755 lines
31 KiB
Plaintext
755 lines
31 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "4c431265",
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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": "026433a4",
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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": "3b1bff64",
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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": "41d26cde",
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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": "de05c5cadee95d63f1acb0ab3f82894f",
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"grade": false,
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"grade_id": "cell-f29a87a28188c3d0",
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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 2__\n",
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"\n",
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"Code from the lecture:"
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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": "cb41d2a1",
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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": "5435cd2800cbe70e733a364b79e86c9b",
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"grade": false,
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"grade_id": "cell-a6520f459483332d",
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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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"from scipy.integrate import quad\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 inversion_sample(f_inverse):\n",
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" '''Obtain an inversion sample based on the inverse-CDF f_inverse.'''\n",
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" return f_inverse(rng.random())\n",
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"\n",
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"def compare_plot(samples,pdf,xmin,xmax,bins):\n",
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" '''Draw a plot comparing the histogram of the samples to the expectation coming from the pdf.'''\n",
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" xval = np.linspace(xmin,xmax,bins+1)\n",
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" binsize = (xmax-xmin)/bins\n",
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" # Calculate the expected numbers by numerical integration of the pdf over the bins\n",
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" expected = np.array([quad(pdf,xval[i],xval[i+1])[0] for i in range(bins)])/binsize\n",
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" measured = np.histogram(samples,bins,(xmin,xmax))[0]/(len(samples)*binsize)\n",
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" plt.plot(xval,np.append(expected,expected[-1]),\"-k\",drawstyle=\"steps-post\")\n",
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" plt.bar((xval[:-1]+xval[1:])/2,measured,width=binsize)\n",
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" plt.xlim(xmin,xmax)\n",
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" plt.legend([\"expected\",\"histogram\"])\n",
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" plt.show()\n",
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" \n",
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"def gaussian(x):\n",
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" return np.exp(-x*x/2)/np.sqrt(2*np.pi)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "3317e002",
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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": "d2c3d8374cf18fd1a12c91353f28dbcf",
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"grade": false,
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"grade_id": "cell-e6c28b1e3e8371c3",
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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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"## Sampling random variables via the inversion method \n",
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"__(35 Points)__\n",
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"\n",
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"Recall from the lecture that for any real random variable $X$ we can construct an explicit random variable via the inversion method that is identically distributed. This random variable is given by $F_X^{-1}(U)$ where $F_X$ is the CDF of $X$ and $U$ is a uniform random variable on $(0,1)$ and \n",
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"\n",
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"$$\n",
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"F_X^{-1}(p) := \\inf\\{ x\\in\\mathbb{R} : F_X(x) \\geq p\\}.\n",
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"$$\n",
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"\n",
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"This gives a very general way of sampling $X$ in a computer program, as you will find out in this exercise.\n",
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"\n",
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"__(a)__ Let $X$ be an **exponential random variable** with **rate** $\\lambda$, i.e. a continuous random variable with probability density function $f_X(x) = \\lambda e^{-\\lambda x}$ for $x > 0$. Write a function `f_inverse_exponential` that computes $F_X^{-1}(p)$. Illustrate the corresponding sampler with the help of the function `compare_plot` above. __(10 pts)__"
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]
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},
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{
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"cell_type": "markdown",
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"id": "6f2c475a",
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"metadata": {
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"deletable": false,
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"nbgrader": {
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"cell_type": "markdown",
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"checksum": "4292b1a356454d496a93ef6555f0a7ae",
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"grade": true,
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"grade_id": "cell-311fd25e116f5066",
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"locked": false,
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"points": 5,
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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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"source": [
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"Reasoning from the PDF, we can find the CDF and invert that as follows.\n",
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"\n",
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"$$\n",
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"f_X(x) = \\lambda{}e^{-\\lambda{}x}\n",
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"$$\n",
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"$$\n",
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"\\implies F_X(x)\n",
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" = \\int_{-\\infty}^x f_X(t)dt\n",
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" = \\int_0^x \\lambda{}e^{-\\lambda{}t}dt\n",
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" = \\left[ -e^{\\lambda{}t} \\right]_{t = 0}^x\n",
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" = 1 - e^{\\lambda{}x}\n",
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" = \\mathbb{P}(X \\leq x) = p\n",
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"$$\n",
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"for $x \\in [0, \\infty)$, otherwise zero.\n",
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"\n",
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"Now we seek $x$ as a function of $p$.\n",
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"\n",
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"$$\n",
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"1 - e^{\\lambda{}x} = p\n",
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"\\iff -\\lambda{}x = \\ln{(1-p)}\n",
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"\\iff x = \\frac{\\ln{(1-p)}}{-\\lambda} = F^{-1}_X(p)\n",
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"$$\n",
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"which works, as $1 - p \\geq 0$ as $p \\in [0, 1]$, allowing $\\ln{0} = -\\infty$."
