9649ce382b
In [commit6072ad4](6072ad4), @KajvanRijn kindly changed all "TO DO" to "TODO" in the code blocks. That's useful. In addition, it should be changed (as here) in the instructions. Then there's no doubt or issue for anyone searching all instances.
508 lines
16 KiB
Plaintext
508 lines
16 KiB
Plaintext
{
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"cells": [
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"colab_type": "text",
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"id": "view-in-github"
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},
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"source": [
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"<a href=\"https://colab.research.google.com/github/udlbook/udlbook/blob/main/Notebooks/Chap17/17_3_Importance_Sampling.ipynb\" target=\"_parent\"><img src=\"https://colab.research.google.com/assets/colab-badge.svg\" alt=\"Open In Colab\"/></a>"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"id": "t9vk9Elugvmi"
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},
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"source": [
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"# **Notebook 17.3: Importance sampling**\n",
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"\n",
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"This notebook investigates importance sampling as described in section 17.8.1 of the book.\n",
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"\n",
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"Work through the cells below, running each cell in turn. In various places you will see the words \"TODO\". Follow the instructions at these places and make predictions about what is going to happen or write code to complete the functions.\n",
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"\n",
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"Contact me at udlbookmail@gmail.com if you find any mistakes or have any suggestions."
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "OLComQyvCIJ7"
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},
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"outputs": [],
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"source": [
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"import numpy as np\n",
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"import matplotlib.pyplot as plt"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"id": "f7a6xqKjkmvT"
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},
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"source": [
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"Let's approximate the expectation\n",
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"\n",
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"\\begin{equation}\n",
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"\\mathbb{E}_{y}\\Bigl[\\exp\\bigl[- (y-1)^4\\bigr]\\Bigr] = \\int \\exp\\bigl[- (y-1)^4\\bigr] Pr(y) dy,\n",
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"\\end{equation}\n",
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"\n",
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"where\n",
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"\n",
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"\\begin{equation}\n",
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"Pr(y)=\\text{Norm}_y[0,1]\n",
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"\\end{equation}\n",
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"\n",
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"by drawing $I$ samples $y_i$ and using the formula:\n",
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"\n",
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"\\begin{equation}\n",
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"\\mathbb{E}_{y}\\Bigl[\\exp\\bigl[- (y-1)^4\\bigr]\\Bigr] \\approx \\frac{1}{I} \\sum_{i=1}^I \\exp\\bigl[-(y_i-1)^4 \\bigr]\n",
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"\\end{equation}"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "VjkzRr8o2ksg"
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},
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"outputs": [],
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"source": [
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"def f(y):\n",
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" return np.exp(-(y-1) *(y-1) *(y-1) * (y-1))\n",
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"\n",
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"\n",
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"def pr_y(y):\n",
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" return (1/np.sqrt(2*np.pi)) * np.exp(-0.5 * y * y)\n",
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"\n",
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"fig,ax = plt.subplots()\n",
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"y = np.arange(-10,10,0.01)\n",
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"ax.plot(y, f(y), 'r-', label='f$[y]$');\n",
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"ax.plot(y, pr_y(y),'b-',label='$Pr(y)$')\n",
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"ax.set_xlabel(\"$y$\")\n",
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"ax.legend()\n",
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"plt.show()"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "LGAKHjUJnWmy"
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},
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"outputs": [],
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"source": [
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"def compute_expectation(n_samples):\n",
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" # TODO -- compute this expectation\n",
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" # 1. Generate samples y_i using np.random.normal\n",
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" # 2. Approximate the expectation of f[y]\n",
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" # Replace this line\n",
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" expectation = 0\n",
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"\n",
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"\n",
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" return expectation"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "nmvixMqgodIP"
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},
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"outputs": [],
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"source": [
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"# Set the seed so the random numbers are all the same\n",
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"np.random.seed(0)\n",
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"\n",
