09_14ans - STAT 410 Examples for Fall 2011 3 Consider two...

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STAT 410 Examples for 09/14/2011 Fall 2011 3. Consider two continuous random variables X and Y with joint p.d.f. f X, Y ( x , y ) = < + > > otherwise 0 1 , 0 , 0 60 2 y x y x y x Consider W = X + Y. Find the p.d.f. of W, f W ( w ) . F W ( w ) = P ( W w ) = - w d x w d x y y x 0 0 2 60 = ( ) - w d x x w x 0 2 2 30 = 10 w 5 – 15 w 5 + 6 w 5 = w 5 , 0 < w < 1. f W ( w ) = 5 w 4 , 0 < w < 1. Fact: Let X and Y be continuous random variables with joint p.d.f. ( ) y x f , . Then ( ) ( ) - = - + dx x w x f w f , Y X ( ) ( ) - = - + dy y y w f w f , Y X (convolution) Proof: ( ) ( ) - - - + = dx dy y x f w x w , Y X F . let u = y + x , then d u = d y , y = u x , , w x w

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( ) ( ) - - - + = dx du x u x f w w , Y X F = ( ) - - - w du dx x u x f , . ( ) w f Y X + = ( ) w ' Y X F + = ( ) - - dx x w x f , . Fact: Let X and Y be independent continuous random variables. Then ( ) ( ) ( ) - = - + dx x w f x f w f Y X Y X ( ) ( ) ( ) - = - + dy y f y w f w f Y X Y X 1. a) Let X and Y be two independent Exponential random variables with mean 1. Find the probability distribution of Z = X + Y. That is, find ( ) z f Z = ( ) z f Y X + . ( ) > - = otherwise 0 0 X x x e x f ( ) < > - - + - + - = = otherwise 0 otherwise 0 0 Y w x e x w e x w f x w x w ( ) w f Y X + = ( ) ( ) - - dx x w f x f Y X = + - - w x w x dx e e 0 = - w w dx e 0 = w e w - , w > 0.
2. a) When a person applies for citizenship in Neverland, first he/she must wait X years for an interview, and then Y more years for the oath ceremony. Thus the total wait is W = X + Y years. Suppose that X and Y are independent, the p.d.f. of X is f X ( x ) = 2 / x 3 , x > 1, zero otherwise, and Y has a Uniform distribution on interval ( 0, 1 ) . Find the p.d.f. of W, f W ( w ) = f X + Y ( w ) . Hint: Consider two cases: 1 < w < 2 and w > 2. f W ( w ) = ( ) ( ) - - dx x w f x f Y X . f X ( x ) = 2 / x 3 , x > 1, zero otherwise. f Y ( w x ) = 1, 0 < w x < 1 OR w – 1 < x < w , zero otherwise. Case 1: 1 < w < 2. 0 < w – 1 < 1. f W ( w ) = w dx x 1 3 1 2 = 2 1 1 w - . Case 2: w > 2. w – 1 > 1. f W ( w ) = - w w dx x 1 3 1 2 = ( ) 2 2 1 1 1 w w - - . Case 3: w < 1. f W ( w ) = 0.

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2 ¼ . Let X be an Exponential random variables with mean 1. Suppose the p.d.f. of Y is f Y ( y ) = 2 y , 0 < y < 1, zero elsewhere. Assume that X and Y are independent. Find the p.d.f. of W = X + Y, f W ( w ) = f X + Y ( w ) . ( ) > = - otherwise 0 0 X x e x f x ( ) < = < otherwise 0 1 0 2 Y y y y f ( ) ( ) ( ) - = - + dy y f y w f w f Y X Y X ( ) ( ) < = > - = - - - - otherwise 0 otherwise 0 0 X w y e y w e y w f w y y w Case 1: 0 < w < 1.
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