638 lines
29 KiB
Plaintext
638 lines
29 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "ac27651a",
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"metadata": {},
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"source": [
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"# Exercise sheet\n",
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"\n",
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"Some general remarks about the exercises:\n",
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"* For your convenience functions from the lecture are included below. Feel free to reuse them without copying to the exercise solution box.\n",
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"* For each part of the exercise a solution box has been added, but you may insert additional boxes. Do not hesitate to add Markdown boxes for textual or LaTeX answers (via `Cell > Cell Type > Markdown`). But make sure to replace any part that says `YOUR CODE HERE` or `YOUR ANSWER HERE` and remove the `raise NotImplementedError()`.\n",
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"* Please make your code readable by humans (and not just by the Python interpreter): choose informative function and variable names and use consistent formatting. Feel free to check the [PEP 8 Style Guide for Python](https://www.python.org/dev/peps/pep-0008/) for the widely adopted coding conventions or [this guide for explanation](https://realpython.com/python-pep8/).\n",
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"* Make sure that the full notebook runs without errors before submitting your work. This you can do by selecting `Kernel > Restart & Run All` in the jupyter menu.\n",
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"* For some exercises test cases have been provided in a separate cell in the form of `assert` statements. When run, a successful test will give no output, whereas a failed test will display an error message.\n",
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"* Each sheet has 100 points worth of exercises. Note that only the grades of sheets number 2, 4, 6, 8 count towards the course examination. Submitting sheets 1, 3, 5, 7 & 9 is voluntary and their grades are just for feedback.\n",
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"\n",
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"Please fill in your name here:"
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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": 1,
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"id": "d8b13e80",
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"metadata": {},
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"outputs": [],
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"source": [
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"NAME = \"Kees van Kempen\"\n",
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"NAMES_OF_COLLABORATORS = \"\""
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]
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},
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{
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"cell_type": "markdown",
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"id": "e21d4bc9",
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"metadata": {},
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"source": [
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"---"
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]
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},
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{
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"cell_type": "markdown",
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"id": "97ec4790",
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"metadata": {
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"deletable": false,
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"editable": false,
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"nbgrader": {
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"cell_type": "markdown",
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"checksum": "ba0b514985ed8d79af63814bda6ee328",
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"grade": false,
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"grade_id": "cell-70094cd1c9529a28",
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"locked": true,
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"schema_version": 3,
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"solution": false,
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"task": false
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}
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},
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"source": [
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"**Exercise sheet 3**\n",
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"\n",
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"Code from the lectures:"
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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": 2,
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"id": "948639fc",
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"metadata": {
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"deletable": false,
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"editable": false,
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"nbgrader": {
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"cell_type": "code",
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"checksum": "1e25d190936687e34fe3ac01d4dfce44",
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"grade": false,
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"grade_id": "cell-9a6bf70e64f84715",
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"locked": true,
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"schema_version": 3,
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"solution": false,
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"task": false
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}
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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.pylab as plt\n",
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"\n",
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"rng = np.random.default_rng()\n",
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"%matplotlib inline\n",
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"\n",
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"def sample_acceptance_rejection(sample_z,accept_probability):\n",
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" while True:\n",
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" x = sample_z()\n",
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" if rng.random() < accept_probability(x):\n",
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" return x \n",
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" \n",
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"def estimate_expectation(sampler,n):\n",
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" '''Compute beste estimate of mean and 1-sigma error with n samples.'''\n",
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" samples = [sampler() for _ in range(n)]\n",
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" return np.mean(samples), np.std(samples)/np.sqrt(n-1)\n",
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"\n",
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"def estimate_expectation_one_pass(sampler,n):\n",
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" sample_mean = sample_square_dev = 0.0\n",
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" for k in range(1,n+1):\n",
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" delta = sampler() - sample_mean\n",
