more chapter 3 progress

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

@@ -4,7 +4,7 @@
 
 total pages=1007
 
-**Currently reading:** chapter 3, page 160
+**Currently reading:** chapter 3, page 162
 
 TODO:
 

study/001_introduction-to-probability-statistics-and-random-processes/ch3/example-problems.ipynb

@@ -67,6 +67,77 @@
     "plt.tight_layout()\n",
     "plt.show()"
    ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f133a554",
+   "metadata": {},
+   "source": [
+    "## Example 3.5\n",
+    "\n",
+    "$$P_Y(y) = \\begin{cases}\n",
+    "    (1 - p)^{y-1}p & \\text{if } y \\in \\mathbb{N} \\\\\n",
+    "    0 & \\text{otherwise }\n",
+    "\\end{cases}$$\n",
+    "\n",
+    "1. Note that $R_Y = \\mathbb{N}$. So\n",
+    "\n",
+    "$$\\sum_{y \\in R_Y}P_Y(y)$$\n",
+    "> Note that therewas a textbook typo\n",
+    "\n",
+    "Turns into\n",
+    "\n",
+    "\\begin{align*}\n",
+    "\\sum_{y \\in \\mathbb{N}}(1 - p)^{y-1}p \\\\\n",
+    "p \\cdot \\sum_{y \\in \\mathbb{N}}(1 - p)^{y-1} \\\\\n",
+    "p \\cdot ((1 - p)^{1-1} + (1 - p)^{2-1} + (1 - p)^{3-1} + \\dots) \\\\\n",
+    "\\end{align*}\n",
+    "\n",
+    "Let $x = 1 - p$, we have\n",
+    "\n",
+    "$$p \\cdot (1 + x + x^2 + x^3 + \\dots)$$\n",
+    "\n",
+    "Is $|x| \\lt 1$? Yes proof:\n",
+    "\n",
+    "\\begin{align*}\n",
+    "0 \\lt p \\lt 1 \\\\\n",
+    "-1 \\lt p - 1 \\lt 0 \\\\\n",
+    "-1 \\lt x \\lt 0 \\implies |x| \\lt 1\n",
+    "\\end{align*}\n",
+    "\n",
+    "So now we have \n",
+    "\n",
+    "\\begin{align*}\n",
+    "p \\cdot (\\frac{1}{1-x}) \\\\\n",
+    "p \\cdot (\\frac{1}{1-(1 - p)}) \\\\\n",
+    "p \\cdot (\\frac{1}{1-(1 - p)}) \\\\\n",
+    "p \\cdot (\\frac{1}{p}) \\\\\n",
+    "1\n",
+    "\\end{align*}\n",
+    "\n",
+    "$\\therefore \\sum_{y \\in R_Y}P_Y(y) = 1$\n",
+    "\n",
+    "2.\n",
+    "\n",
+    "If $p = \\frac{1}{2}$ then $P_Y(y)$ simplifies down to\n",
+    "\n",
+    "$$P_Y(y) = \\begin{cases}\n",
+    "    \\frac{1}{2^y} & \\text{if } y \\in \\mathbb{N} \\\\\n",
+    "    0 & \\text{otherwise }\n",
+    "\\end{cases}$$\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(2 \\leq Y \\lt 5) &= P(Y = 2) + P(Y = 3) + P(Y = 4) \\\\\n",
+    "&= P_Y(2) + P_Y(3) + P_Y(4) \\\\\n",
+    "&= \\frac{1}{2^2} + \\frac{1}{2^3} + \\frac{1}{2^4} = \\frac{7}{16}\n",
+    "\\end{align*}"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "bb2574f0",
+   "metadata": {},
+   "source": []
   }
  ],
  "metadata": {