Upload CS229 homework 2

Author: CJuanvip

Individual-projects/Machine-learning-models/2-EM-convergence.ipynb

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Suppose at $t$-step, $\\theta = \\theta^*$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$$\n",
+    "M = \\sum_i^m\\sum_{z^{(i)}} Q_i^*(z^{(i)}) \\log \\frac{p(x^{(i)}, z^{(i)}; \\theta)}{Q^*(z^{(i)})}\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "where\n",
+    "\n",
+    "$$\n",
+    "Q_i^*(z^{(i)}) = p(z^{(i)} | x^{(i)}; \\theta^*) = \\frac{p(x^{(i)}, z^{(i)}; \\theta^*)}{p(x^{(i)}; \\theta^*)}\n",
+    "$$\n",
+    "\n",
+    "for each $i$, following the E-step. Note $Q_i^*(z^{(i)})$ is a constant indepedent of $\\theta$."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Take derivative of $M$ over $\\theta$\n",
+    "\n",
+    "\\begin{align*}\n",
+    "\\nabla_{\\theta}{M} \n",
+    "&= \\sum_i^m\\sum_{z^{(i)}} Q_i^*(z^{(i)}) \\frac{Q_i^*(z^{(i)})}{p(x^{(i)}, z^{(i)};\\theta)} \\frac{1}{Q_i^*(z^{(i)})} \\nabla_{\\theta}p(x^{(i)}, z^{(i)};\\theta) \\\\\n",
+    "&= \\sum_i^m\\sum_{z^{(i)}} \\frac{p(x^{(i)}, z^{(i)}; \\theta^*)}{p(x^{(i)}; \\theta^*)} \\frac{\\nabla_{\\theta}p(x^{(i)}, z^{(i)};\\theta)}{p(x^{(i)}, z^{(i)};\\theta)} \\\\\n",
+    "\\end{align*}"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Since $\\theta$ has converged to $\\theta^*$, setting $\\theta = \\theta^*$ will make $\\nabla_{\\theta}M = 0$."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "\\begin{align*}\n",
+    "\\nabla_{\\theta}{M}|_{\\theta = \\theta^*} \n",
+    "&= \\sum_i^m\\sum_{z^{(i)}} \\frac{p(x^{(i)}, z^{(i)}; \\theta^*)}{p(x^{(i)}; \\theta^*)}\\frac{\\nabla_{\\theta}p(x^{(i)}, z^{(i)};\\theta)|_{\\theta = \\theta^*}}{p(x^{(i)}, z^{(i)};\\theta^*)} \\\\\n",
+    "&= \\sum_i^m\\sum_{z^{(i)}} \\frac{\\nabla_{\\theta}p(x^{(i)}, z^{(i)};\\theta)|_{\\theta = \\theta^*}}{p(x^{(i)}; \\theta^*)} \\\\\n",
+    "&= \\sum_i^m \\frac{\\nabla_{\\theta}p(x^{(i)}, \\theta)|_{\\theta = \\theta^*}}{p(x^{(i)}; \\theta^*)} \\\\\n",
+    "&= \\sum_i^m \\nabla_{\\theta} \\log p(x^{(i)}; \\theta)|_{\\theta = \\theta^*} \\\\\n",
+    "&= \\nabla_{\\theta}  \\sum_i^m \\log p(x^{(i)}; \\theta)|_{\\theta = \\theta^*} \\\\\n",
+    "&= \\nabla_{\\theta} \\ell(\\theta) \\\\\n",
+    "&= 0\n",
+    "\\end{align*}"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Therefore, upon convergence, Letting $\\nabla_{\\theta}M = 0$ is equivalent to letting $\\nabla_{\\theta} \\ell = 0$."
+   ]
+  }
+ ],
+ "metadata": {
+  "anaconda-cloud": {},
+  "kernelspec": {
+   "display_name": "Python [default]",
+   "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": "3.5.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}