizable, and the energy eigenvalue is continuous.

Scattering states can be approximated by normalizable states by similar method as the classical plane wave.

Scattering States

0.8.3 Non-physical States

Non-physical states are states where the energy eigenvalue E_n is less than the minimum potential energy. This case cannot happen in real physical systems.

Non-physical States

0.9 Continuity Equation and Probability Current

Consider the potential to be real. Define the probability density of a wave function P = \Psi(x, t) as |\Psi(x, t)|^2. Then:

\begin{align} \frac{\partial}{\partial t} P &= \frac{\partial \psi}{\partial t} \psi^* + \psi \frac{\partial \psi}{\partial t}^* \\ &= \frac{1}{i\hbar}\left[ \hat{H}\psi\psi^* - \psi(\hat{H}\psi)^* \right] \\ &= -\frac{\hbar^2}{i\hbar 2m}\left[ \frac{\partial^2 \psi}{\partial x^2}\psi^* - \psi\frac{\partial^2 \psi}{\partial x^2}^* \right] \\ &= -\frac{d}{dx}\left[\frac{\hbar^2}{i\hbar 2m}\left[ \frac{\partial \psi}{\partial x}\psi^* - \frac{\partial \psi}{\partial x}\frac{\partial \psi}{\partial x}^* - \psi\frac{\partial \psi}{\partial x}^* + \frac{\partial \psi}{\partial x}\frac{\partial \psi}{\partial x}^* \right]\right] \\ &= -\frac{d}{dx}\left[\frac{\hbar}{2mi}\left[ -\psi\frac{\partial \psi}{\partial x}^* + \psi^*\frac{\partial \psi}{\partial x} \right]\right] \end{align}

We thus define the probability current J as

J = \frac{\hbar}{2mi}\left[ -\psi\frac{\partial \psi}{\partial x}^* + \psi^*\frac{\partial \psi}{\partial x} \right]

The above equation become:

\frac{\partial P}{\partial t} = -\frac{dJ}{dx}

and is called the continuity equation.

0.10 Quantum Tunnelling

We consider a potential barrier of height V_0 and width 2a.

That is, the potential is given by

V(x) = \begin{cases} V_0 & \text{if } |x| < 2a \\ 0 & \text{otherwise} \end{cases}

Then, the TISE can be divided into three regions:

  1. Region I: x < -a
\frac{d^2 \psi}{dx^2} = -\frac{2m}{\hbar^2}E\psi
  1. Region II: -a < x < a
\frac{d^2 \psi}{dx^2} = -\frac{2m}{\hbar^2}(E-V_0)\psi
  1. Region III: x > a
\frac{d^2 \psi}{dx^2} = -\frac{2m}{\hbar^2}E\psi

0.10.1 Simplified Model of Quantum Tunnelling

For the sake of simplicity, let k = \sqrt{\frac{2mE}{\hbar^2}} we can assume the solution of the TISE in Region I and Region III to be

In region I:

\psi(x) = A e^{ikx} + B e^{-ikx}

In region III:

\psi(x) = F e^{ikx}

Quantum Tunnelling

The wave A e^{ikx} can be interpreted as the wave go toward the barrier, and the wave B e^{-ikx} can be interpreted as the wave being reflected by the barrier and wave F e^{ikx} can be interpreted as the wave go through the barrier.

As there is no probability for the particle to be in region II, the probability current must be same at the point x = a and x = -a.

In region I:

J = \frac{\hbar}{2mi}\left[ \psi\frac{-\partial \psi}{\partial x}^* + \psi^*\frac{\partial \psi}{\partial x} \right] = \frac{\hbar k}{m}(|A|^2 - |B|^2)

In region III:

J = \frac{\hbar}{2mi}\left[ \psi\frac{-\partial \psi}{\partial x}^* + \psi^*\frac{\partial \psi}{\partial x} \right] = \frac{\hbar k}{m}|F|^2

Thus, we have

|A|^2 - |B|^2 = |F|^2

We define the reflection probability R and transmission probability T as

\begin{align} R &= \frac{|B|^2}{|A|^2} \\ T &= \frac{|F|^2}{|A|^2} \end{align}

Then, we have

R + T = 1

0.10.2 Quantum Tunnelling through High and Thin Barrier

We consider a barrier that is infinitely high and thin at origin.

To state this formally, we consider the potential to be the limiting behaviour of the Dirac delta function potential.

\delta_L(x) = \begin{cases} \frac{1}{L} & \text{if } -\frac{L}{2} < x < \frac{L}{2} \\ 0 & \text{otherwise} \end{cases}

and

\delta = \lim_{L \to 0} \delta_L(x)

Also, for all function g(x), we have

\lim_{\epsilon \to 0} \int_{-\epsilon}^{\epsilon} g(x) \delta(x) dx = g(0)

Then, the potential is given by

V(x) = \alpha \delta(x)

We again, consider the simplified model in the previous section.

In region I:

\psi(x) = A e^{ikx} + B e^{-ikx}

In region III:

\psi(x) = F e^{ikx}

As the region II is infinity thin, the right hand side of the region I and the left hand side of the region III must be the same.

Thus, we have

A + B = F

Also, by integrating the TISE over the region [-\epsilon,\epsilon], we get:

\begin{align} \int_{-\epsilon}^{\epsilon} \frac{d^2 \psi}{dx^2} dx &= -\frac{2m}{\hbar^2} \int_{-\epsilon}^{\epsilon} (E-\alpha\delta(x))\psi dx \\ \left[\frac{d \psi}{dx}\right]_{-\epsilon}^{\epsilon} &= -\frac{2m}{\hbar^2} \left[E\psi(\epsilon)-E\psi(-\epsilon) - \alpha\int_{-\epsilon}^{\epsilon} \psi\delta dx\right] \\ \end{align}

Taking the limit \epsilon \to 0, we get

\begin{align} \left[\frac{d \psi}{dx}\right]_{0^+} - \left[\frac{d \psi}{dx}\right]_{0^-} &= \frac{2m}{\hbar^2} \alpha \psi(0) \end{align}

Substituting the solution of the TISE in region I and region III, we get

ik(A-B) - ikF = \frac{2m}{\hbar^2} \alpha F

Solving the above equation, we can get the reflection and transmission probability.

0.11 Functional Analysis of Quantum Mechanics

0.11.1 Hilbert Space

Define the Hilbert space \mathbf{H} as the space of all possible wave functions \Psi(x, t) that are square integrable.

\mathbf{H} = \left\{ \Psi(x, t) \mid \int_{-\infty}^{\infty} |\Psi(x, t)|^2 dx < \infty \right\}

The Hilbert space is a complex vector space.

0.11.2 Inner Product

We define the inner product of two wave functions \Psi_1(x, t) and \Psi_2(x, t) as

\left<\Psi_1, \Psi_2\right> = \int_{-\infty}^{\infty} \Psi_1^*(x, t) \Psi_2(x, t) dx

0.11.2.1 Properties of Inner Product

  1. Linearity in Second Argument: For all \Psi_1, \Psi_2 \in \mathbf{H} and a,b \in \mathbb{C}, we have \left<a\Psi_1 + b\Psi_2, \Psi_3\right> = a\left<\Psi_1, \Psi_3\right> + b\left<\Psi_2, \Psi_3\right>.
  2. Anti-linearity in First Argument: For all \Psi_1, \Psi_2 \in \mathbf{H} and a,b \in \mathbb{C}, we have \left<\Psi_1, a\Psi_2 + b\Psi_3\right> = a^*\left<\Psi_1, \Psi_2\right> + b^*\left<\Psi_1, \Psi_3\right>.
  3. Positive Definite: For all \Psi \in \mathbf{H}, we have \left<\Psi, \Psi\right> \geq 0 and \left<\Psi, \Psi\right> = 0 if and only if \Psi = 0.
  4. Conjugate Symmetric (Skew Symmetric): For all \Psi_1, \Psi_2 \in \mathbf{H}, we have \left<\Psi_1, \Psi_2\right> = \left<\Psi_2, \Psi_1\right>^*.

