STAT_333_Assignment_1_Solutions

STAT_333_Assignment_1_Solutions - STAT 333 Assignment 1...

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STAT 333 Assignment 1 SOLUTIONS 1. Consider rolling a fair 6-sided die n times. Let X represent the number of faces that have NOT been rolled. a. Find the expected value of X. Define indicator variables X i = {1 if the i th face is unrolled after n rolls, {0 otherwise; for i = 1, 2, …, 6 Then X = X 1 + … + X 6 Using the results for indicator variables, E[X i ] = P( i th face is unrolled after n rolls) = (5/6) n Using rules for linear combinations of random variables, E[X] = E[X 1 + … + X 6 ] = E[X 1 ] + … + E[X 6 ] = 6 E[X i ] = 6 (5/6) n b. Find the variance of X. Var(X i ) = (5/6) n (1 – (5/6) n ) Cov(X i , X j ) = E[X i X j ] – E[X i ]E[X j ] = P(both faces i and j are unrolled after n rolls) – (5/6) n (5/6) n = (4/6) n – (5/6) 2 n Var(X) = Var(X 1 + … + X 6 ) = Var(X 1 ) + … + Var(X 6 ) + 2Σ i<j Σ Cov(X i , X j ) = 6 Var(X i ) + 2*15 Cov(X i , X j ) since there are (6C2) = 15 pairs i < j = 6(5/6) n (1 – (5/6) n ) + 30 ((4/6) n – (5/6) 2 n ) You could simplify, but you don’t have to. c. Describe (in words or with a graph) how the mean and variance of X behave for different values of n . Provide a brief logical explanation. The mean makes sense that it decreases from 5 when more rolls are done, since with just one roll, obviously exactly 1 number will come up, but eventually all numbers will come up at least once. For the variance, with one roll or many rolls, there is no variability. But for a few rolls, there are lots of possibilities of having rolled more or fewer different faces than the mean number. That’s why the variance increases a little and then drops off. It peaks at n =7 at about 0.62.
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2. Consider a Negative Binomial random variable Y ~ NB( r , p ). Prove that Y is a proper rv iff p > 0 by the following methods: a. Express Y as the sum of r independent Geometric random variables, and apply a result we know from class about Geometric rvs. Y = X 1 + X 2 + … + X r , where each X i ~ Geo( p ) P(Y < ∞) = P(X 1 + X 2 + … + X r < ∞) = P(X 1 < ∞, X 2 < ∞, … X r < ∞) for the sum to be finite, each X i must be finite = P(X 1 < ∞)P(X 2 < ∞)…P(X r < ∞) by independence = 1 * 1 * … * 1 since Geometrics are proper iff p > 0 = 1 So the Negative Binomial is proper. b.
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This note was uploaded on 12/12/2010 for the course STAT 333 taught by Professor Chisholm during the Spring '08 term at Waterloo.

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STAT_333_Assignment_1_Solutions - STAT 333 Assignment 1...

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