Stat 330 Assignment

# Stat 330 Assignment - 1 Assignment#2 STAT 330 Due in class...

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1 Assignment #2 - STAT 330 Due in class: Tuesday June 14 Important Note: You need to print out this page as the cover page for your assignment. LAST NAME: FIRST NAME: ID. NO.: QUESTION 1. /8 QUESTION 2. /4 QUESTION 3. /8 QUESTION 4. /6 QUESTION 5. /10 QUESTION 6. /4 TOTAL: /40

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2 1. Solution: (a). No. This is because one of the two conditions in the Factorization Theorem for Independence is not met (in fact, both conditions are not true for this problem). Indeed, the support of the joint pmf A XY = { ( x,y ) : 0 x y, 0 y,x and y are integers } 6 = A X × A Y for any sets A X and A Y of real numbers also f ( x,y ) 6 = g ( x ) h ( y ) for any functions g ( x ) and h ( y ). (b). For x = 0 , 1 , 2 ,..., f X ( x ) = X y f ( x,y ) = X y = x e - 2 x !( y - x )! = e - 2 x ! X t =0 1 t ! , set t = y - x y = t + x = e - 1 x ! " Recall: The exponential Series: X t =0 a t t ! = e a ,a R. Here a = 1 # Therefore, f X ( x ) = e - 1 x ! , x = 0 , 1 , 2 ,... i.e., X POI (1). For y = 0 , 1 , 2 ,... , f Y ( y ) = X x f ( x,y ) = y X x =0 e - 2 x !( y - x )! = e - 2 y X x =0 y ! x !( y - x )! · 1 y ! = e - 2 y ! y X x =0 ± y x ² 1 x · 1 y - x = e - 2 y ! (1 + 1) y Recall: Binomial Theorem: ( a + b ) n = n x =0 ( n x ) a x b n - x for a positive integer n Therefore, f Y ( y ) = e - 2 2 y y ! , y = 0 , 1 , 2 ,...
3 i.e., Y POI (2). (c). f Y | X ( y | x ) = f ( x,y ) f X ( x ) = e - 2 x !( y - x )! e

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Stat 330 Assignment - 1 Assignment#2 STAT 330 Due in class...

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