organizang books
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"cells": [
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"id": "206bf674",
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"metadata": {},
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"outputs": [],
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"source": [
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"import os\n",
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"import sys\n",
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"\n",
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"import matplotlib.pyplot as plt\n",
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"import numpy as np\n",
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"import pandas as pd\n",
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"import seaborn as sns\n",
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"\n",
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"sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
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"cell_type": "markdown",
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"id": "612bd02c",
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"metadata": {},
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"source": [
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"# Chapter 1 Summary Notes"
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"cell_type": "markdown",
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"id": "be70f5df",
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"metadata": {},
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"source": [
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"## Intro (1.0.0 - 1.3.1)\n",
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"\n",
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"**Theorem 1.1: De Morgan's law** \\\n",
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"For any sets $A_1, A_2, A_3, \\dots A_n$, we have\n",
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"- $(A_1 \\cup A_2 \\cup A_3 \\cup \\dots A_n)^c = A_1^c \\cap A_2^c \\cap A_3^c \\cap \\dots A_n^c$\n",
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"- $(A_1 \\cap A_2 \\cap A_3 \\cap \\dots A_n)^c = A_1^c \\cup A_2^c \\cup A_3^c \\cup \\dots A_n^c$\n",
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"\n",
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"**Theorem 1.2: Distributive law** \\\n",
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"For any sets $A, B,$ and $C$ we have\n",
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"- $A \\cap (B \\cup C) = (A \\cap B)\\cup(A \\cap C)$\n",
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"- $A \\cup (B \\cap C) = (A \\cup B)\\cap(A \\cup C)$\n",
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"\n",
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"**Inclusion-exclusion principle** \\\n",
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"For a finite collection of sets $A_1, A_2, A_3, \\dots A_n$, we have\n",
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"\n",
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"$\\left| \\bigcup_{i=1}^n A_i \\right| = \\sum_{i=1}^n |A_i| - \\sum_{i < j} |A_i \\cap A_j| + \\sum_{i < j < k} |A_i \\cap A_j \\cap A_k| - \\dots + (-1)^{n-1} |A_1 \\cap A_2 \\cap \\dots \\cap A_n|$\n",
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"\n",
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"$n = 2$ case:\n",
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"\n",
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"$|A \\cup B| = |A| + |B| - |A \\cap B|$\n",
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"\n",
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"$n = 3$ case:\n",
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"\n",
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"$|A \\cup B \\cup C| = |A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C|$\n",
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"\n",
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"## Random experiments (1.3.1 - 1.4)\n",
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"\n",
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"- A **random experiment** is a process by which we observe something uncertain\n",
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"- An **outcome** is a result of a random experiment\n",
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"- The **sample space** $S$ is the set of all possible outcomes\n",
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"- An **event** $A$ is any subset of $S$\n",
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"\n",
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"> In the context of a random experiment, the sample space is our *universal set*\n",
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"\n",
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"**Axioms of Probability**\n",
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"\n",
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"1. For any event $A$, $P(A) \\geq 0$\n",
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"2. $P(S) = 1$\n",
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"3. If $A_1, A_2, A_3, \\dots$ are disjoint events, then $P(A_1 \\cup A_2 \\cup A_3 \\cup \\dots) = P(A_1) + P(A_2) + P(A_3) + \\dots$\n",
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"\n",
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"**Some notation**\n",
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"\n",
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"- $P(A \\cap B) = P(A$ and $B) = P(A,B)$\n",
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"- $P(A \\cup B) = P(A$ or $B)$\n",
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"\n",
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"In a finite sample space $S$, where all outcomes are equally likely, the probability of any event $A$ can be found by\n",
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"\n",
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"\\begin{align*}\n",
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"P(A) = \\frac{|A|}{|S|}\n",
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"\\end{align*}\n",
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"\n",
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"## Conditional probability (1.4.0)\n",