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"# Compute the expectation with a very large number of samples (good estimate)\n",
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"n_samples = 100000000\n",
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"expected_f= compute_expectation(n_samples)\n",
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"print(\"Your value: \", expected_f, \", True value: 0.43160702267383166\")"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"id": "Jr4UPcqmnXCS"
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},
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"source": [
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"Let's investigate how the variance of this approximation decreases as we increase the number of samples $N$.\n",
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"\n",
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"\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "yrDp1ILUo08j"
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},
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"outputs": [],
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"source": [
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"def compute_mean_variance(n_sample):\n",
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" n_estimate = 10000\n",
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" estimates = np.zeros((n_estimate,1))\n",
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" for i in range(n_estimate):\n",
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" estimates[i] = compute_expectation(n_sample.astype(int))\n",
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" return np.mean(estimates), np.var(estimates)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "BcUVsodtqdey"
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},
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"outputs": [],
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"source": [
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"# Compute the mean and variance for 1,2,... 20 samples\n",
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"n_sample_all = np.array([1.,2,3,4,5,6,7,8,9,10,15,20,25,30,45,50,60,70,80,90,100,150,200,250,300,350,400,450,500])\n",
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"mean_all = np.zeros_like(n_sample_all)\n",
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"variance_all = np.zeros_like(n_sample_all)\n",
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"for i in range(len(n_sample_all)):\n",
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" print(\"Computing mean and variance for expectation with %d samples\"%(n_sample_all[i]))\n",
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" mean_all[i],variance_all[i] = compute_mean_variance(n_sample_all[i])"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "feXmyk0krpUi"
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},
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"outputs": [],
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"source": [
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"fig,ax = plt.subplots()\n",
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"ax.semilogx(n_sample_all, mean_all,'r-',label='mean estimate')\n",
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"ax.fill_between(n_sample_all, mean_all-2*np.sqrt(variance_all), mean_all+2*np.sqrt(variance_all))\n",
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"ax.set_xlabel(\"Number of samples\")\n",
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"ax.set_ylabel(\"Mean of estimate\")\n",
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"ax.plot([0,500],[0.43160702267383166,0.43160702267383166],'k--',label='true value')\n",
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"ax.legend()\n",
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"plt.show()\n"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"id": "XTUpxFlSuOl7"
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},
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"source": [
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"As you might expect, the more samples that we use to compute the approximate estimate, the lower the variance of the estimate."
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"id": "6hxsl3Pxo1TT"
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},
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"source": [
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" Now consider the function\n",
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" \\begin{equation}\n",
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" \\mbox{f}[y]= 20.446\\exp\\left[-(y-3)^4\\right],\n",
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" \\end{equation}\n",
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"\n",
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"which decreases rapidly as we move away from the position $y=3$."
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "znydVPW7sL4P"
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},
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"outputs": [],
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"source": [
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"def f2(y):\n",
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" return 20.446*np.exp(- (y-3) *(y-3) *(y-3) * (y-3))\n",
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"\n",
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"fig,ax = plt.subplots()\n",
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"y = np.arange(-10,10,0.01)\n",
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"ax.plot(y, f2(y), 'r-', label='f$[y]$');\n",
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"ax.plot(y, pr_y(y),'b-',label='$Pr(y)$')\n",
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"ax.set_xlabel(\"$y$\")\n",
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"ax.legend()\n",
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"plt.show()"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"id": "G9Xxo0OJsIqD"
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},
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"source": [
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"Let's again, compute the expectation:\n",
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"\n",
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"\\begin{align}\n",
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"\\mathbb{E}_{y}\\left[\\text{f}[y]\\right] &=& \\int \\text{f}[y] Pr(y) dy\\\\\n",
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"&\\approx& \\frac{1}{I} \\text{f}[y]\n",
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"\\end{align}\n",
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"\n",
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"where $Pr(y)=\\text{Norm}_y[0,1]$ by approximating with samples $y_{i}$.\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "l8ZtmkA2vH4y"
