AMS410_EXAMSOL3

AMS410_EXAMSOL3 - AMS 410 Actuarial Mathematics Fall 2009...

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Unformatted text preview: AMS 410 Actuarial Mathematics Fall 2009 Exam 3 solution: Joint distributions 1. Let X and Y be discrete random variables with joint probability function f ( x,y ) = ( 2 x +1- y 9 for x = 1 , 2 and y = 1 , 2 , otherwise. Calculate E ( X Y ) . (A) 8 18 (B) 9 18 (C) 16 18 (D) 25 18 (E) 36 18 Solution : The distribution table is: X = 1 X = 2 Y = 1 2/9 4/9 Y = 2 1/9 2/9 So E ( X Y ) = 2 9 1 + 1 9 1 2 + 4 9 2 + 2 9 1 = 25 18 . 2. Let X and Y be discrete loss random variables with joint probability function f ( x,y ) = ( y 24 x for x = 1 , 2 , 4; y = 2 , 4 , 8; x y otherwise. An insurance policy pays the full amount of loss X and half of loss Y . Find the probability that the total paid by the insurer is no more than 5. (A) 17 24 (B) 5 8 (C) 3 8 (D) 7 24 (E) 1 8 Solution : The distribution table is (Notice that x y in the range of the pmf, So P ( X = 4 ,Y = 2) = 0 ): X = 1 X = 2 X = 4 Y = 2 1/12 1/24 Y = 4 1/6 1/12 1/24 Y = 8 1/3 1/6 1/12 So P ( X + Y 2 5) = 1 12 + 1 24 + 0 + 1 6 + 1 12 + 1 3 = 17 24 . 3. The distribution of Smith's future lifetime is X , an exponential random variable with a mean of 60 years, and the distribution of Brown's future lifetime is Y , an exponential random variable with a mean of 50 years. Smith and Brown have future lifetimes that are independent of one another. Find the probability that Smith outlives Brown. (A) 1 11 (B) 1 6 (C) 1 5 (D) 5 11 (E) 6 11 Solution : X and Y are independent, and f X ( x ) = 1 60 e- x 60 , for < x < , f Y ( y ) = 1 50 e- y 50 , for < y < . So the joint pdf is f ( x,y ) = 1 60 e- x 60 1 50 e- y 50 , for < x < , < y < . The requested probability is P ( X > Y ) = y =0 x = y 1 60 e- x 60 1 50 e- y 50 dxdy = y =0 1 50 e- y 50 e- y 60 = 1...
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This note was uploaded on 03/13/2011 for the course AMS 410 taught by Professor Yang,y during the Spring '08 term at SUNY Stony Brook.

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AMS410_EXAMSOL3 - AMS 410 Actuarial Mathematics Fall 2009...

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