hw1 - 36-226 Summer 2010 Homework 1 Due June 30 1 Math...

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Unformatted text preview: 36-226 Summer 2010 Homework 1 Due June 30 1. Math review. (a) Simplify the following expression as much as possible. The xi are positive integers. There should be no signs in the result but there will be signs. n ln i=1 ea bxi xi ! . (1) (b) Find the value of x which maximizes the following function. Be sure to prove that it is a maximum. f (x | n, a) = a ln x + (n − a) ln(1 − x) ¯ (c) Let X = 1 n n i=1 Xi . 0<a<n (2) Show that n n ¯ ¯ (Xi − X )2 + n(X − µ)2 . (Xi − µ)2 = i=1 (3) i=1 Hint: Start with the left side. The first step is a trick. The rest is easy. Do not work from both sides. Proofs should be linear and sensical. 2. Probability review. Let X1 and X2 be independent, each with pdf f (x) = 1 exp{−x/β }I[0,∞) (x) β β > 0. (4) (a) Find E [X1 ] and V ar[X1 ]. (b) Write down the joint density and integrate to find E [3X1 − 2X2 + 10]. Feel free to cite part (i) if necessary. (c) Use properties of expectation and the answer to (i) to verify your answer in (ii). (d) Find V ar[3X1 − 2X2 + 10] in any manner you wish. 3. The art of statistics. In class, I determined which realization of ones and zeros was random by looking for long strings of successive ones (or zeros). Let’s formalize this idea. Write down a probability model for the random string of heads and tails. To do this, define some random variable(s) and parameter(s) and articulate a distribution for the random variable(s). Also carefully define what it means to see a “long string”. What is complicated about trying to determine the probability of observing at least one “long string”? This question is intentionally open ended and should stimulate thinking about the problem. Do NOT try to actually calculate the probability of “long strings”. 1 ...
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This document was uploaded on 07/14/2011.

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