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solution11

# solution11 - Solutions to Homework 11 1 From Problem 3 of...

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Solutions to Homework 11 1. From Problem 3 of Homework 10, we know that k = 1 and the marginal pdfs of X and Y are given by f X ( x ) = x + 1 2 0 < x < 1 and f Y ( y ) = y + 1 2 0 < y < 1 Therefore f Y ( y | x ) = f X,Y ( x, y ) f X ( x ) = x + y x + 1 2 0 < y < 1 f X ( x | y ) = f X,Y ( x, y ) f Y ( y ) = x + y y + 1 2 0 < x < 1 2. a. P [ N = k ] = Z 0 P [ N = k | R = r ] f R ( r ) dr = Z 0 r k k ! e - r λ ( λr ) α - 1 Γ( α ) e - λr dr = λ α k !Γ( α ) Z 0 r k + α - 1 e (1+ λ ) r dr let t = (1 + λ ) r = λ α k !Γ( α ) 1 (1 + λ ) k + α Z 0 t k + α - 1 e - t dt = λ α k !Γ( α ) 1 (1 + λ ) k + α Γ( k + α ) = Γ( k + α ) k !Γ( α ) λ 1 + λ ! α 1 1 + λ k Note: N is also called the generalized Binomial random variable. b. E [ N ] = Z 0 E [ N | r ] f R ( r ) dr = Z 0 rf R ( r ) dr = E [ R ] = α λ 1

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E [ N 2 ] = Z 0 E [ N 2 | r ] f R ( r ) dr = Z 0 ( r + r 2 ) f R ( r ) dr = E [ R ] + E [ R 2 ] V AR [ N ] = E [ N 2 ] - E [ N ] 2 = E [ R ] + E [ R 2 ] - E [ R ] 2 = E [ R ] + V AR [ R ] = α λ 2 + α λ 3. Denote the probability that a customer is served by clerk i , p i , by P [ C = i ] = p i . a. f T ( t ) = n X i =1 f T ( t | C = i ) P [ C = i ] = n X i =1 α i e - α i t p i b. E [ T ] = n X i =1 E [ T | C = i ) P [ C = i ] = n X i =1 1 α i p i E [ T 2 ] = n X i =1 E [ T 2 | C = i ) P [ C = i ] = n X i =1 2 α 2 i p i V AR [ T ] = E [ T 2 ] - E [ T ] 2 = n X i =1 2 α 2 i
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