moving stuff around

2026-06-22 16:37:57 -07:00
parent 4ab87d74c2
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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "a6732353-51d5-4478-9cf8-5834e57e5a4e",
+   "metadata": {},
+   "source": [
+    "# Chapter 1 Notes"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "48d9ec9e-83da-40ca-ae79-3c45f8af137c",
+   "metadata": {},
+   "source": [
+    "## Main Concepts\n",
+    "\n",
+    "Outcome: A result of a random experiment.\n",
+    "\n",
+    "Sample Space: The set of all possible outcomes.\n",
+    "\n",
+    "Event: A subset of the sample space.\n",
+    "\n",
+    "Inclusion-exclusion principle holds for probability\n",
+    "\n",
+    "Consider a sample space S. If S is a countable set, this refers to a discrete probability\n",
+    "mode\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7ac122be-50b2-423c-b88f-e4b3327b21bd",
+   "metadata": {},
+   "source": [
+    "## Example Problems"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7fb87a35-a470-4d98-935f-80c814e3f95d",
+   "metadata": {},
+   "source": [
+    "Example 1.5 - soln\n",
+    "\n",
+    "- there are 10 people with white shirts and 8 people with red shirts;\n",
+    "- 4 people have black shoes and white shirts\n",
+    "- 3 people have black shoes and red shirts\n",
+    "- the total number of people with white or red shirts or black shoes is 21\n",
+    "\n",
+    "Let A be the set of people with white shirts, B be the set of people with red shirts and let C be the set of people with black shoes.\n",
+    "\n",
+    "\\begin{align*}\n",
+    "|A|=10 \\\\\n",
+    "|B|=8 \\\\\n",
+    "|A \\cap C| = 4 \\\\\n",
+    "|B \\cap C| = 3 \\\\\n",
+    "|A \\cup B \\cup C| = 21\n",
+    "\\end{align*}\n",
+    "\n",
+    "Now we solve for $|C|$:\n",
+    "\n",
+    "\\begin{align*}\n",
+    "|A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C| = 21 \\\\\n",
+    "10 + 8 + |C| - 0 - 4 - 3 - 0 = 21 \\\\\n",
+    "18 + |C| - 7 = 21 \\\\\n",
+    "|C| + 11 = 21 \\\\\n",
+    "|C| = 10\n",
+    "\\end{align*}\n",
+    "\n",
+    "$\\therefore$ number of people with black shoes is 10\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "72b40733-531a-48f3-9879-75601684afc2",
+   "metadata": {},
+   "source": [
+    "Example 1.11 - soln\n",
+    "\n",
+    "Suppose we have the following information:\n",
+    "1. There is a 60 percent chance that it will rain today.\n",
+    "2. There is a 50 percent chance that it will rain tomorrow.\n",
+    "3. There is a 30 percent chance that it does not rain either day.\n",
+    "\n",
+    "T = rains\n",
+    "F = no rain\n",
+    "\n",
+    "$S = \\{(F, F), (F, T), (T, F), (T, T)\\}$\n",
+    "\n",
+    "$P((T, F) \\cup (T, T)) = 0.6$\n",
+    "\n",
+    "$P((F, T) \\cup (T, T)) = 0.5$\n",
+    "\n",
+    "$P((F, F)) = 0.3$\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(S) = 1 \\\\\n",
+    "P(\\{(F, F)\\} \\cup \\{(F, T)\\} \\cup \\{(T, F)\\} \\cup \\{(T, T)\\}) = 1 \\\\\n",
+    "P((F,F)) + P(\\{(F, T)\\} \\cup \\{(T, F)\\} \\cup \\{(T, T)\\}) = 1 \\\\\n",
+    "0.3 + P(\\{(F, T)\\} \\cup \\{(T, F)\\} \\cup \\{(T, T)\\}) = 1 \\\\\n",
