MORE chapter 1 progress

study/001_introduction-to-probability-statistics-and-random-processes/README.md

@@ -2,4 +2,4 @@
 
 total pages=1007
 
-**Currently reading:** chapter 1, page 73
+**Currently reading:** chapter 1, page 80

study/001_introduction-to-probability-statistics-and-random-processes/ch1/notes.ipynb

@@ -543,7 +543,7 @@
    "id": "4790a1dc",
    "metadata": {},
    "source": [
-    "## Example 1.20\n",
+    "### Example 1.20\n",
     "\n",
     "$S = [1:11]$\n",
     "\n",
@@ -560,23 +560,137 @@
     "$P(A|B) = P(A) \\therefore A$ and $B$ are independent. \n"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "b16a508d",
+   "metadata": {},
+   "source": [
+    "### Example 1.21\n",
+    "\n",
+    "$P(HHHHT) = P(H)P(H)P(H)P(H)P(T) = \\frac{1}{2^5} = \\frac{1}{32}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "08fe32c4",
+   "metadata": {},
+   "source": [
+    "### Example 1.22\n",
+    "\n",
+    "Number of plane rides over 20 years $= 20 \\cdot 20 = 400$\n",
+    "\n",
+    "Let $A_i$ be the flight businessman dies where $i \\in [1:401]$\n",
+    "\n",
+    "We want to find $P(A_1 \\cup A_2 \\cup A_3 \\cup \\dots \\cup A_{400})$\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(A_1 \\cup A_2 \\cup A_3 \\cup \\dots \\cup A_n) &= 1 - (1 - P(A_1))(1 - P(A_2))\\cdots(1 - P(A_{400})) \\\\\n",
+    "&= 1 - (1 - \\frac{1}{4\\cdot 10^6})^{400} \\\\\n",
+    "\\approx 10^{-4}\n",
+    "\\end{align*}"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "42589434",
+   "metadata": {},
+   "source": [
+    "### Example 1.23\n",
+    "\n",
+    "a.\n",
+    "\n",
+    "What is the probability someone makes a shot in a round?\n",
+    "\n",
+    "Let $A$ be the event player 1 makes the shot, and $B$ the event player 2 makes the shot. \n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(A \\cup B) &= 1 - (1 - P(A))(1 - P(B)) \\\\\n",
+    "&= 1 - (1 - p_1)(1 - p_2) \\\\\n",
+    "&= p_1 + p_2 - p_1p_2\n",
+    "\\end{align*}\n",
+    "\n",
+    "What is the probability nobody makes a shot in a round?\n",
+    "\n",
+    "$$P((A \\cup B)^c) = 1 - p_1 - p_2 + p_1p_2 = 1 - (1 - p_1)(1 - p_2)$$\n",
+    "\n",
+    "What is the probability player one wins in the first round?\n",
+    "\n",
+    "$$P(W_1(1)) = p_1$$\n",
+    "\n",
+    "What is the probability player one wins in the second round?\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(W_1(2)) = P(R_1(1)) + P(R_1(2)) \\\\\n",
+    "= p_1 + P((A \\cup B)^c)p_1 \\\\\n",
+    "= p_1 + (1 - (1 - p_1)(1 - p_2))p_1\n",
+    "\\end{align*}\n",
+    "\n",
+    "What is the probability player one wins in the third round?\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(W_1(3)) = P(R_1(1)) + P(R_1(2)) + P(R_1(3)) \\\\\n",
+    "= p_1 + P((A \\cup B)^c)p_1 + P((A \\cup B)^c)P((A \\cup B)^c)p_1 \\\\\n",
+    "\\end{align*}\n",
+    "\n",
+    "What is the probability player one wins on round $n$?\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(W_1(n)) = P(R_1(1)) + P(R_1(2)) + P(R_1(3)) + \\dots + P(R_1(n)) \\\\\n",
