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My solutions to Kevin Murphy Machine Learning Book

License: Other

Jupyter Notebook 98.99% TeX 1.01%

machine-learning-a-probabilistic-perspective-solutions's Introduction

Machine-Learning-A-Probabilistic-Perspective-Solutions

Motivation

Hey there. I am recording the solutions of the exercises of the fourth printing of this book in this repository. The only exercises that I do not intend to do in this first run are those which explicit require MATLAB. Any computational exercise will be done in Python using a Jupyter notebook. I will follow a schema where I give a introduction and some insight into the problem, solve it and then make some remarks on the solution. I strongly reccomend the reading of the Intro and Conclusion section of the exercises that you're interested in. I intend to update the solutions in a reasonable pace, starting now (January 2017).

I hope this might help anyone who has an interest in the book and Machine Learning as a whole.

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machine-learning-a-probabilistic-perspective-solutions's Issues

Problem 3.11.d

You derive the likelihood, then say that the resulting distribution is

Ga(theta | N+1, lam+sum(x_i))

but plugging that into the definition given for the Gamma distribution in the textbook gives

Ga(theta | N+1, lam+sum(x_i)) = [lam+sum(x_i)]^(N+1) theta^N e^(- theta [lam+sum(x_i)])

which is not the likelihood that you derived.

Am I missing something here?

Solution to 2.6

When E1 and E2 are conditionally independent given H, we can derive p(e1,e2) as follows:

p(e1,e2) = sum(p(e1,e2|h')p(h')) = sum(p(e1|h')p(e2|h')p(h')) over all h' in support of H.

Thus, (iii) is sufficient for 2.6(b)

An neat approach to problem 2.17

Instead of using brute force to calculate the double integral, we can actually prove the distribution of min(X_1, ..., X_n) is a Beta distribution where X_1,...,X_n are i.i.d. and X_1,...,X_n ~ U(0,1). Then, we can directly apply the Beta mean we derived from 2.16.

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