{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Problem 1\n", "\n", "The likelihood for a coin flipping experiment will be defined by the binomial\n", "\n", "\\begin{equation}\n", "\\mathcal{L}(p) = \\binom{n}{Y}p^Y(1-p)^{n-Y}\n", "\\end{equation}\n", "\n", "where\n", " - $n$: total number of flips\n", " - $Y$: Observed number of Heads\n", " - $p$: Probability of each flip being Heads\n", " \n", "The test statistic, $\\lambda$, is defined as\n", "\n", "\\begin{equation}\n", "\\lambda=\\frac{\\mathcal{L}(p_0)}{\\mathcal{L}(\\hat{p})}\n", "\\end{equation}\n", "\n", "Where $\\hat{p}$ is the value of $p$ that maximizes $\\mathcal{L}$. It can be simply found by setting $\\frac{d}{dp}\\mathcal{L}(p)|_{p=\\hat{p}}=0$ that $\\hat{p}=\\frac{Y}{n}$. From this it follows that\n", "\n", "\\begin{equation}\n", "\\lambda=\\frac{1}{2^n}\\left(\\frac{n}{Y}\\right)^Y \\left(\\frac{n}{n-Y}\\right)^{n-Y}\n", "\\end{equation}\n", "\n", "The likelihood ratio test then is given (for some $\\alpha$), by\n", "\\begin{align}\n", "\\lambda &< k_\\alpha \\\\\n", "\\frac{1}{2^n}\\left(\\frac{n}{Y}\\right)^Y \\left(\\frac{n}{n-Y}\\right)^{n-Y} &< k_\\alpha\n", "\\end{align}" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": true }, "outputs": [], "source": [ "import numpy as np\n", "from scipy.stats import binom\n", "import matplotlib.pyplot as plt\n", "%matplotlib inline\n", "plt.rcParams['figure.dpi']=150" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "scrolled": false }, "outputs": [ { "name": "stderr", "output_type": "stream", "text": [ "/usr/lib/python3.6/site-packages/ipykernel/__main__.py:3: RuntimeWarning: divide by zero encountered in true_divide\n", " app.launch_new_instance()\n" ] }, { "data": { "text/plain": [ "Text(0,0.5,'$\\\\lambda$')" ] }, "execution_count": 2, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "Nflips = 36\n", "def lambda_(n, Y):\n", " return 2.0**-n * (n/Y)**Y * (n/(n-Y))**(n-Y)\n", "\n", "xs = np.array(range(Nflips+1))\n", "\n", "plt.bar(xs, lambda_(Nflips, xs))\n", "plt.xlabel('Y')\n", "plt.ylabel(r'$\\lambda$')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**a)** We have as a given that the exclusion region is $|Y-18| \\ge 4$. This corresponds to $k_\\alpha=0.4080...$, so now let's find the $\\alpha$ value by running some MC experiments." ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "alpha = 0.132401\n" ] } ], "source": [ "k = lambda_(36, 14)\n", "Nexp = 1000000\n", "\n", "Ys = binom.rvs(Nflips, 0.5, size=Nexp)\n", "lambdas = lambda_(Nflips, Ys)\n", "\n", "Npass = np.sum(lambdas < k)\n", "print(f'alpha = {Npass/Nexp}')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "So, for the given acceptance criterion, $\\alpha\\approx 0.13$.\n", "\n", "----\n", "**b)** Now, suppose that $p=0.7$. What is the value of $beta$?" ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "beta = 0.162632\n" ] }, { "name": "stderr", "output_type": "stream", "text": [ "/usr/lib/python3.6/site-packages/ipykernel/__main__.py:3: RuntimeWarning: divide by zero encountered in true_divide\n", " app.launch_new_instance()\n" ] } ], "source": [ "Ys = binom.rvs(Nflips, 0.7, size=Nexp)\n", "lambdas = lambda_(Nflips, Ys)\n", "\n", "Npass = np.sum(lambdas >= k)\n", "print(f'beta = {Npass/Nexp}')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "So, for the chosen value of $p$, $\\beta\\approx0.16$.\n", "\n", "---\n", "**c)** False, it is the probability of our test rejecting $H_0$ when it is true.\n", "\n", "**d)** False, it is the probability of our test failing to reject $H_0$ when $H_a(p=0.7)$ is true.\n", "\n", "**e)** Correct, It was assumed that $H_a(p=0.7)$ was true.\n", "\n", "**f)** True, since the alternative hypotheses is more similar to the null hypotheses, they are more difficult to distinguish, which will lead to a larger $\\beta$ value.\n", "\n", "**g)** False, $\\alpha$ is the probably rejecting $H_0$, when it is true.\n", "\n", "**h)** True, the rejection region growing from $|y-18|\\ge4$ to $|y-18|\\ge2$ would lead to more incorrect rejections.\n", "\n", "**i)** True.\n", "\n", "**j)** False, it would be smaller because a less stringent rejection criteria would make it easier to reject $H_0$." ] } ], "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": "3.6.2" } }, "nbformat": 4, "nbformat_minor": 2 }