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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": "e6b6428c",
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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": "90de5b60de4e43881ab85442cdff704a",
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"grade": false,
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"grade_id": "cell-06ef7d054d38f5c6",
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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",
|
|
"text/plain": [
|
|
"<Figure size 432x288 with 1 Axes>"
|
|
]
|
|
},
|
|
"metadata": {
|
|
"needs_background": "light"
|
|
},
|
|
"output_type": "display_data"
|
|
}
|
|
],
|
|
"source": [
|
|
"def f_inv_exponential(lam,p):\n",
|
|
" return -np.log(1 - p)/lam\n",
|
|
"\n",
|
|
"f_X = lambda x, lam: lam*np.exp(-lam*x) if x >= 0 else 0\n",
|
|
"\n",
|
|
"for lam in [1.5]:\n",
|
|
" pdf = lambda x: f_X(x, lam)\n",
|
|
" samples = [inversion_sample(lambda p: f_inv_exponential(lam, p)) for _ in range(100000)]\n",
|
|
" compare_plot(samples, pdf, -1, 4, 30)"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": 4,
|
|
"id": "804aedbf",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "bce45fa412ba32138080832767338e9d",
|
|
"grade": true,
|
|
"grade_id": "cell-2022e00546cf1bb0",
|
|
"locked": true,
|
|
"points": 5,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"from nose.tools import assert_almost_equal\n",
|
|
"assert_almost_equal(f_inv_exponential(1.0,0.6),0.916,delta=0.001)\n",
|
|
"assert_almost_equal(f_inv_exponential(0.3,0.2),0.743,delta=0.001)"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "d590b09d",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "08fdb1c6ca42806566800f06d7ffb22b",
|
|
"grade": false,
|
|
"grade_id": "cell-f7e0d9b58c948be5",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"__(b)__ Let now $X$ have the **Pareto distribution** of **shape** $\\alpha > 0$ on $(b,\\infty)$, which has probability density function $f_X(x) = \\alpha b^{\\alpha} x^{-\\alpha-1}$ for $x > b$. Write a function `f_inv_pareto` that computes $F_X^{-1}(p)$. Compare a histogram with a plot of $f_X(x)$ to verify your function numerically. __(10 pts)__"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "47c7a42f",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "1d1fc6a16462f0d238005fdb33a99857",
|
|
"grade": true,
|
|
"grade_id": "cell-199713328dcd510d",
|
|
"locked": false,
|
|
"points": 5,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"YOUR ANSWER HERE"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "e177f32d",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "eb07f40a935275cf5883204fc817beaa",
|
|
"grade": false,
|
|
"grade_id": "cell-074f6a1fd6375c22",
|
|
"locked": false,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"### Solution\n",
|
|
"def f_inv_pareto(alpha,b,p):\n",
|
|
" # YOUR CODE HERE\n",
|
|
" raise NotImplementedError()\n",
|
|
"\n",
|
|
"# plotting\n",
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "c0e1426f",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "62920089752d067b0945eb1d6d98135f",
|
|
"grade": true,
|
|
"grade_id": "cell-726b321246679d28",
|
|
"locked": true,
|
|
"points": 5,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"from nose.tools import assert_almost_equal\n",
|
|
"assert_almost_equal(f_inv_pareto(1.0,1.5,0.6),3.75,delta=0.0001)\n",
|
|
"assert_almost_equal(f_inv_pareto(2.0,2.25,0.3),2.689,delta=0.001)"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "66d91446",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "0f3c9abbe9fe756c5cf4bdd6a8a37ac2",
|
|
"grade": false,
|
|
"grade_id": "cell-50306550727804ca",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"__(c)__ Let $X$ be a discrete random variable taking values in $\\{1,2,\\ldots,n\\}$. Write a Python function `f_inv_discrete` that takes the probability mass function $p_X$ as a list `prob_list` given by $[p_X(1),\\ldots,p_X(n)]$ and returns a random sample with the distribution of $X$ using the inversion method. Verify the working of your function numerically on an example. __(15 pts)__"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "210f1302",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "93d51c9c889dd5ba3490e0ee298d4240",
|
|
"grade": false,