+    "P(\\{(F, T)\\} \\cup \\{(T, F)\\} \\cup \\{(T, T)\\}) = 0.7 \\\\\n",
+    "P(\\{(F, T)\\} \\cup \\{(T, T)\\}) + P((T, F)) = 0.7 \\\\\n",
+    "0.5 + P((T, F)) = 0.7 \\\\\n",
+    "P((T, F)) = 0.2 \\\\\n",
+    "P(\\{(T, F)\\} \\cup \\{(T, T)\\}) + P((F, T)) = 0.7 \\\\\n",
+    "P((F, T)) = 0.1\n",
+    "\\end{align*}\n",
+    "\n",
+    "Find the following probabilities:\n",
+    "\n",
+    "a. The probability that it will rain today or tomorrow.\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P((T, F) \\cup (F, T) \\cup (T, T)) = 0.7\n",
+    "\\end{align*}\n",
+    "\n",
+    "b. The probability that it will rain today and tomorrow.\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P((T, T)) = 1 - 0.3 - 0.2 - 0.1 = 0.4\n",
+    "\\end{align*}\n",
+    "\n",
+    "c. The probability that it will rain today but not tomorrow.\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P((T, F)) = 0.2\n",
+    "\\end{align*}\n",
+    "\n",
+    "d. The probability that it either will rain today or tomorrow, but not both.\n",
+    "\n",
+    "\\begin{align*} \n",
+    "P(\\{(T, F)\\} \\cup \\{(F, T)\\}) = P((T, F)) + P((F, T)) = 0.2 + 0.1 = 0.3\n",
+    "\\end{align*}\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "8b5131dd-5ebd-4156-b808-f8df273317fb",
+   "metadata": {},
+   "source": [
+    "Example 1.12 - soln\n",
+    "\n",
+    "$S = \\{ -1, 0, 1, 2, 3, ... \\}$\n",
+    "\n",
+    "$\\forall x \\in S, P(x) = \\frac{1}{2^{x + 2}}$\n",
+    "\n",
+    "What is the probability that I win more than or equal to 1 dollar and less than 4 dollars?\n",
+    "\n",
+    "\\begin{align*} \n",
+    "P({1, 2, 3}) = P(1) + P(2) + P(3) \\\\\n",
+    "= 1/8 + 1/16 + 1/32\n",
+    "\\end{align*}\n",
+    "\n",
+    "What is the probability that I win more than 2 dollars?\n",
+    "\n",
+    "\\begin{align*} \n",
+    "\\sum_{i=3}^{\\infty} P(i) = P(3) + P(4) + P(5) + P(6) + ... \\\\\n",
+    "= 1/32 + 1/64 + 1/128 + 1/256 + ... \\\\\n",
+    "=\\frac{\\frac{1}{32}}{1 - \\frac{1}{2}}\n",
+    "=\\frac{1}{16}\n",
+    "\\end{align*}"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "69962960-18a6-436b-b9eb-9a6dfae7559a",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "11163495-a6cc-43b9-bf41-02748b13d210",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "roadmap (3.14.5)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.14.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
@@ -0,0 +1,16 @@
+# Introduction to Probability, Statistics, and Random Processes - Hossein Pishro-Nik
+
+[pdf](./Introduction%20to%20Probability,%20Statistics,%20and%20Random%20Processes%20-%20Hossein%20Pishro-Nik.pdf)
+
+total pages=1007
+
+**Currently reading:** chapter 3, page 236
+
+TODO:
+
+- 3.2.5 problems
+- ch3 end of chapter problems
+- 4.1.4 problems
+- 4.2.6 problems
+- 4.3.3 problems
+- ch4 end of chapter problems
@@ -0,0 +1,222 @@
+State De Morgan's Law for $n$ sets
+---
+
+For any sets $A_1, A_2, \dots, A_n$:
+
+$$\left(\bigcup_{i=1}^n A_i\right)^c = \bigcap_{i=1}^n A_i^c$$
+
+$$\left(\bigcap_{i=1}^n A_i\right)^c = \bigcup_{i=1}^n A_i^c$$
+
+The complement of a union is the intersection of complements, and vice versa.