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" sample_mean += delta / k\n",
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" sample_square_dev += (k-1)*delta*delta/k \n",
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" return sample_mean, np.sqrt(sample_square_dev / (n*(n-1)))"
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]
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},
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{
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"cell_type": "markdown",
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"id": "bce418c0",
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"metadata": {
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"deletable": false,
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"editable": false,
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"nbgrader": {
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"cell_type": "markdown",
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"checksum": "90c9636c93064db3b008a2e9ec210117",
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"grade": false,
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"grade_id": "cell-8763c7e3277ef0ce",
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"locked": true,
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"schema_version": 3,
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"solution": false,
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"task": false
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}
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},
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"source": [
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"## Acceptance-rejection sampling\n",
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"\n",
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"**(35 points)**\n",
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"\n",
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"The goal of this exercise is to develop a fast sampling algorithm of the discrete random variable $X$ with probability mass function $$p_X(k) = \\frac{6}{\\pi^2} k^{-2}, \\qquad k=1,2,\\ldots$$\n",
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"\n",
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"__(a)__ Let $Z$ be the discrete random variable with $p_Z(k) = \\frac{1}{k} - \\frac{1}{k+1}$ for $k=1,2,\\ldots$. Write a function to compute the inverse CDF $F_Z^{-1}(u)$, such that you can use the inversion method to sample $Z$ efficiently. **(15 pts)**"
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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": 3,
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"id": "564c7f00",
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"metadata": {
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"deletable": false,
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"nbgrader": {
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"cell_type": "code",
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"checksum": "da96fc27e757b00d6e55befc89f5afda",
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"grade": false,
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"grade_id": "cell-b0b3fcf6d9bf35a0",
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"locked": false,
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"schema_version": 3,
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"solution": true,
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"task": false
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}
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},
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"outputs": [],
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"source": [
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"def f_inverse_Z(u):\n",
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" '''Compute the inverse CDF of Z, i.e. F_Z^{-1}(u) for 0 <= u <= 1.'''\n",
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" assert 0 <= u and u <= 1\n",
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" \n",
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" # we need to round to return a natural number (excluding zero, yes).\n",
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" return np.ceil(u/(1 - u)).astype(int)\n",
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"\n",
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"def random_Z():\n",
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" return int(f_inverse_Z(rng.random())) # make sure to return an integer"
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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": 4,
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"id": "2ca9d9ca",
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"metadata": {
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"deletable": false,
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"editable": false,
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"nbgrader": {
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"cell_type": "code",
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"checksum": "3eeabc3f6572ed217a002adefda34876",
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"grade": true,
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"grade_id": "cell-a28cb07f90510b94",
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"locked": true,
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"points": 15,
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"schema_version": 3,
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"solution": false,
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"task": false
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}
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},
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"outputs": [],
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"source": [
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"assert f_inverse_Z(0.2)==1\n",
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"assert f_inverse_Z(0.51)==2\n",
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"assert f_inverse_Z(0.76)==4\n",
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"assert f_inverse_Z(0.991)==111"
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]
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},
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{
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"cell_type": "markdown",
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"id": "9a1b3dcd",
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"metadata": {
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"deletable": false,
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"editable": false,
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"nbgrader": {
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"cell_type": "markdown",
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"checksum": "78f978d46d5f742623cf174e88323ec9",
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"grade": false,
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"grade_id": "cell-576d3991eef6bf11",
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"locked": true,
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"schema_version": 3,
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"solution": false,
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"task": false
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}
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},
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"source": [