|
|
"grade_id": "cell-694eb1261c2dc217",
|
|
"locked": false,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"def f_inv_discrete(prob_list,p):\n",
|
|
" # YOUR CODE HERE\n",
|
|
" raise NotImplementedError()\n",
|
|
"\n",
|
|
"# plotting\n",
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "3c691f0a",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "b11d87e414ba9dfe2741d73dd95a2f12",
|
|
"grade": true,
|
|
"grade_id": "cell-140af6b31464fbef",
|
|
"locked": true,
|
|
"points": 15,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"assert f_inv_discrete([0.5,0.5],0.4)==1\n",
|
|
"assert f_inv_discrete([0.5,0.5],0.8)==2\n",
|
|
"assert f_inv_discrete([0,0,1],0.1)==3"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "47546d37",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "32dd38f0f963c6132fcbe3ef1f5b9682",
|
|
"grade": false,
|
|
"grade_id": "cell-49fd13dc534dfa28",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"## Central limit theorem? \n",
|
|
"__(35 Points)__\n",
|
|
"\n",
|
|
"In this exercise we will have a closer look at central limits of the Pareto distribution, for which you implemented a random sampler in the previous exercise. By performing the appropriate integrals it is straightforward to show that \n",
|
|
"\n",
|
|
"$$ \n",
|
|
"\\mathbb{E}[X] = \\begin{cases} \\infty & \\text{for }\\alpha \\leq 1 \\\\ \\frac{\\alpha b}{\\alpha - 1} & \\text{for }\\alpha > 1 \\end{cases}, \\qquad \\operatorname{Var}(X) = \\begin{cases} \\infty & \\text{for }\\alpha \\leq 2 \\\\ \\frac{\\alpha b^2}{(\\alpha - 1)^2(\\alpha-2)} & \\text{for }\\alpha > 2 \\end{cases}.\n",
|
|
"$$\n",
|
|
"\n",
|
|
"This shows in particular that the distribution is **heavy tailed**, in the sense that some moments $\\mathbb{E}[X^k]$ diverge."
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "ccae582d",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "e6d5659ef88eccfb693b35a088d0d50f",
|
|
"grade": false,
|
|
"grade_id": "cell-a05e255c144ef6c5",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"__(a)__ Write a function `sample_Zn` that produces a random sample for $Z_n= \\frac{\\sqrt{n}}{\\sigma_X}(\\bar{X}_n - \\mathbb{E}[X])$ given $\\alpha>2$, $b>0$ and $n\\geq 1$. Visually verify the central limit theorem for $\\alpha = 4$, $b=1$ and $n=1000$ by comparing a histogram of $Z_n$ to the standard normal distribution (you may use `compare_plot`). __(10 pts)__"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "82fe6efd",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "177917ec75361799067d6c23a28569cd",
|
|
"grade": false,
|
|
"grade_id": "cell-b7186322b09717f8",
|
|
"locked": false,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"def sample_Zn(alpha,b,n):\n",
|
|
" # YOUR CODE HERE\n",
|
|
" raise NotImplementedError()\n",
|
|
"\n",
|
|
"# Plotting\n",
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "b5360d77",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "e50b33644ddd6bce391b36cefcc2e308",
|
|
"grade": true,
|
|
"grade_id": "cell-5d16b014bef9d86f",
|
|
"locked": true,
|
|
"points": 10,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"assert_almost_equal(np.mean([sample_Zn(3.5,2.1,100) for _ in range(100)]),0,delta=0.3)\n",
|
|
"assert_almost_equal(np.std([sample_Zn(3.5,2.1,100) for _ in range(100)]),1,delta=0.3)"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "6192f05d",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "08ece68d59de21d798d9a955f59be690",
|
|
"grade": false,
|
|
"grade_id": "cell-3e7a23657e9b8374",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"__(b)__ Now take $\\alpha = 3/2$ and $b=1$. \n",
|
|
"With some work (which you do not have to do) one can show that the characteristic function of $X$ admits the following expansion around $t=0$,\n",
|
|
"\n",
|
|
"$$\n",
|
|
"\\varphi_X(t) = 1 + 3 i t - (|t|+i t)\\,\\sqrt{2\\pi|t|} + O(t^{2}).\n",
|
|
"$$\n",
|
|
"\n",
|
|
"Based on this, prove the **generalized CLT** for this particular distribution $X$ which states that $Z_n = c\\, n^{1/3} (\\bar{X}_n - \\mathbb{E}[X])$ in the limit $n\\rightarrow\\infty$ converges in distribution, with a to-be-determined choice of overall constant $c$, to a limiting random variable $\\mathcal{S}$ with characteristic function \n",
|
|
"\n",
|
|
"$$\n",
|
|