+
+===
+
+State the Distributive Law for sets $A$, $B$, and $C$
+---
+
+$$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
+
+$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$
+
+===
+
+State the inclusion-exclusion principle for a finite collection of sets $A_1, A_2, A_3, \dots A_n$ where $n=2$
+
+---
+
+$n = 2$ case:
+
+$|A \cup B| = |A| + |B| - |A \cap B|$
+
+---
+
+$n = 3$ case:
+
+$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$
+
+---
+
+For a finite collection of sets $A_1, A_2, A_3, \dots A_n$, we have
+
+$\left| \bigcup_{i=1}^n A_i \right| = \sum_{i=1}^n |A_i| - \sum_{i < j} |A_i \cap A_j| + \sum_{i < j < k} |A_i \cap A_j \cap A_k| - \dots + (-1)^{n-1} |A_1 \cap A_2 \cap \dots \cap A_n|$
+
+===
+
+For the function
+
+$$f: A \to B$$
+
+State the following of $f$:
+
+1. Domain
+2. Co-domain
+3. Range
+
+---
+
+1. $A$
+2. $B$
+3. Set of all possible outputs of $f$ (not necessarily $B$)
+
+===
+
+State definition for each the following:
+
+1. Random experiment
+2. Outcome
+3. Sample space
+4. Event
+
+---
+
+1. A **random experiment** is a process by which we observe something uncertain
+2. An **outcome** is a result of a random experiment
+3. The **sample space** $S$ is the set of all possible outcomes
+4. An **event** is a subset of the sample space
+
+===
+
+State the Axioms of Probability
+
+---
+**Axioms of Probability**
+
+1. For any event $A$, $P(A) \geq 0$
+2. $P(S) = 1$
+3. If $A_1, A_2, A_3, \dots$ are disjoint events, then $P(A_1 \cup A_2 \cup A_3 \cup \dots) = P(A_1) + P(A_2) + P(A_3) + \dots$
+
+===
+
+In a finite sample space $S$, where all outcomes are equally likely, what is the probability of any event $A$?
+
+---
+
+$P(A) = \frac{|A|}{|S|}$
+
+===
+
+What is $P(A^c)$ in terms of $P(A)$?
+---
+
+$$P(A^c) = 1 - P(A)$$
+
+Follows from the axioms: $A$ and $A^c$ are disjoint, and $A \cup A^c = S$, so $P(A) + P(A^c) = P(S) = 1$.
+
+===
+
+Define conditional probability $P(A|B)$
+---
+
+The probability of $A$ given $B$ (when $P(B) > 0$):
+
+$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
+
+In the equally-likely case:
+
+$$P(A|B) = \frac{|A \cap B|}{|B|}$$
+
+===
+
+State the chain rule of probability for $n$ events
+---
+
+$$P(A_1 \cap A_2 \cap \cdots \cap A_n) = P(A_1)\,P(A_2|A_1)\,P(A_3|A_1,A_2)\cdots P(A_n|A_1,A_2,\dots,A_{n-1})$$
+
+Each factor conditions on all previously listed events.
+
+===
+
+What conditions must hold for three events $A$, $B$, $C$ to be independent?
+---
+
+All four conditions must hold:
+
+1. $P(A \cap B) = P(A)P(B)$
+2. $P(A \cap C) = P(A)P(C)$
+3. $P(B \cap C) = P(B)P(C)$
+4. $P(A \cap B \cap C) = P(A)P(B)P(C)$
+
+Pairwise independence alone is **not** sufficient.
+
+===
+
+What does it mean for $n$ events $A_1, A_2, \dots, A_n$ to be independent?
+---
+
+Every subset of the events must satisfy the product rule. That is, for all $i < j < k < \dots$:
+
+$$P(A_i \cap A_j) = P(A_i)P(A_j)$$
+$$P(A_i \cap A_j \cap A_k) = P(A_i)P(A_j)P(A_k)$$
+$$\vdots$$
+$$P(A_1 \cap A_2 \cap \cdots \cap A_n) = \prod_{i=1}^n P(A_i)$$
+
+===
+
+If $A_1, A_2, \dots, A_n$ are independent, what is $P(A_1 \cup A_2 \cup \cdots \cup A_n)$?
+---
+
+$$P(A_1 \cup A_2 \cup \cdots \cup A_n) = 1 - \prod_{i=1}^n (1 - P(A_i))$$
+
+Derived by taking the complement (none of the events occur) and using independence.
+
+===
+
+What is the difference between disjoint and independent?
