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# 7 - (b We have using the chain rule E[Y = d MY(s ds = s=0 d...

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(b) We have using the chain rule E [ Y ] = d ds M Y ( s ) s =0 = d ds M X ( s ) s =0 · λe λ ( M X ( s ) 1) s =0 = 1 2 · λ = λ 2 , where we have used the fact that M X (0) = 1. (c) From the law of iterated expectations we obtain E [ Y ] = E E [ Y | N ] = E N E [ X ] = E [ N ] E [ X ] = λ 2 . Solution to Problem 4.42. Take X and Y to be normal with means 1 and 2, respectively, and very small variances. Consider the random variable that takes the value of X with some probability p and the value of Y with probability 1 p . This random variable takes values near 1 and 2 with relatively high probability, but takes values near its mean (which is 3 2 p ) with relatively low probability. Thus, this random variable is not normal. Now let N be a random variable taking only the values 1 and 2 with probabilities p and 1 p , respectively. The sum of a number N of independent normal random variables with mean equal to 1 and very small variance is a mixture of the type discussed above, which is not normal. Solution to Problem 4.43. (a) Using the total probability theorem, we have P ( X > 4) = 4 k =0 P ( k lights are red) P ( X > 4 | k lights are red) . We have P ( k lights are red) = 4 k 1 2 4 . The conditional PDF of X given that k lights are red, is normal with mean k minutes and standard deviation (1 / 2) k . Thus, X is a mixture of normal random variables and the transform associated with its (unconditional) PDF is the corresponding mixture of the transforms associated with the (conditional) normal PDFs. However, X is not normal, because a mixture of normal PDFs need not be normal. The probability P ( X > 4 | k lights are red) can be computed from the normal tables for each k , and P ( X > 4) is obtained by substituting the results in the total probability formula above. (b) Let K be the number of traﬃc lights that are found to be red. We can view X as the sum of K independent normal random variables. Thus the transform associated with X can be found by replacing in the binomial transform M K ( s ) = (1 / 2+(1 / 2) e s ) 4 the occurrence of e s by the normal transform corresponding to μ = 1 and σ = 1 / 2. Thus M X ( s ) = 1 2 + 1 2 e (1 / 2) 2 s 2 2 + s 4 . Note that by using the formula for the transform, we cannot easily obtain the proba- bility P ( X > 4). 61

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Solution to Problem 4.44. (a) Using the random sum formulas, we have E [ N ] = E [ M ] E [ K ] , var( N ) = E [ M ] var( K ) + ( E [ K ] ) 2 var( M ) . (b) Using the random sum formulas and the results of part (a), we have E [ Y ] = E [ N ] E [ X ] = E [ M ] E [ K ] E [ X ] , var( Y ) = E [ N ] var( X ) + ( E [ X ] ) 2 var( N ) = E [ M ] E [ K ] var( X ) + ( E [ X ] ) 2 E [ M ] var( K ) + ( E [ K ] ) 2 var( M ) . (c) Let N denote the total number of widgets in the crate, and let X i denote the weight of the i th widget. The total weight of the crate is Y = X 1 + · · · + X N , with N = K 1 + · · · + K M , so the framework of part (b) applies. We have E [ M ] = 1 p , var( M ) = 1 p p 2 , (geometric formulas) , E [ K ] = μ, var( M ) = μ, (Poisson formulas) , E [ X ] = 1 λ , var( M ) = 1 λ 2 , (exponential formulas) .
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7 - (b We have using the chain rule E[Y = d MY(s ds = s=0 d...

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