+    "= p_1 + P((A \\cup B)^c)p_1 + P((A \\cup B)^c)P((A \\cup B)^c)p_1 + \\dots + P((A \\cup B)^c)^np_1 \\\\\n",
+    "= p_1 + (1 - p_1 - p_2 + p_1p_2)p_1 + (1 - p_1 - p_2 + p_1p_2)^2p_1 + \\dots + (1 - p_1 - p_2 + p_1p_2)^np_1 \\\\\n",
+    "= p_1 (1 + (1 - p_1 - p_2 + p_1p_2) + (1 - p_1 - p_2 + p_1p_2)^2 + \\dots + (1 - p_1 - p_2 + p_1p_2)^n) \n",
+    "\\end{align*}\n",
+    "\n",
+    "Let $x = 1 - p_1 - p_2 + p_1p_2$, substitute for \n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(W_1(n)) = p_1 (1 + x + x^2 + \\dots + x^n) \n",
+    "\\end{align*}\n",
+    "\n",
+    "Notice that $|x| \\lt 1$. Proof\n",
+    "\n",
+    "$$x = 1 - p_1 - p_2 + p_1p_2 = (1 - p_1)(1 - p_2)$$\n",
+    "\n",
+    "Since $0<p_1,p_2<1$, we have\n",
+    "\n",
+    "$0 < (1 - p_1) < 1$ and $0 < (1 - p_2) < 1$ which implies $0 < (1 - p_1)(1 - p_2) < 1$ so $0 < x < 1$ which implies $|x| < 1$\n",
+    "\n",
+    "Now we consider \n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(W_1) &= p_1 \\cdot \\lim_{n \\to \\infty} (1 + x + x^2 + \\dots + x^n) \\\\\n",
+    "&= p_1 \\cdot (\\frac{1}{1 - x}) \\\\\n",
+    "&= p_1 \\cdot (\\frac{1}{1 - (1 - p_1 - p_2 + p_1p_2)}) \\\\\n",
+    "&= \\frac{p_1}{p_1 + p_2 - p_1p_2} \\\\\n",
+    "\\end{align*}\n",
+    "\n",
+    "b.\n",
+    "\n",
+    "If $P(W_1) = \\frac{1}{2}$, find the values of $p_1$ and $p_2$.\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(W_1) &= \\frac{1}{2} \\\\\n",
+    "\\frac{p_1}{p_1 + p_2 - p_1p_2} &= \\frac{1}{2} \\\\\n",
+    "2p_1 &= p_1 + p_2 - p_1p_2 \\\\\n",
+    "0 &= p_1 + p_2 - p_1p_2 - 2p_1 \\\\\n",
+    "0 &= p_2 - p_1p_2 - p_1 \\\\\n",
+    "\\end{align*}\n",
+    "\n",
+    "In terms of $p_1$\n",
+    "\n",
+    "$$p_2 = \\frac{p_1}{1 - p_1}$$\n",
+    "\n",
+    "In terms of $p_2$\n",
+    "\n",
+    "$$p_1 = \\frac{p_2}{1 + p_2}$$"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,
-   "id": "ef40221f",
+   "id": "d283ef80",
    "metadata": {},
-   "outputs": [
+   "outputs": [],
-    {
-     "data": {
-      "text/plain": [
-       "0.8559987631416203"
-      ]
-     },
-     "execution_count": 1,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
    "source": []
   }
  ],

study/001_introduction-to-probability-statistics-and-random-processes/ch1/summary.ipynb

@@ -23,7 +23,7 @@
    "id": "612bd02c",
    "metadata": {},
    "source": [
-    "# Summary of Introduction to Probability, Statistics, and Random Processes (Hossein Pishro-Nik) chapter 1"
+    "# Chapter 1 Summary"
    ]
   },
   {
@@ -99,11 +99,46 @@
     "- $P(A \\cup B|C) = P(A|C) + P(B|C) - P(A \\cap B|C)$\n",
     "- if $A \\subset B$ then $P(A|C) \\leq P(B|C)$\n",
     "\n",
-    "![](./public/conditional_prob_tree.png)\n",
+    "![](../public/conditional_prob_tree.png)"
-    "\n",
+   ]
+  },
+  {
+   "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 $P(A|B) = P(A)$"
+    "**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": "c33df800",
+   "metadata": {},
+   "source": [
+    "\n"
    ]
   }
  ],