+
+---
+
+- Disjoint: $A$ and $B$ cannot occur at the same time. $A \cap B = \empty$
+- Independent: $A$ does not give any information about $B$
+
+===
+
+State the Law of Total Probability using event $B$ and its complement
+---
+
+$$P(A) = P(A|B)\,P(B) + P(A|B^c)\,P(B^c)$$
+
+===
+
+State the general Law of Total Probability for a partition of the sample space
+---
+
+If $B_1, B_2, B_3, \dots$ is a partition of $S$, then for any event $A$:
+
+$$P(A) = \sum_i P(A \cap B_i) = \sum_i P(A|B_i)\,P(B_i)$$
+
+===
+
+State Bayes' Rule for two events $A$ and $B$
+---
+
+For $P(A) \neq 0$:
+
+$$P(B|A) = \frac{P(A|B)\,P(B)}{P(A)}$$
+
+===
+
+State Bayes' Rule when $B_1, B_2, \dots$ form a partition of $S$
+---
+
+For any event $A$ with $P(A) \neq 0$:
+
+$$P(B_j|A) = \frac{P(A|B_j)\,P(B_j)}{\displaystyle\sum_i P(A|B_i)\,P(B_i)}$$
+
+The denominator expands $P(A)$ via the Law of Total Probability.
+
+===
+
+State the meaning of conditionally independent between events $A$ and $B$ given event $C$. Note $P(C) \gt 0$.
+
+---
+
+$$P(A \cap B \mid C) = P(A \mid C)P(B \mid C)$$
+
+===
+
+State the multiplication rule for $P(A \cap B)$
+---
+
+$$P(A \cap B) = P(A|B)\,P(B) = P(B|A)\,P(A)$$
+
+Rearrangement of the definition of conditional probability.
@@ -0,0 +1,17 @@
+For any event $A$, prove $P(A^c) = 1 - P(A)$
+
+---
+
+$$
+1 = P(S)
+= P(A \cup A^c)
+= P(A) + P(A^c)
+$$
+
+===
+
+Prove $P(A \setminus B) = P(A) - P(A \cap B)$
+
+---
+
+TODO
@@ -0,0 +1,216 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "206bf674",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import os\n",
+    "import sys\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "import numpy as np\n",
+    "import pandas as pd\n",
+    "import seaborn as sns\n",
+    "\n",
+    "sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "612bd02c",
+   "metadata": {},
+   "source": [
+    "# Chapter 1 Summary Notes"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "be70f5df",
+   "metadata": {},
+   "source": [
+    "## Intro (1.0.0 - 1.3.1)\n",
+    "\n",
+    "**Theorem 1.1: De Morgan's law** \\\n",
+    "For any sets $A_1, A_2, A_3, \\dots A_n$, we have\n",
+    "- $(A_1 \\cup A_2 \\cup A_3 \\cup \\dots A_n)^c = A_1^c \\cap A_2^c \\cap A_3^c \\cap \\dots A_n^c$\n",
+    "- $(A_1 \\cap A_2 \\cap A_3 \\cap \\dots A_n)^c = A_1^c \\cup A_2^c \\cup A_3^c \\cup \\dots A_n^c$\n",
+    "\n",
+    "**Theorem 1.2: Distributive law** \\\n",
+    "For any sets $A, B,$ and $C$ we have\n",
+    "- $A \\cap (B \\cup C) = (A \\cap B)\\cup(A \\cap C)$\n",
+    "- $A \\cup (B \\cap C) = (A \\cup B)\\cap(A \\cup C)$\n",
+    "\n",
+    "**Inclusion-exclusion principle** \\\n",
+    "For a finite collection of sets $A_1, A_2, A_3, \\dots A_n$, we have\n",
+    "\n",
+    "$\\left| \\bigcup_{i=1}^n A_i \\right| = \\sum_{i=1}^n |A_i| - \\sum_{i < j} |A_i \\cap A_j| + \\sum_{i < j < k} |A_i \\cap A_j \\cap A_k| - \\dots + (-1)^{n-1} |A_1 \\cap A_2 \\cap \\dots \\cap A_n|$\n",
+    "\n",
+    "$n = 2$ case:\n",
+    "\n",
+    "$|A \\cup B| = |A| + |B| - |A \\cap B|$\n",
+    "\n",
+    "$n = 3$ case:\n",
+    "\n",
+    "$|A \\cup B \\cup C| = |A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C|$\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "c7475d4f",
+   "metadata": {},
+   "source": [
+    "## Random experiments (1.3.1 - 1.4)\n",
+    "\n",
+    "- A **random experiment** is a process by which we observe something uncertain\n",
+    "- An **outcome** is a result of a random experiment\n",
+    "- The **sample space** $S$ is the set of all possible outcomes\n",
+    "- An **event** $A$ is any subset of $S$\n",
+    "\n",
+    "> In the context of a random experiment, the sample space is our *universal set*\n",
+    "\n",
+    "**Axioms of Probability**\n",
+    "\n",
+    "1. For any event $A$, $P(A) \\geq 0$\n",
+    "2. $P(S) = 1$\n",
+    "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",
+    "\n",
+    "**Some notation**\n",
+    "\n",