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"__(b)__ Implement a sampler for $X$ using acceptance-rejection based on the sampler of $Z$. For this you need to first determine a $c$ such that $p_X(k) \\leq c\\,p_Z(k)$ for all $k=1,2,\\ldots$, and then consider an acceptance probability $p_X(k) / (c p_Z(k))$. Verify the validity of your sampler numerically (e.g. for $k=1,\\ldots,10$). **(20 pts)**"
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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": 5,
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"id": "8e201507",
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"metadata": {
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"deletable": false,
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"nbgrader": {
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"cell_type": "code",
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"checksum": "862ecc31c9ee75e459dafe5262ddc776",
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"grade": false,
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"grade_id": "cell-20fd55e7a4c4c531",
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"locked": false,
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"schema_version": 3,
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"solution": true,
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"task": false
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}
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},
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"outputs": [
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{
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"data": {
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"image/png": 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\n",
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"text/plain": [
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"<Figure size 432x288 with 1 Axes>"
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|
]
|
|
},
|
|
"metadata": {
|
|
"needs_background": "light"
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},
|
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"output_type": "display_data"
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}
|
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],
|
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"source": [
|
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"def accept_probability_X(k):\n",
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" '''Return the appropriate acceptance probability on the event Z=k.'''\n",
|
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" \n",
|
|
" # The resulting c is found by doing algebraic things, and seeing that\n",
|
|
" # 6/pi^2*(1 + 1/k) <= c should hold for all allowed k. For k = 1, c is\n",
|
|
" # the largest, so: tada.\n",
|
|
" c = 12*np.pi**(-2)\n",
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" \n",
|
|
" # But we do not need this variable, we only need the acceptance probability.\n",
|
|
" return .5*(1 + 1/k)\n",
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" \n",
|
|
"def random_X():\n",
|
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" return sample_acceptance_rejection(random_Z,accept_probability_X)\n",
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|
"\n",
|
|
"# Verify numerically\n",
|
|
"samples = [random_X() for _ in range(1000000)]\n",
|
|
"k = np.array(range(1,11))\n",
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|
"\n",
|
|
"# The theoretical densities to compare the densities from the\n",
|
|
"# acceptance-rejection densities to:\n",
|
|
"p_X = lambda k: 6/(np.pi*k)**2\n",
|
|
"\n",
|
|
"plt.figure()\n",
|
|
"plt.hist(samples, k - .5, density=True, color=\"lightblue\", label=\"density of acceptance-rejection samples\")\n",
|
|
"plt.scatter(k[:-1], [p_X(i) for i in k[:-1]], s=800, marker=\"_\", color=\"black\", label=\"theoretical\")\n",
|
|
"plt.yscale(\"log\")\n",
|
|
"plt.ylabel(\"$p_X(k)$\")\n",
|
|
"plt.xlabel(\"$k$\")\n",
|
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"plt.legend()\n",
|
|
"plt.show()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": 6,
|
|
"id": "1b0e1d47",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "641adac2c6462be641ae2d45a4d80e7d",
|
|
"grade": true,
|
|
"grade_id": "cell-04bad69f7cc2ad18",
|
|
"locked": true,
|
|
"points": 20,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"from nose.tools import assert_almost_equal\n",
|
|
"assert min([random_X() for _ in range(10000)]) >= 1\n",
|
|
"assert_almost_equal([random_X() for _ in range(10000)].count(1),6079,delta=400)\n",
|
|
"assert_almost_equal([random_X() for _ in range(10000)].count(3),675,delta=75)"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "d010b3c8",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "ec5e4c478f45388c302b82613e45a0c0",
|
|
"grade": false,
|
|
"grade_id": "cell-f831d5b44932cf2f",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"## Monte Carlo integration & Importance sampling\n",
|
|
"\n",
|
|
"**(30 Points)**\n",
|
|
"\n",
|
|
"Consider the integral \n",
|
|
"\n",
|
|
"$$\n",
|
|
"I = \\int_0^1 \\sin(\\pi x(1-x))\\mathrm{d}x = \\mathbb{E}[X], \\quad X=g(U), \\quad g(U)=\\sin(\\pi U(1-U)),\n",
|
|
"$$ \n",
|
|
"\n",
|
|
"where $U$ is a uniform random variable in $(0,1)$.\n",
|
|
"\n",
|
|
"__(a)__ Use Monte Carlo integration based on sampling $U$ to estimate $I$ with $1\\sigma$ error at most $0.001$. How many samples do you need? (It is not necessary to automate this: trial and error is sufficient.) **(10 pts)**"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "bbf5c843",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "febbb705202776d2af26bd95c4c08625",
|
|
"grade": true,
|
|
"grade_id": "cell-47eab0537d21690e",
|
|
"locked": false,
|
|
"points": 10,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "26b8ff78",
|
|
"metadata": {},
|
|
"source": [
|
|
"__(b)__ Choose a random variable $Z$ on $(0,1)$ whose density resembles the integrand of $I$ and which you know how to sample efficiently (by inversion method, acceptance-rejection, or a built-in Python function). Estimate $I$ again using importance sampling, i.e. $I = \\mathbb{E}[X']$ where $X' = g(Z) f_U(Z)/f_Z(Z)$, with an error of at most 0.001. How many samples did you need this time? **(20 pts)**"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "5793420a",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "d546b8c253155efa96a399ec15c24fb8",
|
|
"grade": true,
|
|
"grade_id": "cell-467d7240beab5b29",
|
|
"locked": false,
|
|
"points": 20,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"def sample_nice_Z():\n",
|
|
" '''Sample from the nice distribution Z'''\n",
|
|
" # YOUR CODE HERE\n",
|
|
" raise NotImplementedError()\n",
|
|
" \n",
|
|
"def sample_X_prime():\n",
|
|
" '''Sample from X'.'''\n",
|
|
" # YOUR CODE HERE\n",
|
|
" raise NotImplementedError()\n",
|
|
" \n",
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "e2a5b934",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "8fd70da242e4db50dd7a7db9851e839a",
|
|
"grade": false,
|
|
"grade_id": "cell-59ae6e6f3f476d2a",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"## Direct sampling of Dyck paths\n",
|
|
"\n",
|
|
"**(35 points)**\n",
|
|
"\n",
|
|
"Direct sampling of random variables in high dimensions requires some luck and/or ingenuity. Here is an example of a probability distribution on $\\mathbb{Z}^{2n+1}$ that features prominently in the combinatorial literature and can be sampled directly in an efficient manner. A sequence $\\mathbf{x}\\equiv(x_0,x_1,\\ldots,x_{2n})\\in\\mathbb{Z}^{2n+1}$ is said to be a **Dyck path** if $x_0=x_{2n}=0$, $x_i \\geq 0$ and $|x_{i}-x_{i-1}|=1$ for all $i=1,\\ldots,2n$. Dyck paths are counted by the Catalan numbers $C(n) = \\frac{1}{n+1}\\binom{2n}{n}$. Let $\\mathbf{X}=(X_0,\\ldots,X_n)$ be a **uniform Dyck path**, i.e. a random variable with probability mass function $p_{\\mathbf{X}}(\\mathbf{x}) = 1/C(n)$ for every Dyck path $\\mathbf{x}$. Here is one way to sample $\\mathbf{X}$."
