One call e 4 4 0 e 4 4 1 1 5 e 4 9 16 6 8 points find

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(one call) = e - 4 4 0 0! + e - 4 4 1 1! = 5 e - 4 9 . 16% . 6. [8 points] Find the joint density of the 3rd and 5th arrival times of a Poisson arrival process with rate λ . Solution: Let T 3 and T 5 be the 3rd and 5th arrival times, respectively. For 0 < s < t , P ( T 3 ds and T 5 dt ) = P (two arrivals in (0 , s ) , an arrival in ds, an arrival in ( s, t ) , an arrival in dt ) = e - λs ( λs ) 2 2! [ λ ds ] e - λ ( t - s ) λ ( t - s ) 1! [ λ dt ] = e - λt λ 5 2 s 2 ( t - s ) ds dt. Therefore, the joint density of T 3 and T 5 is f ( s, t ) = e - λt λ 5 2 s 2 ( t - s ), for 0 < s < t . 7. Let X be uniformly distributed on (0 , 1). Given X = x , let Y be uniformly distributed on (0 , 1 /x ). (a) [6 points] What is E ( Y | X )? Solution: The mean of a uniform distribution on an interval is the midpoint, so E ( Y | X = x ) = 1 2 x . More compactly, E ( Y | X ) = 1 2 X . (b) [6 points] What is the joint density of X and Y ? Solution: We have f X ( x ) = 1, for x (0 , 1), f Y ( y | X = x ) = x , for x (0 , 1) and y (0 , 1 /x ), so f X,Y ( x, y ) = f X ( x ) f Y ( y | X = x ) = x , for x (0 , 1) and y (0 , 1 /x ). (c) [6 points] Find P ( Y < X ). Page 4
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Math 431: Final Exam Solution: Using (b) and the fact that if x < 1, then x < 1 /x , P ( Y < X ) = Z 1 0 Z x 0 f X,Y ( x, y ) dy dx = Z 1 0 Z x 0 x dy dx = 1 3 . 8. [8 points] Let X and Y be independent random variables, each uniformly distributed on (0 , 1). Let Z = XY . Find Corr ( X, Z ). Solution: E ( X ) = 1 2 , E ( X 2 ) = Z 1 0 x 2 dx = 1 3 , and V ar ( X ) = E ( X 2 ) - ( E ( X )) 2 = 1 12 , and similarly for Y . Because X and Y are independent: Cov ( X, Z ) = E ( X 2 Y ) - E ( X ) E ( XY ) = E ( X 2 ) E ( Y ) - E ( X ) 2 E ( Y ) = 1 24 , V ar ( Z ) = V ar ( XY ) = E ( X 2
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