+    "- $P(A \\cap B) = P(A$ and $B) = P(A,B)$\n",
+    "- $P(A \\cup B) = P(A$ or $B)$\n",
+    "\n",
+    "In a finite sample space $S$, where all outcomes are equally likely, the probability of any event $A$ can be found by\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(A) = \\frac{|A|}{|S|}\n",
+    "\\end{align*}"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "b705ef32",
+   "metadata": {},
+   "source": [
+    "## Conditional probability (1.4.0)\n",
+    "\n",
+    "If $A$ and $B$ are twos events in sample space $S$, then the **conditional probability of $A$ given $B$** is defined as\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(A|B) = \\frac{P(A \\cap B)}{P(B)}, \\text{when } P(B) > 0\n",
+    "\\end{align*}\n",
+    "\n",
+    "or\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(A|B) = \\frac{|A \\cap B|}{|B|}, \\text{when } P(B) > 0\n",
+    "\\end{align*}\n",
+    "\n",
+    "For events $A, B,$ and $C$, with $P(C) \\gt 0$, we have\n",
+    "\n",
+    "- $P(A^c|C) = 1 - P(A|C)$\n",
+    "- $P(\\empty|C) = 0$\n",
+    "- $P(A|C) \\leq 1$\n",
+    "- $P(A \\setminus B|C) = P(A|C) - P(A \\cap B|C)$\n",
+    "- $P(A \\cup B|C) = P(A|C) + P(B|C) - P(A \\cap B|C)$\n",
+    "- if $A \\subset B$ then $P(A|C) \\leq P(B|C)$\n",
+    "\n",
+    "![](../public/conditional_prob_tree.png)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "188a8fc2",
+   "metadata": {},
+   "source": [
+    "## Independence (1.4.1)\n",
+    "\n",
+    "**Definition.** Two events $A$ and $B$ are *independent* if $P(A \\cap B) = P(A)P(B)$. AKA $P(A|B) = P(A)$\n",
+    "\n",
+    "**Definition.** Three events $A, B,$ and $C$ are independent if **all** of the following conditions hold:\n",
+    "- $P(A \\cap B) = P(A)P(B)$\n",
+    "- $P(A \\cap C) = P(A)P(C)$\n",
+    "- $P(B \\cap C) = P(B)P(C)$\n",
+    "- $P(A \\cap B \\cap C) = P(A)P(B)P(C)$\n",
+    "\n",
+    "**Definition.** $N$ events $A_1, A_2, A_3, \\dots, A_n$ are independent if **all** the following conditions holds:\n",
+    "- $P(A_i \\cap B_j) = P(A_i)P(A_j)$\n",
+    "- $P(A_i \\cap A_j \\cap A_k) = P(A_i)P(A_j)P(A_k)$ where $i \\in [1:n+1]$, $j \\in [i:n+1]$, $k \\in [j:n+1]$\n",
+    "- $\\dots$\n",
+    "- $P(A_1 \\cap A_2 \\cap A_3 \\cap \\dots \\cap A_n) = \\prod_{i=1}^nP(A_i)$\n",
+    "\n",
+    "**Lemma.** \\\n",
+    "If $A$ and $B$ are independent then\n",
+    "- $A$ and $B^c$ are independent\n",
+    "- $A^c$ and $B$ are independent\n",
+    "- $A^c$ and $B^c$ are independent\n",
+    "\n",
+    "**Definition.** If $A_1, A_2, \\dots, A_n$ are independent then\n",
+    "$$P(A_1 \\cup A_2 \\cup \\dots \\cup A_n) = 1 - (1 - P(A_1))(1 - P(A_2))\\dots(1 - P(A_n))$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "90cf5b00",
+   "metadata": {},
+   "source": [
+    "## Law of Total Probability (1.4.2)\n",
+    "\n",
+    "$$P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)$$\n",
+    "\n",
+    "**Definition.** Law of Total Probability: \\\n",
+    "If $B_1, B_2, B_3, \\dots $ is a partition of the sample space $S$, then for any event $A$ we have\n",
+    "\n",
+    "$$P(A) = \\sum_i P(A \\cap B_i) = \\sum_i P(A|B_i)P(B_i)$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "c33df800",
+   "metadata": {},
+   "source": [
+    "## Bayes' Rule (1.4.3)\n",
+    "\n",
+    "**Definition.** Bayes' Rule\n",
+    "\n",
+    "- For any two events $A$ and $B$, where $P(A) \\neq 0$, we have\n",
+    "\n",
+    "$$P(B|A) = \\frac{P(A|B)P(B)}{P(A)}$$\n",
+    "\n",
+    "- 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",
+    "\n",
+    "$$P(B_j|A) = \\frac{P(A|B_j)P(B_j)}{\\sum_i P(A|B_i)P(B_i)}$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "849d7c99",
+   "metadata": {},
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "roadmap (3.14.5)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.14.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