0.11.3 Norm

The norm of a wave function \Psi(x, t) is defined as

||\Psi|| = \sqrt{\left<\Psi, \Psi\right>}

0.11.4 Orthogonality

Two wave functions \Psi_1(x, t) and \Psi_2(x, t) are said to be orthogonal if

\left<\Psi_1, \Psi_2\right> = 0

0.11.5 Angle

The angle between two wave functions \Psi_1(x, t) and \Psi_2(x, t) is defined as

\cos\theta = \frac{\left<\Psi_1, \Psi_2\right>}{||\Psi_1|| ||\Psi_2||}

0.11.6 Orthonormal Basis

A set of wave functions \{\Psi_1(x, t), \Psi_2(x, t), \ldots\} is said to be an orthonormal basis if

  1. The set is orthogonal.
  2. The set is normalized.
  3. The set spans the Hilbert space. That is, for all \Psi(x, t) \in \mathbf{H}, there exists a set of complex numbers \{c_1, c_2, \ldots\} such that
\Psi(x, t) = \sum_{n=1}^{\infty} c_n \Psi_n(x, t)

0.11.7 Operator

An operator is a function that maps a wave function to another wave function.

0.11.8 Hermitian Conjugate

The Hermitian conjugate of an operator \hat{A} is denoted by \hat{A}^\dagger and is the unique operator that satisfies

\left<\hat{A}\Psi_1, \Psi_2\right> = \left<\Psi_1, \hat{A}^\dagger\Psi_2\right>

0.11.9 Hermitian Operator

An Hermitian operator is an operator that satisfies

\left<\hat{A}\Psi_1, \Psi_2\right> = \left<\Psi_1, \hat{A}\Psi_2\right>

for all \Psi_1, \Psi_2 \in \mathbf{H}.

An equivalent definition is that the operator is equal to its Hermitian conjugate.

0.11.10 Spectral Theorem

The spectral theorem states that for all Hermitian operators \hat{A}, there exists an orthonormal basis \{\Psi_1(x, t), \Psi_2(x, t), \ldots\} such that

\hat{A}\Psi_n(x, t) = a_n\Psi_n(x, t)

where a_n are the eigenvalues of \hat{A} and is real.

0.11.10.1 Hamiltonian Operator is Hermitian

The Hamiltonian operator \hat{H} is Hermitian.

Proof:

\begin{align} \left<\hat{H}\Psi_1, \Psi_2\right> &= \int_{-\infty}^{\infty} \Psi_1^*(x, t) \hat{H}\Psi_2(x, t) dx \\ &= \int_{-\infty}^{\infty} \Psi_1^*(x, t) \left[-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi_2}{\partial x^2} + V(x)\Psi_2(x, t)\right] dx \quad \text{By Integration By Part} \\ &= \int_{-\infty}^{\infty} \left[-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi_1^*}{\partial x^2} + V(x)\Psi_1^*(x, t)\right] \Psi_2(x, t) dx \\ &= \left<\Psi_1, \hat{H}\Psi_2\right> \end{align}

0.11.11 Positivity Operators

An operator \hat{A} is said to be positive definite if give any non-zero wave function \Psi(x, t), we have

\left<\hat{A}\Psi, \Psi\right> > 0

An operator \hat{A} is said to be positive semi-definite if give any wave function \Psi(x, t), we have

\left<\hat{A}\Psi, \Psi\right> \geq 0

Any operator given by \hat{A} = \hat{B}^\dagger \hat{B} is positive semi-definite.

0.12 Measurement Postulate

Given a orthonormal basis \{\Psi_1(x, t), \Psi_2(x, t), \ldots\}, the measurement postulate states that the probability of measuring the normalized wave function \Psi(x, t) to be in the state \Psi_n(x, t) is given by

P_n = \left|\left<\Psi_n, \Psi\right>\right|^2

0.12.1 Post Measurement State

After the measurement, the state of the wave function will collapse to the state that is measured.

0.13 Measurement of Observables

Given an observable \hat{A}, it is postulated that \hat{A} is Hermitian and has an orthonormal basis \{\Psi_1(x, t), \Psi_2(x, t), \ldots\}.

The expectation value of the observable \hat{A} of a wave function \psi is given by

\left<\hat{A}\right> = \sum_{n=1}^{\infty} a_n P_n = \left<\psi|\hat{A}\psi\right>

0.14 Commutators and Lie Bracket

In general, two operators \hat{A} and \hat{B} do not commute.

We define the commutator (Lie Bracket) of two operators \hat{A} and \hat{B} as

[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}

0.14.1 Properties of Commutators

  1. Linearity: [a\hat{A} + b\hat{B}, \hat{C}] = a[\hat{A}, \hat{C}] + b[\hat{B}, \hat{C}]
  2. Anti-linearity: [\hat{A}, a\hat{B} + b\hat{C}] = a[\hat{A}, \hat{B}] + b[\hat{A}, \hat{C}]
  3. Linearity in Second Argument: [\hat{C}, a\hat{A} + b\hat{B}] = a[\hat{C},\hat{A}] + b[\hat{C},\hat{B}]
  4. Distributivity: [\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}]

0.14.1.1 Commutator of Position and Momentum Operators

The commutator of position operator \hat{x} and momentum operator \hat{p} is given by

\begin{align} [\hat{x},\hat{p}] &= \hat{x}\hat{p} - \hat{p}\hat{x} \\ &= x\left(-i\hbar\frac{\partial}{\partial x}\right) - \left(-i\hbar\frac{\partial}{\partial x}\right)x \\ &= x\left(-i\hbar\frac{\partial}{\partial x}\right) - \left(-i\hbar\right) - x\left(-i\hbar\frac{\partial}{\partial x}\right) \\ &= i\hbar \end{align}

0.15 Compatibility of Observables

Two observables \hat{A} and \hat{B} are said to be compatible if they commute.

0.16 Robertson Inequality

The Robertson Inequality states that for any two observables \hat{A} and \hat{B}, we have

\Delta \hat{A} \Delta \hat{B} \ge \frac{1}{2} \left|\left<[\hat{A}, \hat{B}]\right>\right|

Proof:

Without loss of generality, we can assume \left<\hat{A}\right> = 0 and \left<\hat{B}\right> = 0. Otherwise we replace \hat{A} with \hat{A} - \left<\hat{A}\right> and \hat{B} with \hat{B} - \left<\hat{B}\right>.

Then, we have

\begin{align} \Delta \hat{A} \Delta \hat{B} &= \sqrt{\left<\hat{A}^2\right>\left<\hat{B}^2\right>} \\ &= \sqrt{\left<\hat{A}\psi|\hat{A}\psi\right>\left<\hat{B}\psi|\hat{B}\psi\right>} \\ &\ge \left|\left<\hat{A}\psi|\hat{B}\psi\right>\right| \quad \text{Cauchy-Schwartz}\\ &= \left|\left<\psi|\hat{A}\hat{B}\psi\right>\right| \\ \end{align}

By similar argument, we have

\Delta \hat{A} \Delta \hat{B} \ge \left|\left<\psi|\hat{B}\hat{A}\psi\right>\right|

Thus,

\begin{align} \Delta \hat{A} \Delta \hat{B} &= \frac{1}{2}\left(\Delta \hat{A} \Delta \hat{B} + \Delta \hat{A} \Delta \hat{B}\right) \\ &\ge \frac{1}{2}\left( \left|\left<\psi|\hat{A}\hat{B}\psi\right>\right| + \left|\left<\psi|\hat{B}\hat{A}\psi\right>\right| \right)\\ &\ge \frac{1}{2}\left| \left<\psi|\hat{A}\hat{B}\psi\right> - \left<\psi|\hat{B}\hat{A}\psi\right> \right| \\ &= \frac{1}{2}\left| \left<\psi|[\hat{A}, \hat{B}]\psi\right> \right| \\ \end{align}

0.16.1 Robertson Inequality for Position and Momentum Operators

Given the position operator \hat{x} and momentum operator \hat{p}, we have

\Delta \hat{x} \Delta \hat{p} \ge \frac{\hbar}{2}

0.17 Quantum Harmonic Oscillator

A quantum harmonic oscillator is a system where the potential energy is given by

V(x) = \frac{1}{2}m\omega^2x^2

0.17.1 Annihilation and Creation Operators

Given the Hamiltonian operator \hat{H} of the quantum harmonic oscillator, we can define the annihilation operator \hat{a} and creation operator \hat{a}^\dagger as

\begin{align} \hat{a} &= u\hat{x} + v\hat{p} i \\ \hat{a}^\dagger &= u\hat{x} - v\hat{p} i \end{align}

