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 · 1 2 =1 , since ² 0 λ e - λ x dx 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 ² 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 ±nds 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 ±rst 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 , we have W and hence P [ W x | N = 0] = 1 (because x 0 ), when N , 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 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 ³ 2 x r 2 if 0 x r 0 otherwise.
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hw7_sol - ECE 3100 Homework#7 Solutions 1 We have Fall 2009...

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