@@ -0,0 +1,139 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "c58309b2",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import os\n",
+    "import sys\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "import numpy as np\n",
+    "import pandas as pd\n",
+    "import seaborn as sns\n",
+    "\n",
+    "sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a6732353-51d5-4478-9cf8-5834e57e5a4e",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 Notes"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "9f0046c2",
+   "metadata": {},
+   "source": [
+    "## Counting (2.0.0 - 2.1.5)\n",
+    "\n",
+    "***Definition.*** Multiplication Principle:  \\\n",
+    "Suppose that we perform $r$ experiments such that the $k\\text{th}$ experiment has $n_k$ possible outcomes, for $k=1,2,\\dots,r$. Then there are a total of $n_1 \\times n_2 \\times n_3 \\times \\dots \\times n_r$ possible outcomes for the sequence of $r$ experiments\n",
+    "\n",
+    "### Terminology\n",
+    "\n",
+    "- **Sampling**: Sampling from a set means choosing an element from that set. We\n",
+    "often **draw** a sample at random from a given set in which each element of the\n",
+    "set has equal chance of being chosen\n",
+    "\n",
+    "- **With or without replacement**: Usually we draw multiple samples from a set. If\n",
+    "we put each object back after each draw, we call this sampling with\n",
+    "replacement. In this case a single object can be possibly chosen multiple times.\n",
+    "For example, if A = {a1, a2, a3, a4} and we pick 3 elements with replacement, a\n",
+    "possible choice might be (a3, a1, a3). Thus \"with replacement\" means \"repetition\n",
+    "is allowed.\" On the other hand, if repetition is not allowed, we call it sampling\n",
+    "without replacement\n",
+    "\n",
+    "- **Ordered or unordered**: If ordering matters (i.e.: a1, a2, a3 ≠ a2, a3, a1), this is\n",
+    "called ordered sampling. Otherwise, it is called unordered\n",
+    "\n",
+    "### Counting Formulas\n",
+    "\n",
+    "- **ordered sampling with replacement:** $n^k$\n",
+    "\n",
+    "- **ordered sampling without replacement:** $n$ permute $k$ $\\quad$ ie $P^n_k = \\frac{n!}{(n - k)!}$\n",
+    "\n",
+    "- **unordered sampling without replacement:** $n$ choose $k$ $\\quad$ ie $\\binom{n}{k} = \\frac{n!}{k!(n-k)!}$\n",
+    "\n",
+    "- **unordered sampling with replacement:** $\\binom{n + k - 1}{k}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "2e93e0fe",
+   "metadata": {},
+   "source": [
+    "## Problem Solving Principles"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "87279e59",
+   "metadata": {},
+   "source": [
+    "When solving a combinatorics problem, consider:\n",
+    "1. Does order matter?\n",
+    "    - Yes → Permutations\n",
+    "    - No → Combinations\n",
+    "\n",
+    "    - \"Are HHHTT and THHHT the same outcome to me?\"\n",
+    "\n",
+    "2. Are we sampling with of without replacement?\n",
+    "    - Without replacement → Hypergeometric (phone problem)\n",
+    "    - With replacement → Binomial (coin flips)\n",
+    "    \n",
+    "    \"Can the same item be chosen twice?\"\n",
+    "3. Are the \"groups\" labeled or unlabeled?\n",
+    "    - Labeled/distinguishable → Just multiply combinations\n",
+    "    - Unlabeled/interchangeable → Divide by k!\n",
+    "\n",
+    "    \"Does it matter which group is called group 1?\"\n",
+    "4. Are the items distinguishable?\n",
+    "    - Distinguishable → Each item is unique, classical probability applies\n",
+    "    - Indistinguishable → Outcomes are not equally likely, be careful\n",
+    "\n",
+    "      \"Could I label these items 1 to n?\"\n",
+    "5. Is complement of inclusion-exclusion easier?\n",
+    "    - Complement → When \"at least\" or \"at most\" language appears\n",
+    "    - Inclusion-Exclusion → When events overlap\n",
+    "\n",
+    "    \"Is the opposite event simpler to count?\"\n",
+    "6. Am I counting each outcome exactly once? \n",
