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 10 independent coin flips having probability p of landing heads. a. Find the expected number of changeovers. Let X i = 1 if there is a changeover between trial i and trial i+1, 0 otherwise, i = 1, 2, …, 9. Then X, the number of changeovers is X 1 + X 2 + … + X 9 E[X i ] = P(Changeover) = P(HT or TH) = p (1 – p ) + (1 – p ) p = 2( p p 2 ), i = 1, 2, …, 9. So E[X] = E[X 1 + X 2 +…+ X 9 ] = E[X 1 ] + E[X 2 ] + … + E[X 9 ] = 9[2( p p 2 )] = 18( p p 2 ) b. Find the variance of the number of changeovers. Var(X i ) = P(Changeover)(1 – P(Changeover)) = 2( p p 2 ) (1 – 2( p p 2 )) = 2(– 2 p 4 + 4 p 3 – 3 p 2 + p ), i = 1, 2, …, 9. Cov(X i , X j ) will be 0 unless i and j are adjacent. We want i < j so let j = i + 1. E[X i X i+1 ] = P(2 changeovers) = P(HTH or THT) = p (1 – p ) p + (1 – p ) p (1 – p ) = p p 2 Cov(X i , X i+1 ) = E[X i X i+1 ] – E[X i ]E[X i+1 ] = ( p p 2 ) – (2( p p 2 )) 2 = – 4 p 4 + 8 p 3 – 5 p 2 + p , i = 1, 2, …, 8. So Var(X) = ∑ Var(X i ) + 2 ∑ Cov(X i , X i+1 ) = 9[2(– 2 p 4 + 4 p 3 – 3 p 2 + p )] + 2*8[– 4 p 4 + 8 p 3 – 5 p 2 + p ] = – 100 p 4 + 200 p 3 – 134 p 2 + 34 p c. Describe how the mean and variance of the number of changeovers behave for different values of p . Provide a brief logical explanation. Using Excel to plot E[X] and Var(X) for different values of p yields the following result: 0 1 2 3 4 5 0.01 0.05 0.09 0.13 0.17 0.21 0.25 0.29 0.33 0.37 0.41 0.45 0.49 0.53 0.57 0.61 0.65 0.69 0.73 0.77 0.81 0.85 0.89 0.93 0.97 E[X] Var(X) Explanation: the expected value makes sense because with a higher likelihood of getting either all T or all H, we expect fewer changeovers. For the variance, as p goes to 0 or 1, the number of changeovers approaches 0 with certainty, and hence the variance approaches 0 too. In the middle, things are a little weird: variance is highest when p is around 0.78 and 0.22 and dips lower when p is closer to 0.5. It makes sense if you remember that variance is the average squared distance from the mean. When p is close to 0.5, the mean is 4.5, and the farthest away the actual number of changeovers could possibly be is 0 or 9 (a distance of 4.5). When p is slightly higher or lower, the mean drifts down, so the possibility of being really far away (e.g. 9 when the mean is 3, a distance of 6) drags the variance up.
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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:
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STAT_333_Assignment_1_Solutions - STAT 333 Assignment 1...

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