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},
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"outputs": [],
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"source": [
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"def compute_expectation2(n_samples):\n",
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" y = np.random.normal(size=(n_samples,1))\n",
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" expectation = np.mean(f2(y))\n",
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"\n",
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" return expectation"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "dfUQyJ-svZ6F"
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},
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"outputs": [],
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"source": [
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"# Set the seed so the random numbers are all the same\n",
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"np.random.seed(0)\n",
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"\n",
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"# Compute the expectation with a very large number of samples (good estimate)\n",
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"n_samples = 100000000\n",
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"expected_f2= compute_expectation2(n_samples)\n",
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"print(\"Expected value: \", expected_f2)"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"id": "2sVDqP0BvxqM"
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},
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"source": [
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"I deliberately chose this function, because it's expectation is roughly the same as for the previous function.\n",
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"\n",
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"Again, let's look at the mean and the variance of the estimates"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "mHnILRkOv0Ir"
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},
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"outputs": [],
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"source": [
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"def compute_mean_variance2(n_sample):\n",
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" n_estimate = 10000\n",
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" estimates = np.zeros((n_estimate,1))\n",
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" for i in range(n_estimate):\n",
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" estimates[i] = compute_expectation2(n_sample.astype(int))\n",
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" return np.mean(estimates), np.var(estimates)\n",
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"\n",
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"# Compute the variance for 1,2,... 20 samples\n",
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"mean_all2 = np.zeros_like(n_sample_all)\n",
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"variance_all2 = np.zeros_like(n_sample_all)\n",
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"for i in range(len(n_sample_all)):\n",
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" print(\"Computing variance for expectation with %d samples\"%(n_sample_all[i]))\n",
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" mean_all2[i], variance_all2[i] = compute_mean_variance2(n_sample_all[i])"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "FkCX-hxxAnsw"
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},
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"outputs": [],
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"source": [
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"fig,ax1 = plt.subplots()\n",
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"ax1.semilogx(n_sample_all, mean_all,'r-',label='mean estimate')\n",
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"ax1.fill_between(n_sample_all, mean_all-2*np.sqrt(variance_all), mean_all+2*np.sqrt(variance_all))\n",
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"ax1.set_xlabel(\"Number of samples\")\n",
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"ax1.set_ylabel(\"Mean of estimate\")\n",
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"ax1.plot([1,500],[0.43160702267383166,0.43160702267383166],'k--',label='true value')\n",
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"ax1.set_ylim(-5,6)\n",
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"ax1.set_title(\"First function\")\n",
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"ax1.legend()\n",
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"\n",
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"fig2,ax2 = plt.subplots()\n",
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"ax2.semilogx(n_sample_all, mean_all2,'r-',label='mean estimate')\n",
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"ax2.fill_between(n_sample_all, mean_all2-2*np.sqrt(variance_all2), mean_all2+2*np.sqrt(variance_all2))\n",
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"ax2.set_xlabel(\"Number of samples\")\n",
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"ax2.set_ylabel(\"Mean of estimate\")\n",
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"ax2.plot([0,500],[0.43160428638892556,0.43160428638892556],'k--',label='true value')\n",
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"ax2.set_ylim(-5,6)\n",
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"ax2.set_title(\"Second function\")\n",
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"ax2.legend()\n",
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"plt.show()"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"id": "EtBP6NeLwZqz"
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},
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"source": [
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"You can see that the variance of the estimate of the second function is considerably worse than the estimate of the variance of estimate of the first function\n",
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"\n",
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"TODO: Think about why this is."
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {
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"id": "_wuF-NoQu1--"
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},
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"source": [
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" Now let's repeat this experiment with the second function, but this time use importance sampling with auxiliary distribution:\n",
