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03: Fetch week 3

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{
"cells": [
{
"cell_type": "markdown",
"id": "ac27651a",
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"source": [
"# Exercise sheet\n",
"\n",
"Some general remarks about the exercises:\n",
"* For your convenience functions from the lecture are included below. Feel free to reuse them without copying to the exercise solution box.\n",
"* 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",
"* 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",
"* 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",
"* 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",
"* 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",
"\n",
"Please fill in your name here:"
]
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"NAME = \"\"\n",
"NAMES_OF_COLLABORATORS = \"\""
]
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"source": [
"---"
]
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"source": [
"**Exercise sheet 3**\n",
"\n",
"Code from the lectures:"
]
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"source": [
"import numpy as np\n",
"import matplotlib.pylab as plt\n",
"\n",
"rng = np.random.default_rng()\n",
"%matplotlib inline\n",
"\n",
"def sample_acceptance_rejection(sample_z,accept_probability):\n",
" while True:\n",
" x = sample_z()\n",
" if rng.random() < accept_probability(x):\n",
" return x \n",
" \n",
"def estimate_expectation(sampler,n):\n",
" '''Compute beste estimate of mean and 1-sigma error with n samples.'''\n",
" samples = [sampler() for _ in range(n)]\n",
" return np.mean(samples), np.std(samples)/np.sqrt(n-1)\n",
"\n",
"def estimate_expectation_one_pass(sampler,n):\n",
" sample_mean = sample_square_dev = 0.0\n",
" for k in range(1,n+1):\n",
" delta = sampler() - sample_mean\n",
" sample_mean += delta / k\n",
" sample_square_dev += (k-1)*delta*delta/k \n",
" return sample_mean, np.sqrt(sample_square_dev / (n*(n-1)))"
]
},
{
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"grade_id": "cell-8763c7e3277ef0ce",
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"source": [
"## Acceptance-rejection sampling\n",
"\n",
"**(35 points)**\n",
"\n",
"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",
"\n",
"__(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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"source": [
"def f_inverse_Z(u):\n",
" '''Compute the inverse CDF of Z, i.e. F_Z^{-1}(u) for 0 <= u <= 1.'''\n",
" # YOUR CODE HERE\n",
" raise NotImplementedError()\n",
"\n",
"def random_Z():\n",
" return int(f_inverse_Z(rng.random())) # make sure to return an integer"
]
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"source": [
"assert f_inverse_Z(0.2)==1\n",
"assert f_inverse_Z(0.51)==2\n",
"assert f_inverse_Z(0.76)==4\n",
"assert f_inverse_Z(0.991)==111"
]
},
{
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"id": "9a1b3dcd",
"metadata": {
"deletable": false,
"editable": false,
"nbgrader": {
"cell_type": "markdown",
"checksum": "78f978d46d5f742623cf174e88323ec9",
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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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"execution_count": null,
"id": "8e201507",
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"source": [
"def accept_probability_X(k):\n",
" '''Return the appropriate acceptance probability on the event Z=k.'''\n",
" # YOUR CODE HERE\n",
" raise NotImplementedError()\n",
" \n",
"def random_X():\n",
" return sample_acceptance_rejection(random_Z,accept_probability_X)\n",
"\n",
"# Verify numerically\n",
"# YOUR CODE HERE\n",
"raise NotImplementedError()"
]
},
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"execution_count": null,
"id": "1b0e1d47",
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"deletable": false,
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"locked": true,
"points": 20,
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"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)"
]
},
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"id": "d010b3c8",
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"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)**"
]
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"locked": false,
"points": 10,
"schema_version": 3,
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"outputs": [],
"source": [
"# YOUR CODE HERE\n",
"raise NotImplementedError()"
]
},
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"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)**"
]
},
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"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()"
]
},
{
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"id": "e2a5b934",
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"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 $\\mathbf{X}$."
]
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"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()"
]
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"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)**"
]
},
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"source": [
"# Collect height estimates\n",
"n_values = [2**k for k in range(5,11+1)]\n",
"# YOUR CODE HERE\n",
"raise NotImplementedError()"
]
},
{
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"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))"
]
},
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"source": [
"# Plotting\n",
"# YOUR CODE HERE\n",
"raise NotImplementedError()"
]
},
{
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"id": "142aa5cb",
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"__(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)**"
]
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"# YOUR CODE HERE\n",
"raise NotImplementedError()"
]
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