Thus, we have

\begin{align} \hat{a}\hat{a}^\dagger &= u^2 \hat{x}^2 - v^2 \hat{p}^2 - uv \hat{x}\hat{p} i + uv \hat{p}\hat{x} i \\ &= u^2 \hat{x}^2 - v^2 \hat{p}^2 + uv\hbar \\ \end{align}

If we take:

\begin{align} u^2 &= \frac{1}{2\hbar}m\omega \\ v^2 &= \frac{1}{2 m\omega\hbar} \\ \end{align}

Then, we have

\begin{align} \hat{a}\hat{a}^\dagger &= \frac{1}{2\hbar}m\omega \hat{x}^2 + \frac{1}{2 m\omega\hbar} \hat{p}^2 + \frac{1}{2} \\ &= \frac{1}{\hbar\omega}\hat{H} + \frac{1}{2} \end{align}

0.17.2 Bosonic Commutation Relation

Given the annihilation operator \hat{a} and creation operator \hat{a}^\dagger of the quantum harmonic oscillator, we have

[\hat{a}, \hat{a}^\dagger] = 2uvh = 1

0.17.3 Number Operator

Given the annihilation operator \hat{a} and creation operator \hat{a}^\dagger of the quantum harmonic oscillator, we can define the number operator \hat{N} as

\hat{N} = \hat{a}^\dagger\hat{a}

Thus, we have

\begin{align} \hat{H} = \hbar\omega\left(\hat{N} + \frac{1}{2}\right) \end{align}

0.17.4 Eigenstates of Number Operator

As the number operator \hat{N} is Hermitian, we can find an orthonormal basis \{\Psi_0(x), \Psi_1(x), \ldots\} such that

\hat{N}\Psi_n(x) = E_n\Psi_n(x)

We next prove that E_n can only be non-negative integers.

Given \Psi_n an eigenvector of \hat{N} with eigenvalue E_n, we have

\begin{align} \hat{N}\hat{a}\Psi_n &= \hat{a}^\dagger\hat{a}^2\Psi_n \\ &= (\hat{a}\hat{a}^\dagger - 1)\hat{a}\Psi_n \\ &= \hat{a}\hat{N} - \hat{a}\Psi_n \\ &= (E_n-1)\hat{a}\Psi_n \\ \end{align}

Thus, if \hat{a}\Psi_n is non-zero, then \hat{a}\Psi_n is also an eigenvector of \hat{N} with eigenvalue E_n-1.

We next prove that \hat{a}\Psi_n = 0 if and only if E_n = 0.

If E_n = 0, then

\begin{align} \left<\hat{a}\Psi_n|\hat{a}\Psi_n\right> &= \left<\Psi_n|\hat{a}^\dagger\hat{a}\Psi_n\right> \\ &= \left<\Psi_n|0\right> \\ &= 0 \end{align}

Thus, \hat{a}\Psi_n = 0.

If \hat{a}\Psi_n = 0, then

\begin{align} \left<\hat{a}\Psi_n|\hat{a}\Psi_n\right> &= \left<\Psi_n|\hat{a}^\dagger\hat{a}\Psi_n\right> \\ &= \left<\Psi_n|\hat{N}\Psi_n\right> \\ &= E_n\left<\Psi_n|\Psi_n\right> \\ &= 0 \end{align}

Thus, E_n = 0.

By proceeding the previous argument, we conclude that \hat{a}^k\Psi_n is an eigenvector of \hat{N} with eigenvalue E_n-k.

If En is not an integer, then k can be any positive integer as the annihilation process can be repeated indefinitely when E_n - k never hit zero. And there exits k such that E_n - k < 0, which is not possible, as \hat{N} = \hat{a}^\dagger\hat{a}, and is positive semi-definite.

Thus, E_n is a non-negative integer.

0.17.5 Ground State of Quantum Harmonic Oscillator

By discussion in the previous section, we can find an normalised eigenstate \Psi_0(x) of the number operator \hat{N} with eigenvalue E_0 = 0.

In this section, we prove that it is unique, up to a complex constant.

By previous discussion, we see that \hat{a}\Psi_0 = 0.

\begin{align} u\hat{x}\Psi_0 + v\hat{p} i \Psi_0 &= 0 \\ u\hat{x}\Psi_0 + v\hbar \frac{d}{dx}\Psi_0 &= 0 \\ \end{align}

As the above equation is a first oder homogeneous ODE, and the solution is unique up to a complex constant.

0.17.6 Creation Operator and Eigenstates of Number Operator

Given the ground state \Psi_0(x) of the quantum harmonic oscillator, and eigenstate \Psi_n(x) of the number operator \hat{N} with eigenvalue E_n = n. Then:

\hat{N}\hat{a}^\dagger\Psi_n = (E_n+1) \hat{a}^\dagger\Psi_n

and

\begin{align} \left<\hat{a}^\dagger\Psi_n|\hat{a}^\dagger\Psi_n\right> &= \left<\Psi_n|\hat{a}\hat{a}^\dagger\Psi_n\right> \\ &= \left<\Psi_n|[\hat{a},\hat{a}^\dagger]+\hat{a}^\dagger\hat{a}\Psi_n\right> \\ &= \left<\Psi_n|\Psi_n\right> + \left<\Psi_n|\hat{N}\Psi_n\right> \\ &= E_n + 1 \end{align}

Thus, we could define the recursive relation:

\Psi_{n+1} = \frac{\hat{a}^\dagger\Psi_n}{\sqrt{E_n+1}}

Repeating the formula, we can get all the eigenstates of the number operator \hat{N}. And could be defined by the ground state \Psi_0(x).

\Psi_n(x) = \frac{(\hat{a}^\dagger)^n\Psi_0(x)}{\sqrt{n!}}

0.17.7 Statistics of Quantum Harmonic Oscillator

As the creation operator \hat{a}^\dagger and annihilation operator \hat{a} are linear combinations of position operator \hat{x} and momentum operator \hat{p}, we can derive the position operator \hat{x} and momentum operator \hat{p} in terms of \hat{a} and \hat{a}^\dagger.

\begin{align} \hat{x} &= \frac{1}{\sqrt{2m\hbar\omega}}(\hat{a} + \hat{a}^\dagger) \\ \hat{p} &= \frac{i}{\sqrt{2m\hbar\omega}}(\hat{a} - \hat{a}^\dagger) \end{align}

Thus, we have

\begin{align} \left<\Psi_n|\hat{x}\Psi_n\right> &= \frac{1}{\sqrt{2m\hbar\omega}}\left<\Psi_n|(\hat{a} + \hat{a}^\dagger)\Psi_n\right> = 0 \\ \end{align}

and

\begin{align} \left<\Psi_n|\hat{p}\Psi_n\right> &= \frac{i}{\sqrt{2m\hbar\omega}}\left<\Psi_n|(\hat{a} - \hat{a}^\dagger)\Psi_n\right> = 0 \\ \end{align}

0.18 Constant of Motion and Commutators

Given an operator \hat{A}, which is probably time dependent, then

\begin{align} \frac{d}{dt}\left<A\right> &= \frac{d}{dt}\left<\Psi|\hat{A}\Psi\right> \\ &= \frac{d}{dt} \int_{-\infty}^{\infty} \Psi^*\hat{A}\Psi dx \\ &= \int_{-\infty}^{\infty} \left(\frac{\partial \Psi^*}{\partial t}\hat{A}\Psi + \Psi^*\frac{\partial \hat{A}}{\partial t}\Psi + \Psi^*\hat{A}\frac{\partial \Psi}{\partial t}\right) dx \\ &= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right> +\int_{-\infty}^{\infty} \left(\frac{\partial \Psi^*}{\partial t}\hat{A}\Psi + \Psi^*\hat{A}\frac{\partial \Psi}{\partial t}\right) dx \\ &= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right> + \left<\frac{\partial \Psi}{\partial t}|\hat{A}\Psi\right> + \left<\Psi|\hat{A}\frac{\partial \Psi}{\partial t}\right> \\ &= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right> + \left<\left(\frac{1}{i\hbar}\hat{H}\Psi\right)|\hat{A}\Psi\right> + \left<\Psi|\hat{A}\left(\frac{1}{i\hbar}\hat{H}\Psi\right)\right> \\ &= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right> + \frac{1}{i\hbar}\left[ - \left<\Psi|\hat{H}\hat{A}\Psi\right> + \left<\Psi|\hat{A}\hat{H}\Psi\right> \right] \\ &= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right> + \frac{1}{i\hbar} \left<[\hat{A},\hat{H}]\right> \end{align}]]>https://alex1222.com/posts/2024-05-27-Quantum-Mechanics-Note.mdhttps://alex1222.com/posts/2024-05-27-Quantum-Mechanics-Note.mdMon, 27 May 2024 00:00:00 GMT<![CDATA[Probability Note]]>

0.1 Contents

0.2 Random Variables and Probability Distributions

0.2.1 Random Variables

A random variable is a function that maps the outcomes \Omega of a random process to numerical values \mathbb{R}. If the \Omega is discrete, the random variable is called a discrete random variable. If the \Omega is continuous, the random variable is called a continuous random variable.

