{ "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 }