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02_Exam_Solution

# 02_Exam_Solution - Exam 2 Solution ORIE 3500/5500...

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Exam 2 - Solution ORIE 3500/5500: Engineering Probability and Statistics II Question 1 (i) (a) Conditional PDF of X given Y f X | Y ( x | y ) is a function with f X | Y ( x | y ) 0 and -∞ f X | Y ( x | y ) dx = 1 such that P ( X a | Y = y ) = a -∞ f X | Y ( x | y ) dx . Also f X | Y ( x | y ) = f X,Y ( x, y ) f Y ( y ) , where f X,Y is the joint PDF of X and Y and f Y is the marginal PDF of Y . (b) X and Y will be independent if f X,Y ( x, y ) = f X ( x ) f Y ( y ) , where f X,Y is the joint PDF of X and Y and f X , f Y are the marginal PDF of X and Y . (c) Moment-generating function (transform) of a random variable X is M X ( s ) = E e s · X . (d) Moment-generating function (transform) of two random variables X and Y is a func- tion M X,Y ( s 1 , s 2 ) such that M X,Y ( s 1 , s 2 ) = E e s 1 · X + s 2 · Y . (ii) (a) M Y ( s ) = E e s · Y = E e s · ( a + bX ) = E e sa e s · bX = e sa M X ( bs ) . (b) M X ( s ) = E [ e sX ]. So d 2 ds 2 M X ( s ) = E d 2 ds 2 e sX = E X 2 e sX E ( X 2 ) = d 2 ds 2 M X ( s ) | s =0 . Question 2 (i) (a) If X Poisson ( λ ) then p X ( k ) = e - λ λ k /k ! for k 0. So M X ( s ) = k =0 e sk e - λ λ k /k !

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