0.2.2 Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of a random variable X is defined as

F(x) = P(X \leq x)

0.2.3 Probability Mass Function (PMF) for Discrete Random Variables

The probability mass function (PMF) of a discrete random variable X is defined as

p(x) = P(X = x)

0.2.4 Probability Density Function (PDF) for Continuous Random Variables

The probability density function (PDF) of a continuous random variable X is defined as

f(x) = \frac{dF(x)}{dx}

where F(x) is the CDF of X.

0.2.5 Expectation and Variance of Random Variables

The expectation of a random variable X is defined as

E[X] = \sum_{x} x p(x) \quad \text{for discrete random variables}

where p(x) is the PMF of X.

E[X] = \int_{-\infty}^{\infty} x f(x) dx \quad \text{for continuous random variables}

where f(x) is the PDF of X.

The variance of a random variable X is defined as

\text{Var}(X) = E[(X - E[X])^2]

By the definition of variance, the variance is always non-negative.

Alternatively, the variance can be calculated as

\text{Var}(X) = E[X^2] - E[X]^2

0.2.5.1 Properties of Expectation and Variance

  1. E[aX + b] = aE[X] + b
  2. E[aX + bY] = aE[X] + bE[Y]
  3. \text{Var}(aX + b) = a^2 \text{Var}(X)
  4. \text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X, Y)

where a and b are constants, and \text{Cov}(X, Y) is the covariance between X and Y.

0.2.6 Moments of Random Variables

The n-th moment of a random variable X is defined as

E[X^n] = \sum_{x} x^n p(x) \quad \text{for discrete random variables}

0.2.7 Standard Deviation of Random Variables

The standard deviation of a random variable X is defined as

\text{SD}(X) = \sqrt{\text{Var}(X)}

0.2.8 Standard Discrete Distributions

0.2.8.1 Bernoulli Distribution

The Bernoulli distribution is a discrete distribution with two possible outcomes: 0 and 1. The PMF of a Bernoulli random variable X is defined as

p(x) = \begin{cases} p & \text{if } x = 1 \\ 1 - p & \text{if } x = 0 \end{cases}

where p is the probability of success.

The expectation and variance of a Bernoulli random variable X are

E[X] = p \quad \text{and} \quad \text{Var}(X) = p(1 - p)

0.2.8.2 Binomial Distribution

The Binomial distribution is a discrete distribution that models the number of successes in a fixed number of independent Bernoulli trials. The PMF of a Binomial random variable X is defined as

p(x) = \binom{n}{x} p^x (1 - p)^{n - x}

where n is the number of trials, x is the number of successes, and p is the probability of success.

The expectation and variance of a Binomial random variable X are

E[X] = np \quad \text{and} \quad \text{Var}(X) = np(1 - p)

0.2.8.3 Poisson Distribution

The Poisson distribution is a discrete distribution that models the number of events occurring in a fixed interval of time or space. The PMF of a Poisson random variable X is defined as

p(x) = \frac{\lambda^x e^{-\lambda}}{x!}

where \lambda is the average rate of events.

The expectation and variance of a Poisson random variable X are

E[X] = \lambda \quad \text{and} \quad \text{Var}(X) = \lambda

0.2.8.4 Geometric Distribution

The Geometric distribution is a discrete distribution that models the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials. The PMF of a Geometric random variable X is defined as

p(x) = (1 - p)^{x - 1} p

where x is the number of trials needed to achieve the first success, and p is the probability of success.

The expectation and variance of a Geometric random variable X are

E[X] = \frac{1}{p} \quad \text{and} \quad \text{Var}(X) = \frac{1 - p}{p^2}

0.2.9 Standard Continuous Distributions

0.2.9.1 Uniform Distribution

The Uniform distribution is a continuous distribution with a constant probability density function (PDF) over a fixed interval. The PDF of a Uniform random variable X is defined as

f(x) = \begin{cases} \frac{1}{b - a} & \text{if } a \leq x \leq b \\ 0 & \text{otherwise} \end{cases}

where a and b are the lower and upper bounds of the interval.

The expectation and variance of a Uniform random variable X are

E[X] = \frac{a + b}{2} \quad \text{and} \quad \text{Var}(X) = \frac{(b - a)^2}{12}

0.2.9.2 Exponential Distribution

The Exponential distribution is a continuous distribution that models the time between events in a Poisson process. The PDF of an Exponential random variable X is defined as

f(x) = \lambda e^{-\lambda x}

where \lambda is the rate parameter.

The expectation and variance of an Exponential random variable X are

E[X] = \frac{1}{\lambda} \quad \text{and} \quad \text{Var}(X) = \frac{1}{\lambda^2}

0.2.9.3 Normal Distribution

The Normal distribution is a continuous distribution that is symmetric and bell-shaped. The PDF of a Normal random variable X is defined as

f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x - \mu)^2}{2\sigma^2}}

where \mu is the mean and \sigma is the standard deviation.

The expectation and variance of a Normal random variable X are

E[X] = \mu \quad \text{and} \quad \text{Var}(X) = \sigma^2

We usually write X \sim N(\mu, \sigma^2) to denote that X follows a Normal distribution with mean \mu and variance \sigma^2.

0.2.9.4 Gamma Distribution

The Gamma distribution is a continuous distribution that generalizes the Exponential distribution. The PDF of a Gamma random variable X is defined as

f(x) = \frac{\lambda^k x^{k - 1} e^{-\lambda x}}{\Gamma(k)}

where \lambda is the rate parameter, k is the shape parameter, and \Gamma(k) is the gamma function.

\Gamma(k) = \int_{0}^{\infty} x^{k - 1} e^{-x} dx

The expectation and variance of a Gamma random variable X are

E[X] = \frac{k}{\lambda} \quad \text{and} \quad \text{Var}(X) = \frac{k}{\lambda^2}

0.2.9.5 Beta Distribution

The Beta distribution is a continuous distribution that is defined on the interval [0, 1]. The PDF of a Beta random variable X is defined as

f(x) = \frac{x^{\alpha - 1} (1 - x)^{\beta - 1}}{B(\alpha, \beta)}

where \alpha and \beta are the shape parameters, and B(\alpha, \beta) is the beta function.

B(\alpha, \beta) = \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha + \beta)}

The expectation and variance of a Beta random variable X are

E[X] = \frac{\alpha}{\alpha + \beta} \quad \text{and} \quad \text{Var}(X) = \frac{\alpha \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)}

0.3 Discrete Bivariate Distributions

A bivariate distribution is a probability distribution that describes the joint behaviour of two random variables.

0.3.1 Joint Probability Mass Function (PMF)

Given two random variables X and Y, the joint probability mass function (PMF) for discrete random variables is defined as

p(x, y) = P(X = x, Y = y)

0.3.2 Marginal Probability Mass Function (PMF)

The marginal probability mass function (PMF) of a random variable X is defined as

p_X(x) = \sum_{y} p(x, y)

0.3.3 Conditional Probability Mass Function (PMF)

The conditional probability mass function (PMF) of a random variable X given Y = y is defined as

p_{X|Y}(x|y) = \frac{p(x, y)}{p_Y(y)}

0.3.4 Expectation and Variance of Bivariate Distributions

The expectation of a bivariate distribution is defined as

E[g(X, Y)] = \sum_{x} \sum_{y} g(x, y) p(x, y)

The covariance of two random variables X and Y is defined as

\text{Cov}(X, Y) = E[(X - E[X])(Y - E[Y])]

Covariance can sometimes be negative, zero, or positive.