"\\varphi_{\\mathcal{S}}(t) = \\exp\\big(-(|t|+it)\\sqrt{|t|}\\big).\n",
|
|
"$$\n",
|
|
"\n",
|
|
"__(15 pts)__"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "9735cd88",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "dfd8683eea5663baa81f138a2809722b",
|
|
"grade": true,
|
|
"grade_id": "cell-b25551eca32c4807",
|
|
"locked": false,
|
|
"points": 15,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"YOUR ANSWER HERE"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "5b1d9f54",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "342020128f929d47eabfdf9c075ff20c",
|
|
"grade": false,
|
|
"grade_id": "cell-d1701433c3c77172",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"__(c)__ The random variable $\\mathcal{S}$ has a [stable Lévy distribution](https://en.wikipedia.org/wiki/Stable_distribution) with index $\\alpha = 3/2$ and skewness $\\beta = 1$. Its probability density function $f_{\\mathcal{S}}(x)$ does not admit a simple expression, but can be accessed numerically using SciPy's `scipy.stats.levy_stable.pdf(x,1.5,1.0)`. Verify numerically that the generalized CLT of part (b) holds by comparing an appropriate histogram to this PDF. __(10 pts)__"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "b06896e5",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "c6fe081427f342c354ee8a9b3b3331e7",
|
|
"grade": true,
|
|
"grade_id": "cell-e08d054985cfa762",
|
|
"locked": false,
|
|
"points": 10,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"from scipy.stats import levy_stable\n",
|
|
"\n",
|
|
"# YOUR CODE HERE\n",
|
|
"raise NotImplementedError()"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "f49856d8",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "d8c57e5a527eaad8318e7d31dba01694",
|
|
"grade": false,
|
|
"grade_id": "cell-bc80caacda124bf9",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"## Joint probability density functions and sampling the normal distribution \n",
|
|
"__(30 Points)__\n",
|
|
"\n",
|
|
"Let $\\Phi$ be a uniform random variable on $(0,2\\pi)$ and $R$ an independent continuous random variable with probability density function $f_R(r) = r\\,e^{-r^2/2}$ for $r>0$. Set $X = R \\cos \\Phi$ and $Y = R \\sin \\Phi$. This is called the **Box-Muller transform**.\n",
|
|
"\n",
|
|
"__(a)__ Since $\\Phi$ and $R$ are independent, the joint probability density of $\\Phi$ and $R$ is $f_{\\Phi,R}(\\phi,r) = f_\\Phi(\\phi)f_R(r) = \\frac{1}{2\\pi}\\, r\\,e^{-r^2/2}$. Show by change of variables that $X$ and $Y$ are also independent and both distributed as a standard normal distribution $\\mathcal{N}$. __(15 pts)__"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "aa3821de",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "2514e6664aeb4e24a9e881522a8f3a0f",
|
|
"grade": true,
|
|
"grade_id": "cell-4f20e3b730ba0d23",
|
|
"locked": false,
|
|
"points": 15,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"YOUR ANSWER HERE"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "5d064cef",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"editable": false,
|
|
"nbgrader": {
|
|
"cell_type": "markdown",
|
|
"checksum": "1af73334332fe512ef7d0edb5803a58d",
|
|
"grade": false,
|
|
"grade_id": "cell-2f07fdb2a906bb71",
|
|
"locked": true,
|
|
"schema_version": 3,
|
|
"solution": false,
|
|
"task": false
|
|
}
|
|
},
|
|
"source": [
|
|
"__(b)__ Write a function to sample a pair of independent normal random variables using the Box-Muller transform. Hint: to sample $R$ you can use the inversion method of the first exercise. Produce a histogram to check the distribution of your normal variables. __(15 pts)__"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"id": "e4023f99",
|
|
"metadata": {
|
|
"deletable": false,
|
|
"nbgrader": {
|
|
"cell_type": "code",
|
|
"checksum": "86173970c865da7b0cb8ab78ec4a87b6",
|
|
"grade": true,
|
|
"grade_id": "cell-9bf8873cce1d179c",
|
|
"locked": false,
|
|
"points": 15,
|
|
"schema_version": 3,
|
|
"solution": true,
|
|
"task": false
|
|
}
|
|
},
|
|
"outputs": [],
|
|
"source": [
|
|
"def random_normal_pair():\n",
|
|
" '''Return two independent normal random variables.'''\n",
|
|
" # YOUR CODE HERE\n",
|
|
" raise NotImplementedError()\n",
|
|
" return x, y\n",
|
|
"\n",
|
|
"# Plotting\n",
|
|
"# 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
|
|
}
|