+    "  - If yes, done. Otherwhise we are overcounting or undercounting"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "9c5783dc",
+   "metadata": {},
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "roadmap (3.14.5)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.14.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
@@ -0,0 +1,101 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "206bf674",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import os\n",
+    "import sys\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "import numpy as np\n",
+    "import pandas as pd\n",
+    "import seaborn as sns\n",
+    "\n",
+    "sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "612bd02c",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 Example Problems"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "5ce47b88",
+   "metadata": {},
+   "source": [
+    "4.2 \n",
+    "\n",
+    "a. \n",
+    "\n",
+    "$\\int_{-\\infty}^{\\infty}f_X(u)du = 1$\n",
+    "\n",
+    "ie\n",
+    "\n",
+    "\\begin{align*}\n",
+    "\\int_{-\\infty}^{0^+}f_X(u)du + \\int_{0}^{\\infty}f_X(x)dx = 1 \\\\\n",
+    "0 + \\int_{0}^{\\infty}ce^{-x}dx = 1 \\\\\n",
+    "c\\cdot\\lim_{t \\to \\infty }\\int_{0}^{t}e^{-x}dx = 1 \\\\\n",
+    "c\\cdot\\lim_{t \\to \\infty } [-e^{-x}]_0^t = 1 \\\\\n",
+    "c\\cdot\\lim_{t \\to \\infty } ((-e^{-t}) - (-e^{-0})) = 1 \\\\\n",
+    "c\\cdot\\lim_{t \\to \\infty } (-e^{-t} + 1) = 1 \\\\\n",
+    "c\\cdot 1 = 1 \\\\\n",
+    "c = 1\n",
+    "\\end{align*}\n",
+    "\n",
+    "b.\n",
+    "\n",
+    "Note that\n",
+    "\n",
+    "$$F_X(x) = \\int_{-\\infty}^x f_X(u)du$$\n",
+    "\n",
+    "and $c=1$\n",
+    "\n",
+    "so\n",
+    "\n",
+    "\\begin{align*}\n",
+    "F_X(x) = \\begin{cases} \n",
+    "1-e^{-x}     & \\text{for } x \\geq 0 \\\\  \n",
+    "0     & \\text{otherwise} \n",
+    "\\end{cases}\n",
+    "\\end{align*}\n",
+    "\n",
+    "c. $F_X(3) - F_X(1) = $"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "95a3b21c",
+   "metadata": {},
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "roadmap (3.14.5)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.14.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
@@ -0,0 +1,176 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "c58309b2",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import os\n",
+    "import sys\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "import numpy as np\n",
+    "import pandas as pd\n",
+    "import seaborn as sns\n",
+    "\n",
+    "sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a6732353-51d5-4478-9cf8-5834e57e5a4e",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 Notes"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "9f0046c2",
+   "metadata": {},
+   "source": [
+    "## Continuous Random Variables and their Distributions (4.1.0)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ea9b1f96",
+   "metadata": {},
+   "source": [
+    "***Definition*** A random variable $X$ with CDF $F_X(x)$ is said to be continuous if $F_X(x)$ is a continuous for all $x \\in \\mathbb{R}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a96da4d3",
+   "metadata": {},
+   "source": [
+    "## Probability Density Function (PDF) (4.1.1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "c62129a5",
+   "metadata": {},
+   "source": [
+    "***Definition*** Consider a continuous random variable $X$ with an absolutely continuous CDF $F_X(x)$. The function $f_X(x)$ defined by\n",
+    "\n",
+    "$$f_X(x) = \\frac{dF_X(x)}{dx} = F'_X(x) \\quad \\text{if } F_X(x) \\text{ is differentiable at } x$$\n",
+    " \n",
+    "is called the probability density function (PDF) of $X$."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "df411869",
+   "metadata": {},
+   "source": [
+    "NOTE: The PDF being constant implies uniformity\n",
+    "\n",
+    "NOTE: For small values of $\\delta$,\n",
+    "\n",
+    "$$P(x \\lt X \\leq x + \\delta) \\approx  f_X(x)\\delta$$\n",