And can be calculated as

\text{Cov}(X, Y) = E[XY] - E[X]E[Y]

The correlation coefficient of two random variables X and Y is defined as

\rho(X, Y) = \frac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X) \text{Var}(Y)}}

We can prove that -1 \leq \rho(X, Y) \leq 1.

0.3.5 Independent Random Variables

Two random variables X and Y are independent if and only if

p(x, y) = p_X(x) p_Y(y)

for all x and y.

If X and Y are independent, then

E[XY] = E[X]E[Y] \quad \text{and} \quad \text{Cov}(X, Y) = 0

0.3.6 Uncorrelated Random Variables

Two random variables X and Y are uncorrelated if and only if

\text{Cov}(X, Y) = 0

If X and Y are uncorrelated, then

E[XY] = E[X]E[Y]

Note: Uncorrelated random variables are not necessarily independent.

0.4 Continuous Bivariate Distributions

0.4.1 Joint Cumulative Distribution Function (CDF)

Given two random variables X and Y, the joint cumulative distribution function (CDF) for continuous random variables is defined as

F_{X,Y}(x, y) = P(X \leq x, Y \leq y)

0.4.2 Marginal Cumulative Distribution Function (CDF)

The marginal cumulative distribution function (CDF) of a random variable X is defined as

F_X(x) = P(X \leq x) = P(X \leq x, Y < \infty) = F_{X,Y}(x, \infty)

0.4.3 Joint Probability Density Function (PDF)

Given two random variables X and Y, if there exists a function f(x, y) such that

P((X, Y) \in A) = \iint_{A} f(x, y) dx dy

for all Lebesgue-measurable sets A, then f(x, y) is the joint probability density function (PDF) of X and Y. And X and Y are called jointly continuous random variables.

By the definition of the joint PDF, we have

F_{X,Y}(x, y) = \int_{-\infty}^{x} \int_{-\infty}^{y} f(u, v) du dv

And

f(x, y) = \frac{\partial^2 F_{X,Y}(x, y)}{\partial x \partial y}

0.4.4 Marginal Probability Density Function (PDF)

The marginal probability density function (PDF) of a random variable X is defined as

f_X(x) = \frac{dF_X(x)}{dx} = \int_{-\infty}^{\infty} f(x, y) dy

0.5 Continuous Multivariate Distributions

0.5.1 Joint Cumulative Distribution Function (CDF)

Given n random variables X_1, X_2, \ldots, X_n, let the vector \mathbf{X} = (X_1, X_2, \ldots, X_n), the joint cumulative distribution function (CDF) for continuous random variables is defined as

F_\mathbf{X}(\mathbf{x}) = P(\mathbf{X} < \mathbf{x})

0.5.2 Joint Probability Density Function (PDF)

Given n random variables X_1, X_2, \ldots, X_n, let the vector \mathbf{X} = (X_1, X_2, \ldots, X_n), if there exists a function f(x_1, x_2, \ldots, x_n) such that

P(\mathbf{X} \in A) = \int_{A} f(\mathbf{x}) d\mathbf{X}

for all Lebesgue-measurable sets A, then f(\mathbf{x}) is the joint probability density function (PDF) of \mathbf{X}. And \mathbf{X} are called jointly continuous random variables.

By the definition of the joint PDF, we have

F_\mathbf{X}(\mathbf{x}) = \int_{-\infty}^{x_1} \int_{-\infty}^{x_2} \ldots \int_{-\infty}^{x_n} f(u_1, u_2, \ldots, u_n) du_1 du_2 \ldots du_n

And

f(\mathbf{x}) = \frac{\partial^n F_\mathbf{X}(\mathbf{x})}{\partial x_1 \partial x_2 \ldots \partial x_n}

0.5.3 Marginal Probability Density Function (PDF)

The marginal probability density function (PDF) of a random variable X_i is defined as

F_{X_{k_1}, X_{k_2}, \ldots, X_{k_m}}(x_{k_1}, x_{k_2}, \ldots, x_{k_m}) = \int_{-\infty}^{\infty} \ldots \int_{-\infty}^{\infty} F(x_1, x_2, \ldots, x_n) \prod_{j \neq k_i}dx_j

0.6 Independence of Random Variables

Two random variables X and Y are independent if and only if

F(x, y) = F_X(x) F_Y(y)

for all x and y.

This can be thought as the joint behaviour of X and Y is the product of the marginal behaviour of X and Y.

The definition can also be formulated in terms of the joint PDF:

f(x, y) = f_X(x) f_Y(y)

To show that two random variables are not independent, we only need to find one pair of x and y such that the equation does not hold.

0.6.1 Functions of Independent Random Variables

Given two independent random variables X and Y, and a function g(X), and a function h(Y), the random variables Z = g(X) and W = h(Y) are also independent.

0.6.2 Mutual Independence of Random Variables

A set of random variables X_1, X_2, \ldots, X_n are mutually independent if and only if

F(x_1, x_2, \ldots, x_n) = F_{X_1}(x_1) F_{X_2}(x_2) \ldots F_{X_n}(x_n)

Note: Mutual independence implies pairwise independence, however, the converse is not true.

0.6.3 Identically Independent Random Variables (IID)

A set of random variables X_1, X_2, \ldots, X_n are identically independent (IID) if and only if

  1. They are mutually independent.
  2. They have the same distribution.

0.6.4 Sum of Random Variables

Given two independent random variables X and Y, the sum of X and Y is defined as

Z = X + Y

Then, the CDF of Z can be calculated as

\begin{align} F_Z(z) &= P(Z \leq z) \\ &= P(X + Y \leq z) \\ &= \int_{-\infty}^{\infty} P(X + Y \leq z | X = x) f_X(x) dx \\ &= \int_{-\infty}^{\infty} P(Y \leq z - x) f_X(x) dx \\ &= \int_{-\infty}^{\infty} F_Y(z - x) f_X(x) dx \end{align}

The PDF of Z can be calculated as

f_Z(z) = \int_{-\infty}^{\infty} f_Y(z - x) f_X(x) dx

This is called the convolution of the PDFs of X and Y.

0.7 Expectation, Covariance and Correlation of Multiple Random Variables

0.7.1 Expectation of Multiple Random Variables

Let the vector \mathbf{X} = (X_1, X_2, \ldots, X_n) be a set of random variables, and for Lebesgue-measurable functions g: \mathbb{R}^n \rightarrow \mathbb{R} the expectation of \mathbf{X} is defined as

E[g(\mathbf{X})] = \int_{\mathbb{R}^n} g(\mathbf{x}) f(\mathbf{x}) d\mathbf{X}

0.7.2 Properties of Expectation of Multiple Random Variables

  1. E[a g(\mathbf{X}) + bh(\mathbf{X}) + c] = a E[g(\mathbf{X})] + b E[h(\mathbf{X})] + c
  2. If X and Y are independent, then E[g(X)h(Y)] = E[g(X)]E[h(Y)] for any functions g and h.

0.7.3 Covariance of Multiple Random Variables

The covariance of two random variables X and Y is defined as

\text{Cov}(X, Y) = E[(X - E[X])(Y - E[Y])]

The covariance can be calculated as

\text{Cov}(X, Y) = E[XY] - E[X]E[Y]

0.7.3.1 Properties of Covariance of Multiple Random Variables

  1. \text{Cov}(X, Y) = \text{Cov}(Y, X)
  2. \text{Cov}(X, X) = \text{Var}(X)
  3. \text{Cov}(aX + b, cY + d) = ac \text{Cov}(X, Y)
  4. \text{Cov}(\sum a_iX_{i}, \sum b_iY_{i}) = \sum a_i b_j \text{Cov}(X_i,Y_j)
  5. \text{Cov}(X,Y) = E(XY) - E(X)E(Y)

0.7.3.2 Cauchy-Schwarz Inequality In Terms of Covariance

|\text{Cov}(X, Y)| \leq \sqrt{\text{Var}(X) \text{Var}(Y)}

This equality holds if and only if X and Y are linearly related.

0.7.3.3 Correlation Coefficient of Multiple Random Variables

The correlation coefficient of two random variables X and Y is defined as

\rho(X, Y) = \frac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X) \text{Var}(Y)}}

By Cauchy-Schwarz Inequality, we have -1 \leq \rho(X, Y) \leq 1.