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"\n",
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" \\begin{equation}\n",
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" q(y)=\\text{Norm}_{y}[3,1]\n",
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" \\end{equation}\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "jPm0AVYVIDnn"
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},
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"outputs": [],
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"source": [
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"def q_y(y):\n",
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" return (1/np.sqrt(2*np.pi)) * np.exp(-0.5 * (y-3) * (y-3))\n",
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"\n",
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"def compute_expectation2b(n_samples):\n",
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" # TODO -- complete this function\n",
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" # 1. Draw n_samples from auxiliary distribution\n",
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" # 2. Compute f2[y] for those samples\n",
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" # 3. Scale the results by pr_y / q_y\n",
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" # 4. Compute the mean of these weighted samples\n",
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" # Replace this line\n",
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" expectation = 0\n",
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"\n",
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" return expectation"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "No2ByVvOM2yQ"
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},
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"outputs": [],
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"source": [
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"# Set the seed so the random numbers are all the same\n",
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"np.random.seed(0)\n",
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"\n",
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"# Compute the expectation with a very large number of samples (good estimate)\n",
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"n_samples = 100000000\n",
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"expected_f2= compute_expectation2b(n_samples)\n",
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"print(\"Your value: \", expected_f2,\", True value: 0.43163734204459125 \")"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "6v8Jc7z4M3Mk"
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},
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"outputs": [],
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"source": [
|
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"def compute_mean_variance2b(n_sample):\n",
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" n_estimate = 10000\n",
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" estimates = np.zeros((n_estimate,1))\n",
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" for i in range(n_estimate):\n",
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" estimates[i] = compute_expectation2b(n_sample.astype(int))\n",
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" return np.mean(estimates), np.var(estimates)\n",
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"\n",
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"# Compute the variance for 1,2,... 20 samples\n",
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"mean_all2b = np.zeros_like(n_sample_all)\n",
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"variance_all2b = np.zeros_like(n_sample_all)\n",
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"for i in range(len(n_sample_all)):\n",
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" print(\"Computing variance for expectation with %d samples\"%(n_sample_all[i]))\n",
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" mean_all2b[i], variance_all2b[i] = compute_mean_variance2b(n_sample_all[i])"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "C0beD4sNNM3L"
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},
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"outputs": [],
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"source": [
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"fig,ax1 = plt.subplots()\n",
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"ax1.semilogx(n_sample_all, mean_all,'r-',label='mean estimate')\n",
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"ax1.fill_between(n_sample_all, mean_all-2*np.sqrt(variance_all), mean_all+2*np.sqrt(variance_all))\n",
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"ax1.set_xlabel(\"Number of samples\")\n",
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"ax1.set_ylabel(\"Mean of estimate\")\n",
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"ax1.plot([1,500],[0.43160702267383166,0.43160702267383166],'k--',label='true value')\n",
|
|
"ax1.set_ylim(-5,6)\n",
|
|
"ax1.set_title(\"First function\")\n",
|
|
"ax1.legend()\n",
|
|
"\n",
|
|
"fig2,ax2 = plt.subplots()\n",
|
|
"ax2.semilogx(n_sample_all, mean_all2,'r-',label='mean estimate')\n",
|
|
"ax2.fill_between(n_sample_all, mean_all2-2*np.sqrt(variance_all2), mean_all2+2*np.sqrt(variance_all2))\n",
|
|
"ax2.set_xlabel(\"Number of samples\")\n",
|
|
"ax2.set_ylabel(\"Mean of estimate\")\n",
|
|
"ax2.plot([0,500],[0.43160428638892556,0.43160428638892556],'k--',label='true value')\n",
|
|
"ax2.set_ylim(-5,6)\n",
|
|
"ax2.set_title(\"Second function\")\n",
|
|
"ax2.legend()\n",
|
|
"plt.show()\n",
|
|
"\n",
|
|
"fig2,ax2 = plt.subplots()\n",
|
|
"ax2.semilogx(n_sample_all, mean_all2b,'r-',label='mean estimate')\n",
|
|
"ax2.fill_between(n_sample_all, mean_all2b-2*np.sqrt(variance_all2b), mean_all2b+2*np.sqrt(variance_all2b))\n",
|
|
"ax2.set_xlabel(\"Number of samples\")\n",
|
|
"ax2.set_ylabel(\"Mean of estimate\")\n",
|
|
"ax2.plot([0,500],[ 0.43163734204459125, 0.43163734204459125],'k--',label='true value')\n",
|
|
"ax2.set_ylim(-5,6)\n",
|
|
"ax2.set_title(\"Second function with importance sampling\")\n",
|
|
"ax2.legend()\n",
|
|
"plt.show()"
|
|
]
|
|
},
|
|
{
|
|
"attachments": {},
|
|
"cell_type": "markdown",
|
|
"metadata": {
|
|
"id": "y8rgge9MNiOc"
|
|
},
|
|
"source": [
|
|
"You can see that the importance sampling technique has reduced the amount of variance for any given number of samples."
|
|
]
|
|
}
|
|
],
|
|
"metadata": {
|
|
"colab": {
|
|
"authorship_tag": "ABX9TyNecz9/CDOggPSmy1LjT/Dv",
|
|
"include_colab_link": true,
|
|
"provenance": []
|
|
},
|
|
"kernelspec": {
|
|
"display_name": "Python 3",
|
|
"name": "python3"
|
|
},
|
|
"language_info": {
|
|
"name": "python"
|
|
}
|
|
},
|
|
"nbformat": 4,
|
|
"nbformat_minor": 0
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}
|