+    "\n",
+    "Thus if $f_X(x_1) \\gt f_X(x_2)$, we can say $P(x_1 \\lt X \\leq x_1 + \\delta) \\gt P(x_2 \\lt X \\leq x_2 + \\delta)$, ie the value of $X$ is more likely to be around $x_1$ then $x_2$\n",
+    "\n",
+    "NOTE: The CDF can be obtained from the PDF via (assuming absolute continuity)\n",
+    "\n",
+    "$$F_X(x) = \\int_{-\\infty}^x f_X(u)du$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f29a3bfb",
+   "metadata": {},
+   "source": [
+    "***Properties*** Consider a continuous random variable $X$ with PDF $f_X(x)$. We have\n",
+    "- $f_X(x) \\geq 0, \\forall x \\in \\mathbb{R}$\n",
+    "- $\\int_{-\\infty}^{\\infty}f_X(u)du = 1$\n",
+    "- $P(a \\lt X \\leq b) = F_X(b) - F_X(a) = \\int_a^bf_X(u)du$\n",
+    "- For a set $A$, $P(X \\in A) = \\int_Af_X(u)du$. However, set $A$ must satisfy:"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "6f546fed",
+   "metadata": {},
+   "source": [
+    "***Definition*** If $X$ is a continuous random variable, we can write the range of $X$ as\n",
+    "\n",
+    "$$R_X = \\{ x \\mid f_X(x) \\gt 0 \\}$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "170db3a0",
+   "metadata": {},
+   "source": [
+    "***Property*** The expected value if a continuous random variable $X$ is\n",
+    "\n",
+    "$$E[X] = \\int_{-\\infty}^{\\infty}xf_X(x)dx$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "deccea51",
+   "metadata": {},
+   "source": [
+    "***Property*** Law of the unconscious statistician (LOTUS) for continuous random variables\n",
+    "\n",
+    "$$E[g(X)] = \\int_{-\\infty}^{\\infty}g(x)f_X(x)dx$$\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "c0a20195",
+   "metadata": {},
+   "source": [
+    "***Property*** Variance for a continuous random variable $X$, we can write\n",
+    "\n",
+    "\\begin{align*}\n",
+    "\\text{Var}(X) &= E[(X - E[x])^2] = \\int_{-\\infty}^{\\infty}(x - E[X])^2f_X(x)dx \\\\\n",
+    "&= E[X^2] - E[X]^2 = \\int_{-\\infty}^{\\infty}x^2f_X(x)dx - E[X]^2\n",
+    "\\end{align*}"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "6f2f6b72",
+   "metadata": {},
+   "source": [
+    "## Functions of Continuous Random Variables"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f4a95738",
+   "metadata": {},
+   "source": [
+    "***Theorem***"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "roadmap (3.14.5)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.14.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
@@ -0,0 +1,7 @@
+# Algorithmic Trading - Winning Strategies and their rationale
+
+[pdf](./Algorithmic%20Trading%20Winning%20Strategies%20and%20their%20rationale.pdf)
+
+total pages=225
+
+**Currently reading:** chapter 1, page 19
@@ -0,0 +1,51 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "93b4302b",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import os\n",
+    "import sys\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "import numpy as np\n",
+    "import pandas as pd\n",
+    "import seaborn as sns\n",
+    "\n",
+    "sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "bdd46cba",
+   "metadata": {},
+   "source": [
+    "# Chapter 1 Notes"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "roadmap (3.14.5)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.14.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
@@ -0,0 +1,51 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "93b4302b",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import os\n",
+    "import sys\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "import numpy as np\n",
+    "import pandas as pd\n",
+    "import seaborn as sns\n",
+    "\n",
+    "sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "bdd46cba",
+   "metadata": {},
+   "source": [
+    "# Notes"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "roadmap (3.14.5)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.14.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}