0.7.3.4 Moments of Multiple Random Variables

The n-th (raw) moment of a random variable X is defined as E[X^n], and the n-th central moment of a random variable X is defined as E[(X - E[X])^n].

The joint (raw) moment of random variables X and Y is defined as E[X^iY^j], and the joint central moment of random variables X and Y is defined as E[(X - E[X])^i(Y - E[Y])^j].

0.7.3.5 Expectation and Variance of Multiple Random Variables

Given a set of random variables \mathbf{X} = (X_1, X_2, \ldots, X_n)^T,

E[\mathbf{X}] = (E[X_1], E[X_2], \ldots, E[X_n])^T

which is a vector of n\times1 dimensions vector.

The covariance matrix of \mathbf{X} is defined as

\text{Cov}(\mathbf{X}) = E[(\mathbf{X} - E[\mathbf{X}])(\mathbf{X} - E[\mathbf{X}])^T]

which is a n\times n matrix.

0.8 Conditional Distribution and Expectation

0.8.1 Conditional Distribution

0.8.1.1 Conditional Probability Mass Function (PMF)

Given two random variables X and Y, the conditional probability mass function (PMF) of X given Y = y is defined as

p_{X|Y}(x|y) = \frac{p(x, y)}{p_Y(y)}

0.8.1.2 Conditional Commutative Distribution Function (CDF) for Discrete Random Variables

Given two random variables X and Y, the conditional commutative distribution function (CDF) of X given Y = y is defined as

F_{X|Y}(x|y) = P(X \leq x | Y = y)

0.8.1.3 Conditional Probability Density Function (PDF) for Continuous Random Variables

Given two random variables X and Y, the conditional probability density function (PDF) of X given Y = y is defined as

f_{X|Y}(x|y) = \frac{f(x, y)}{f_Y(y)}

0.8.1.4 Conditional Commutative Distribution Function (CDF) for Continuous Random Variables

Given two random variables X and Y, the conditional commutative distribution function (CDF) of X given Y = y is defined as

F_{X|Y}(x|y) = P(X \leq x | Y = y) = \int_{-\infty}^{x} f_{X|Y}(u|y) du

0.8.2 Conditional Expectation

Given two random variables X and Y, the conditional expectation of X given Y = y is defined as

E[X|Y = y] = \sum_{x} x p_{X|Y}(x|y) \quad \text{for discrete random variables} E[X|Y = y] = \int_{-\infty}^{\infty} x f_{X|Y}(x|y) dx \quad \text{for continuous random variables}

We can also define function of Y as

\psi(y) = E[X|Y = y]

This is a random variable, and we call this the conditional expectation of X given Y.

0.8.2.1 The Law of Iterated Expectations (The Tower Law)

Given two random variables X and Y, the law of iterated expectations states that

E[E[X|Y]] = E[X]

Proof:

\begin{align} E[E[X|Y]] &= \int_{-\infty}^{\infty} E[X|Y = y] f_Y(y) dy \\ &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} x f_{X|Y}(x|y) dx \right) f_Y(y) dy \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x f(x, y) dx dy \\ &= E[X] \end{align}

0.8.2.2 Law of Total Probability

Given a random variable X and an event A, the law of total probability states that

P(A) = \int_{-\infty}^{\infty} P(A|X = x) f_X(x) dx

0.8.2.3 Wald's Equation

Given a random variable X and a stopping time N which is an integer valued random variable, then

E[\sum_{i=1}^{N}X] = E[X]E[N]

0.8.2.4 Properties of Conditional Expectation

  1. E[aX + bY + c|Z] = aE[X|Z] + bE[Y|Z] + c
  2. If Y > 0, then E[X|Y] > 0.
  3. If X and Y are independent, then E[X|Y] = E[X].
  4. For any function g and h, E[g(X)h(Y)|Y] = h(Y)E[g(X)|Y].

0.8.3 Conditional Variance for Multiple Random Variables

Given two random variables X and Y, the conditional variance of X given Y is defined as

\text{Var}(X|Y) = E[(X - E[X|Y])^2|Y]

The conditional variance can be calculated as

\text{Var}(X|Y) = E[X^2|Y] - E[X|Y]^2

Note that the conditional variance is a random variable of Y.

0.8.3.1 Law of Total Variance

Given a random variable X and an event A, the law of total variance states that

\text{Var}(X) = E[\text{Var}(X|Y)] + \text{Var}(E[X|Y])

0.9 Transformations of Random Variables

0.9.1 Support of Probability Density Function (PDF)

Given a random variable X with a PDF f_X(x), the support of f_X(x) is the set of values of x where f_X(x) > 0.

0.9.2 Monotonic Transformations

Given a random variable X with a PDF f_X(x), and a function Y = g(X), if g is a monotonic function, then the CDF of Y is

F_Y(y) = \begin{cases} F_X(g^{-1}(y)) & \text{if } g \text{ is increasing} \\ 1 - F_X(g^{-1}(y)) & \text{if } g \text{ is decreasing} \end{cases}

Then, the PDF of Y is

f_Y(y) = f_X(g^{-1}(y)) \left| \frac{d}{dy} g^{-1}(y) \right|

For non-monotonic transformations, we can break the transformation into monotonic parts.

0.9.3 Transformation of Bivariate Random Variables

Given random variables X_1 and X_2 with a joint PDF f_{X_1, X_2}(x_1, x_2), and functions (Y_1, Y_2) = T(X_1, X_2), where T: \mathbb{R}^2 \rightarrow \mathbb{R}^2 is a one-to-one transformation, and let H = T^{-1}.

We define J_{H}, which is the Jacobian determinate of H as

J_{H} = \left| \frac{\partial (H_1, H_2)}{\partial (x_1, x_2)} \right| = \det\begin{bmatrix} \frac{\partial H_1}{\partial x_1} & \frac{\partial H_1}{\partial x_2} \\ \frac{\partial H_2}{\partial x_1} & \frac{\partial H_2}{\partial x_2} \end{bmatrix}

Then, the joint PDF of (Y_1, Y_2) is

f_{Y_1, Y_2}(y_1, y_2) = f_{X_1, X_2}(H_1(y_1, y_2), H_2(y_1, y_2)) |J_{H}|

Note: The Jacobian determinate satisfy: J_{H} = J_{H^{-1}}^{-1} = J_T^{-1}.

0.9.4 Transformation of Multivariate Random Variables

The theorem from Transformation of Bivariate Random Variables can be generalized to multiple random variables.

Given random variables X_1, X_2, \ldots, X_n with a joint PDF f_{X_1, X_2, \ldots, X_n}(x_1, x_2, \ldots, x_n), and functions (Y_1, Y_2, \ldots, Y_n) = T(X_1, X_2, \ldots, X_n), where T: \mathbb{R}^n \rightarrow \mathbb{R}^n is a one-to-one transformation, and let H = T^{-1}.

We define J_{H}, which is the Jacobian determinate of H as

J_{H} = \left| \frac{\partial (H_1, H_2, \ldots, H_n)}{\partial (x_1, x_2, \ldots, x_n)} \right| = \det\begin{bmatrix} \frac{\partial H_1}{\partial x_1} & \frac{\partial H_1}{\partial x_2} & \ldots & \frac{\partial H_1}{\partial x_n} \\ \frac{\partial H_2}{\partial x_1} & \frac{\partial H_2}{\partial x_2} & \ldots & \frac{\partial H_2}{\partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial H_n}{\partial x_1} & \frac{\partial H_n}{\partial x_2} & \ldots & \frac{\partial H_n}{\partial x_n} \end{bmatrix}

Then, the joint PDF of (Y_1, Y_2, \ldots, Y_n) is

f_{Y_1, Y_2, \ldots, Y_n}(y_1, y_2, \ldots, y_n) = f_{X_1, X_2, \ldots, X_n}(H_1(y_1, y_2, \ldots, y_n), H_2(y_1, y_2, \ldots, y_n), \ldots, H_n(y_1, y_2, \ldots, y_n)) |J_{H}|

0.10 Generating Functions

0.10.1 The Moment Generating Function (MGF)

Given a random variable X, the moment generating function (MGF) M_X(t) of X is defined as

M_X(t) = E[e^{tX}]

The domain of the MGF is the set of t such that M_X(t) exists and is finite.