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"\n",
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"If $A$ and $B$ are twos events in sample space $S$, then the **conditional probability of $A$ given $B$** is defined as\n",
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"\n",
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"\\begin{align*}\n",
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"P(A|B) = \\frac{|A \\cap B|}{|B|}, \\text{when } P(B) > 0\n",
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"\\end{align*}\n",
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"\n",
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"For events $A, B,$ and $C$, with $P(C) \\gt 0$, we have\n",
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"\n",
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"- $P(A^c|C) = 1 - P(A|C)$\n",
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"- $P(\\empty|C) = 0$\n",
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"- $P(A|C) \\leq 1$\n",
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"- $P(A \\setminus B|C) = P(A|C) - P(A \\cap B|C)$\n",
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"- $P(A \\cup B|C) = P(A|C) + P(B|C) - P(A \\cap B|C)$\n",
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"- if $A \\subset B$ then $P(A|C) \\leq P(B|C)$\n",
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"\n",
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""
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]
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"cell_type": "markdown",
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"source": [
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"## Independence (1.4.1)\n",
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"\n",
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"**Definition.** Two events $A$ and $B$ are *independent* if $P(A \\cap B) = P(A)P(B)$. AKA $P(A|B) = P(A)$\n",
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"\n",
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"**Definition.** Three events $A, B,$ and $C$ are independent if **all** of the following conditions hold:\n",
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"- $P(A \\cap B) = P(A)P(B)$\n",
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"- $P(A \\cap C) = P(A)P(C)$\n",
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"- $P(B \\cap C) = P(B)P(C)$\n",
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"- $P(A \\cap B \\cap C) = P(A)P(B)P(C)$\n",
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"\n",
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"**Definition.** $N$ events $A_1, A_2, A_3, \\dots, A_n$ are independent if **all** the following conditions holds:\n",
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"- $P(A_i \\cap B_j) = P(A_i)P(A_j)$\n",
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"- $P(A_i \\cap A_j \\cap A_k) = P(A_i)P(A_j)P(A_k)$ where $i \\in [1:n+1]$, $j \\in [i:n+1]$, $k \\in [j:n+1]$\n",
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"- $\\dots$\n",
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"- $P(A_1 \\cap A_2 \\cap A_3 \\cap \\dots \\cap A_n) = \\prod_{i=1}^nP(A_i)$\n",
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"\n",
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"**Lemma.** \\\n",
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"If $A$ and $B$ are independent then\n",
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"- $A$ and $B^c$ are independent\n",
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"- $A^c$ and $B$ are independent\n",
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"- $A^c$ and $B^c$ are independent\n",
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"\n",
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"**Definition.** If $A_1, A_2, \\dots, A_n$ are independent then\n",
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"$$P(A_1 \\cup A_2 \\cup \\dots \\cup A_n) = 1 - (1 - P(A_1))(1 - P(A_2))\\dots(1 - P(A_n))$$"
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]
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},
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"cell_type": "markdown",
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"id": "90cf5b00",
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"metadata": {},
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"source": [
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"## Law of Total Probability (1.4.2)\n",
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"\n",
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"$$P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)$$\n",
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"\n",
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"**Definition.** Law of Total Probability: \\\n",
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"If $B_1, B_2, B_3, \\dots $ is a partition of the sample space $S$, then for any event $A$ we have\n",
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"\n",
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"$$P(A) = \\sum_i P(A \\cap B_i) = \\sum_i P(A|B_i)P(B_i)$$"
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]
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"cell_type": "markdown",
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"source": [
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"## Bayes' Rule (1.4.3)\n",
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"\n",
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"**Definition.** Bayes' Rule\n",
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"\n",
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"- For any two events $A$ and $B$, where $P(A) \\neq 0$, we have\n",
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"\n",
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"$$P(B|A) = \\frac{P(A|B)P(B)}{P(A)}$$\n",
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"\n",
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"- If $B_1, B_2, B_3, \\dots$ form a partition of the sample space $S$, and $A$ is any event with $P(A) \\neq 0$, we have\n",
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"\n",
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"$$P(B_j|A) = \\frac{P(A|B_j)P(B_j)}{\\sum_i P(A|B_i)P(B_i)}$$"
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]
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