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hw7_sol - ECE 3100 Homework#7 Solutions 1 We have Fall 2009...

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ECE 3100 Fall 2009 Homework #7 Solutions 1. We have -∞ f X ( x ) dx = -∞ λ 2 e - λ | x | dx = 0 -∞ λ 2 e - λ | x | dx + 0 λ 2 e - λ | x | dx = 2 · 1 2 0 λ e - λ x dx = 2 · 1 2 = 1 , since 0 λ e - λ x dx = 1 by the normalization property of the exponential PDF. Since the Laplace PDF is symmetric about x = 0 , we have E [ X ] = 0 . Also, E [ X 2 ] = -∞ x 2 λ 2 e - λ | x | dx = 0 x 2 λ e - λ x dx = 2 λ 2 , since 0 x 2 λ e - λ x dx = 2 λ 2 is the second moment of the exponential PDF. Thus Var [ X ] = E [ X 2 ] - ( E [ X ]) 2 = 2 λ 2 - 0 = 2 λ 2 . 2. Denote by W the waiting time for Calamity Jane and let N denote the number of customers that she finds ahead of her, N takes the value 1 with probability 1 2 and the value 0 with probability 1 2 , we are interested in F W ( x ) = P [ W x ] . Note first that if x < 0 , then F W ( x ) = 0 because the random variable W is nonnegative ( W 0 ). For x 0 , using the total law of probability we obtain F W ( x ) = P [ W x ] = P [ W x | N = 1] P [ N = 1] + P [ W x | N = 0] P [ N = 0] = 1 2 ( P [ W x | N = 1] + P [ W x | N = 0]) . When N = 0 , we have W = 0 and hence P [ W x | N = 0] = 1 (because x 0 ), when N = 1 , we have W is exponentially distributed with parameter λ , consequently, P [ W x | N = 1] = 1 - e - λ x , hence we obtain that for x 0 , F W ( x ) = 1 2 (2 - e - λ x ) . 3. (a) We calculate the CDF of X as follows. For x [0 , r ] , F X ( x ) = P ( X x ) = π x 2 π r 2 = x 2 r 2 . For x < 0 , F X ( x ) = 0 and for x > r , F X ( x ) = 1 . The PDF is obtained by differentiating the CDF. Thus f X ( x ) = 2 x r 2
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