If the domain does not contain an open neighbourhood of 0, then we say the MGF does not exist.

0.10.1.1 Example: The Moment Generating Function of the Standard Normal Distribution

Given a random variable X that follows the standard normal distribution,

f_X(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}

By definition,

\begin{align} M_X(t) &= E[e^{tX}] \\ &= \int_{-\infty}^{\infty} e^{tx} f_X(x) dx \\ &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{tx-\frac{x^2}{2}} dx \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{tx-\frac{x^2}{2}} dx \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}(x^2 - 2tx)} dx \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}(x - t)^2 + \frac{t^2}{2}} dx \\ &= e^{\frac{t^2}{2}} \end{align}

0.10.1.2 Example: The Moment Generating Function of the Exponential Distribution

Given a random variable X that follows the exponential distribution,

f_X(x) = \lambda e^{-\lambda x}

By definition,

\begin{align} M_{X}(t) &= E[e^{tX}] \\ &= \int_{0}^{\infty} e^{tx} \lambda e^{-\lambda x} dx \\ &= \lambda \int_{0}^{\infty} e^{tx-\lambda x} dx \\ &= \frac{\lambda}{t-\lambda} \int_{0}^{\infty} (t-\lambda) e^{(t-\lambda)x} dx \\ &= \frac{\lambda}{t-\lambda} \end{align}

0.10.1.3 Properties of the Moment Generating Function

M_X(0) = E[1] = 1

The n-th derivative of the MGF at t = 0 is:

M_X^{(n)}(0) = E[X^{n}e^{tX}](0) = E[X^{n}]

By the previous property, the Maclaurin series of the MGF is:

M_X(t) = \sum_{n=0}^{\infty} \frac{E[X^{n}]}{n!} t^{n}

Also, if X have MGF M_X(t) and Y = aX + b, then Y have MGF M_Y(t) = e^{tb}M_X(at).

0.10.1.4 Uniqueness of the Moment Generating Function

Given two random variables X and Y with MGF M_X(t) and M_Y(t), if M_X(t) = M_Y(t) for all t in an open neighbourhood of 0, then X and Y have the same distribution.

0.10.2 Joint Moment Generating Function (JMGF)

The joint moment generating function (JMGF) of random variables X_1, X_2, \ldots, X_n is defined as a function from \mathbb{R}^n to \mathbb{R}:

M_\mathbf{X}(\mathbf{t}) = E[e^{\mathbf{t}^T \mathbf{X}}]

where \mathbf{t} = (t_1, t_2, \ldots, t_n)^T and \mathbf{X} = (X_1, X_2, \ldots, X_n)^T.

If the JMGF exists and is finite on a open neighbourhood of \mathbf{0}, then we say the JMGF exists.

0.10.2.1 Properties of the Joint Moment Generating Function

If the JMGF exists and is finite on a open neighbourhood of \mathbf{0}, then it uniquely determines the joint distribution of X_1, X_2, \ldots, X_n.

The MGF of X_i can be expressed as:

M_{X_i}(t_i) = M_{\mathbf{X}}(0, \ldots, 0, t_i, 0, \ldots, 0)

The joint moment of X_1, X_2, \ldots, X_n can be expressed as:

E[X_1^{i_1}X_2^{i_2}\ldots X_n^{i_n}] = \left. \frac{\partial^{i_1 + i_2 + \ldots + i_n} M_{\mathbf{X}}(\mathbf{t})}{\partial t_1^{i_1} \partial t_2^{i_2} \ldots \partial t_n^{i_n}} \right|_{\mathbf{t} = \mathbf{0}}

0.10.3 Relation Between the Joint Moment Generating Function and Moment Generating Function

Given random variables X_1, X_2, \ldots, X_n with MGF M_{X_i}(t_i), and JMGF M_{X_1, X_2, \ldots, X_n}(\mathbf{t}), then X_1, X_2, \ldots, X_n are mutually independent if and only if

M_{X_1, X_2, \ldots, X_n}(\mathbf{t}) = M_{X_1}(t_1)M_{X_2}(t_2)\ldots M_{X_n}(t_n)

0.10.4 Sums of Independent Random Variables

Given random variables X_1, X_2, \ldots, X_n that are independent, and S = a_1X_1 + a_2X_2 + \ldots + a_nX_n, then the MGF of S is

M_S(t) = M_{X_1}(a_1t)M_{X_2}(a_2t)\ldots M_{X_n}(a_nt)

0.10.5 Probability Generating Function (PGF)

Given a random variable X that takes non-negative integer values, the probability generating function (PGF) G_X(z) of X is defined as

\phi_X(z) = E[z^X] = \sum_{x=0}^{\infty} z^x P(X = x)

0.10.5.1 Properties of the Probability Generating Function

\phi_X(1) = 1

The PMF of X is uniquely determined by \phi_X(z).

The n-th factorial moment of X is

E[X(X-1)\ldots(X-n+1)] = \left. \frac{d^n \phi_X(z)}{dz^n} \right|_{z=1}

Random variables X_1, X_2, \ldots, X_n are mutually independent if and only if the joint PGF \phi_{X_1, X_2, \ldots, X_n}(z_1, z_2, \ldots, z_n) is

\phi_{X_1, X_2, \ldots, X_n}(z_1, z_2, \ldots, z_n) = E[z_1^{X_1}z_2^{X_2}\ldots z_n^{X_n}] =\phi_{X_1}(z_1)\phi_{X_2}(z_2)\ldots \phi_{X_n}(z_n)

The PGF of sum of independent random variables X_1, X_2, \ldots, X_n is

\phi_{X_1 + X_2 + \ldots + X_n}(z) = \phi_{X_1}(z)\phi_{X_2}(z)\ldots \phi_{X_n}(z)

0.10.5.2 Relation of PGF and MGF

Given a random variable X that takes non-negative integer values, and the PGF \phi_X(z) and MGF M_X(t) of X, then

\begin{align} \phi_X(e^t) = M_X(t) \\ M_X(\ln(t)) = \phi_X(t) \end{align}

0.11 Markov and Chebyshev Inequalities

0.11.1 Markov Inequality

Given a non-negative random variable X and a > 0, then

P(X \geq a) \leq \frac{E[X]}{a}

Proof:

\begin{align} P(X \geq c) &= \int_c^{\infty} f_X(x) dx \\ &= \frac{1}{c} \int_c^{\infty} cf_X(x) dx \\ &\le \frac{1}{c} \int_c^{\infty} xf_X(x) dx \\ &\le \frac{1}{c} \int_0^{\infty} xf_X(x) dx &= \frac{E[X]}{c} \end{align}

0.11.2 Chebyshev Inequality

Given a random variable X with mean \mu and variance \sigma^2, and a > 0, then

P(|X - \mu| \geq a) \leq \frac{\sigma^2}{a^2}

Proof:

Define Y = (X - \mu)^2, then Y is a non-negative random variable, and E[Y] = \text{Var}[X] = \sigma^2.

By Markov Inequality,

P(Y \geq a^2) \leq \frac{E[Y]}{a^2} = \frac{\sigma^2}{a^2}

Then,

P(|X - \mu| \geq a) = P((X - \mu)^2 \geq a^2) = P(Y \geq a^2) \leq \frac{\sigma^2}{a^2}

0.12 Multivariate Normal Distribution

We define the higher dimensional normal distribution as an analog of the one dimensional normal distribution.

We say a random vector \mathbf{X} = (X_1, X_2, \ldots, X_n)^T follows a multivariate normal distribution if it can be expressed as

\mathbf{X} = \mathbf{\mu} + \mathbf{A}\mathbf{Z}

where l\le n and \mathbf{\mu} is a vector of means, \mathbf{A} is a n\times l matrix of constants, and \mathbf{Z} is a l\times 1 vector of independent standard normal random variables.