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "d11a0889",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "0ea1008043f2711810b9da3c2cf96a02",
|
|
"grade": false,
|
|
"grade_id": "cell-3598f5c8aabad8b9",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"def random_dyck_path(n):\n",
|
|
" '''Returns a uniform Dyck path of length 2n as an array [x_0, x_1, ..., x_{2n}] of length 2n.'''\n",
|
|
" # produce a (2n+1)-step random walk from 0 to -1\n",
|
|
" increments = [1]*n +[-1]*(n+1)\n",
|
|
" rng.shuffle(increments)\n",
|
|
" unconstrained_walk = np.cumsum(increments)\n",
|
|
" # determine the first time it reaches its minimum\n",
|
|
" argmin = np.argmin(unconstrained_walk)\n",
|
|
" # cyclically permute the increments to ensure walk stays non-negative until last step\n",
|
|
" rotated_increments = np.roll(increments,-argmin)\n",
|
|
" # turn off the superfluous -1 step\n",
|
|
" rotated_increments[0] = 0\n",
|
|
" # produce dyck path from increments\n",
|
|
" walk = np.cumsum(rotated_increments)\n",
|
|
" return walk\n",
|
|
"\n",
|
|
"\n",
|
|
"plt.plot(random_dyck_path(50))\n",
|
|
"plt.show()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "3e6ad6ad",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "e88c23c1f0a6e2d23e0cd8ab4e2a20fd",
|
|
"grade": false,
|
|
"grade_id": "cell-8c2933e84038cd5e",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"__(a)__ Let $H$ be the (maximal) height of $X$, i.e. $H=\\max_i X_i$. Estimate the expected height $\\mathbb{E}[H]$ for $n = 2^5, 2^6, \\ldots, 2^{11}$ (including error bars). Determine the growth $\\mathbb{E}[H] \\approx a\\,n^\\beta$ via an appropriate fit. *Hint*: use the `scipy.optimize.curve_fit` function with the option `sigma = ...` to incorporate the standard errors on $\\mathbb{E}[H]$ in the fit. Note that when you supply the errors appropriately, fitting on linear or logarithmic scale should result in the same answer. **(25 pts)**"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "2b79f35a",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "54439c27bcec7884b6168c5bf92c27fd",
|
|
"grade": true,
|
|
"grade_id": "cell-310664cb8eaaa831",
|
|
"locked": false,
|
|
"points": 10,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"# Collect height estimates\n",
|
|
"n_values = [2**k for k in range(5,11+1)]\n",
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "77d3723b",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "960be7ade957300785f4b69233a8f462",
|
|
"grade": true,
|
|
"grade_id": "cell-0ca47ced5547d67c",
|
|
"locked": false,
|
|
"points": 10,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"from scipy.optimize import curve_fit\n",
|
|
"\n",
|
|
"# Fitting\n",
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()\n",
|
|
"print(\"Fit parameters: a = {}, beta = {}\".format(a_fit,beta_fit))"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "edd467e4",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "421cc39b43f6f9ebc46886d7ca0ab1e4",
|
|
"grade": true,
|
|
"grade_id": "cell-7ebce3629edd06a2",
|
|
"locked": false,
|
|
"points": 5,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"# Plotting\n",
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "142aa5cb",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "1df181e550da9f9c1edff06f45d51984",
|
|
"grade": false,
|
|
"grade_id": "cell-2dcc8f8e438aad86",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"__(b)__ Produce a histogram of the height $H / \\sqrt{n}$ for $n = 2^5, 2^6, \\ldots, 2^{11}$ and $3000$ samples each and demonstrate with a plot that it appears to converge in distribution as $n\\to\\infty$. *Hint*: you could call `plt.hist(...,density=True,histtype='step')` for each $n$ to plot the densities on top of each other. **(10 pts)**"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "eed4339c",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "cb48ed2a17571debd5f778f87f34a2ab",
|
|
"grade": true,
|
|
"grade_id": "cell-b2bb42d9970a7271",
|
|
"locked": false,
|
|
"points": 10,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()"
|
|
]
|
|
}
|
|
],
|
|
"metadata": {
|
|
"kernelspec": {
|
|
"display_name": "Python 3",
|
|
"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": "3.9.12"
|
|
}
|
|
},
|
|
"nbformat": 4,
|
|
"nbformat_minor": 5
|
|
}
|