In convention, we write \Sigma = A A^T, and we denote the multivariate normal distribution as

\mathbf{X} \sim N_n(\mathbf{\mu}, \Sigma)

0.12.1 Joint PDF of Multivariate Normal Distribution

If we assume that \Sigma has full rank, We can use multivariate transformation to derive the joint PDF of \mathbf{X}:

\begin{align} f_X(x) &= f_Z(A^{-1}(x-\mu))|\det(A^{-1})| \\ &= |\det(A^{-1})| \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}z_i^2} \\ &= |\det(A^{-1})| \frac{1}{\sqrt{2\pi}^n} e^{-\frac{1}{2}Z^TZ} \\ &= |\det(A^{-1})| \frac{1}{\sqrt{2\pi}^n} e^{-\frac{1}{2}(A^{-1}(x-\mu))^TA^{-1}(x-\mu)} \\ &= |\det(A^{-1})| \frac{1}{\sqrt{2\pi}^n} e^{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)} \\ \end{align}

0.12.2 Joint Moment Generating Function of Multivariate Normal Distribution

If we assume that \Sigma has full rank, the joint moment generating function of \mathbf{X} is

\begin{align} M_{X}(t) &= E[e^{t^TX}] \\ &= \int_{\mathbb{R}^n} e^{t^Tx} f_X(x) dx \\ &= \int_{\mathbb{R}^n} e^{t^Tx} |\det(A^{-1})| \frac{1}{\sqrt{2\pi}^n} e^{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)} dx \\ &= \frac{|\det(A^{-1})|}{\sqrt{2\pi}^n} \int_{\mathbb{R}^n} e^{-\frac{1}{2}\left[ 2t^Tx +x^T\Sigma^{-1}x -x^T\Sigma^{-1}\mu -\mu^T\Sigma^{-1}x +\mu^T\Sigma^{-1}\mu \right]} dx \\ &= \frac{|\det(A^{-1})|}{\sqrt{2\pi}^n} \int_{\mathbb{R}^n} e^{-\frac{1}{2}\left[ 2(\Sigma^{T}t)^T \Sigma^{-1}x +x^T\Sigma^{-1}x -2\mu^T\Sigma^{-1}x +\mu^T\Sigma^{-1}\mu \right]} dx \\ &= \frac{|\det(A^{-1})|}{\sqrt{2\pi}^n} \int_{\mathbb{R}^n} e^{-\frac{1}{2}\left[ +x^T\Sigma^{-1}x -2(\mu-\Sigma^{T}t)^T\Sigma^{-1}x +\mu^T\Sigma^{-1}\mu \right]} dx \\ &= \frac{|\det(A^{-1})|}{\sqrt{2\pi}^n} e^{-\frac{1}{2}\left[ -(\mu-\Sigma^{T}t)^T\Sigma^{-1}(\mu-\Sigma^{T}t) +\mu^T\Sigma^{-1}\mu \right]} \int_{\mathbb{R}^n} e^{-\frac{1}{2}\left[ (x-\mu-\Sigma^{T}t)^T\Sigma^{-1}(x-\mu-\Sigma^{T}t) \right]} dx \\ &= e^{-\frac{1}{2}\left[ -(\mu-\Sigma^{T}t)^T\Sigma^{-1}(\mu-\Sigma^{T}t) +\mu^T\Sigma^{-1}\mu \right]} \\ &= e^{t^T\mu + \frac{1}{2}t^T\Sigma t} \end{align}

0.12.3 Moments of Multivariate Normal Distribution

By the joint moment generating function of multivariate normal distribution,

E[X_{1}^{k_1}X_{2}^{k_2}\ldots X_{n}^{k_n}] = \left. \frac{\partial^{k_1 + k_2 + \ldots + k_n} M_{X}(\mathbf{t})}{\partial t_1^{k_1} \partial t_2^{k_2} \ldots \partial t_n^{k_n}} \right|_{\mathbf{t} = \mathbf{0}}

Especially, as,

\begin{align} \frac{\partial}{\partial t_i} M_{X}(\mathbf{t}) &= \frac{\partial}{\partial t_i} e^{t^T\mu + \frac{1}{2}t^T\Sigma t} \\ &= \left[ \frac{\partial t}{\partial t_i}^T\mu + \frac{1}{2} \frac{\partial t}{\partial t_i}^T\Sigma t + t^T\Sigma \frac{\partial t}{\partial t_i} \right] M_{X}(\mathbf{t}) &= e_i^T \left[ \mu + \Sigma t \right] M_{X}(\mathbf{t}) \end{align} \begin{align} \frac{\partial^2}{\partial t_j\partial t_i} &= \frac{\partial}{\partial t_j} e_i^T \left[ \mu + \Sigma t \right] M_{X}(\mathbf{t}) \\ &= e_i^T \Sigma e_j^T M_{X}(\mathbf{t}) + e_i^T \left[ \mu + \Sigma t \right] e_j^T \left[ \mu + \Sigma t \right] M_{X}(\mathbf{t}) \\ &= e_i^T \Sigma e_j^T M_{X}(\mathbf{t}) + e_i^T \left[ \mu + \Sigma t \right] \left[ \mu + \Sigma t \right]^T e_j M_{X}(\mathbf{t}) \\ &= e_i^T \Sigma e_j^T M_{X}(\mathbf{t}) + e_i^T \left[ \mu \mu^T + \mu (\Sigma t)^T + \Sigma t \mu^T + \Sigma t (\Sigma t)^T \right] e_j M_{X}(\mathbf{t}) \\ \end{align}

where e_i is the i-th unit vector.

Then, we can calculate the moments of the multivariate normal distribution.

E(X_i) = \frac{\partial}{\partial t_i} M_{X}(0) = e_i^T \left[ \mu \right] M_{X}(0) = \mu_i \begin{align} E(X_iX_j) &= \frac{\partial^2}{\partial t_j\partial t_i} M_{X}(0) \\ &= e_i^T \Sigma e_j^T M_{X}(0) + e_i^T \left[ \mu \mu^T + \mu (\Sigma 0)^T + \Sigma 0 \mu^T + \Sigma 0 (\Sigma 0)^T \right] e_j M_{X}(0) \\ &= e_i^T \Sigma e_j^T M_{X}(0) \\ &= \Sigma_{ij} + \mu_i\mu_j \end{align}

And the covariance:

\text{Cov}(X_i, X_j) = E(X_iX_j) - E(X_i)E(X_j) = \Sigma_{ij}

Thus, the covariance matrix of \mathbf{X} is \Sigma.

0.12.4 Bivariate Normal Distribution

Given n=2 in multivariate normal distribution, we have the bivariate normal distribution.

The joint PDF of bivariate normal distribution is

f_{X_1, X_2}(x_1, x_2) = \frac{|\det A^{-1}|}{2\pi} e^{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)}

where \mu = (\mu_1, \mu_2)^T and \Sigma = \begin{bmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22} \end{bmatrix}.

By the moments of multivariate normal distribution, \Sigma can also be expressed as

\Sigma = \begin{bmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22} \end{bmatrix} = \begin{bmatrix} \text{Var}(X_1) & \text{Cov}(X_1, X_2) \\ \text{Cov}(X_1, X_2) & \text{Var}(X_2) \end{bmatrix}

0.12.5 Properties of Multivariate Normal Distribution

0.12.5.1 Affine Transformation of Multivariate Normal Distribution

Given X a multivariate normal distribution, and Y = AX+b, then Y is also a multivariate normal distribution.

0.12.5.2 Marginal Distribution of Multivariate Normal Distribution

Given X a multivariate normal distribution, to get the marginal distribution of Y = (X_{k_1},X_{k_2},\ldots,X_{k_i}), we can let A be a i\times n matrix with 1 at the k_1, k_2, \ldots, k_i-th row, and b be a i\times 1 vector of zeros.

Then Y= AX + b.

0.12.6 Condition of Independence of Multivariate Normal Distribution

Given X a multivariate normal distribution, X_i and X_j are independent if and only if \text{Cov}(X_i, X_j) = 0.

0.12.7 Degenerate Multivariate Normal Distribution

Given X a multivariate normal distribution, if \Sigma is a singular matrix, then X is a degenerate multivariate normal distribution.

Suppose x is a eigenvector of \Sigma with eigenvalue 0, then let Y = x^T X, then the mean of Y is x^T \mu, and the variance of Y is x^T \Sigma x = 0.

Then Y is the distribution of a constant.

0.13 Limiting Behaviors of Sums of Random Variables

In this section, we assume X to be IID random variables. And \mu = E[X], \sigma^2 = \text{Var}(X).

Let S_{n} = \sum_{i=1}^{n}X.

Then,