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Unformatted text preview: AMS 311 (Fall, 2010) Joe Mitchell PROBABILITY THEORY Homework Set # 7 Solution Notes (1). (15 points) (a). If E (3 X ) = var ( X/ 2) and var (2 X ) = 3 , find (i). E [(2 + X ) 2 ] and (ii). var (4 + 3 X ) . If E (3 X ) = var ( X/ 2) = (1 / 4) var ( X ) and 3 = var (2 X ) = 4 var ( X ), then we know that var ( X ) = 3 / 4 and E ( X ) = (1 / 3)(1 / 4) var ( X ) = 1 / 16. Thus, since var ( X ) = E ( X 2 ) [ E ( X )] 2 , we know that 3 / 4 = E ( X 2 ) [1 / 16] 2 , so E ( X 2 ) = 193 / 256. Thus, E [(2 + X ) 2 ] = E [4 + 4 X + X 2 ] = 4 + 4 E ( X ) + E ( X 2 ) = 4 + 4 ( 1 16 ) + 193 256 = 1281 256 and var (4 + 3 X ) = 3 2 var ( X ) = 9 (3 / 4) = 27 / 4 (b). Suppose that X and W are independent and that var ( W ) = 7 , E (( X W )( X + W )) = 100 , E (3 W ) = 30 , E ( X 3 ) = 60 , and E ( X + W ) = 12 . Compute (i) var ( X ) and (ii) cov ( X 3 ,W 2 ) . We are told that E (3 W ) = 30; thus, E (3 W ) = 3 E ( W ) = 30, so E ( W ) = 10. We are told that E ( X + W ) = 12; thus, E ( X + W ) = E ( X ) + E ( W ) = E ( X ) + 10 = 12. So we conclude that E ( X ) = 2. We are told that var ( W ) = 7; thus, var ( W ) = E ( W 2 ) [ E ( W )] 2 = E ( W 2 ) 10 2 = 7, so we conclude that E ( W 2 ) = 107. We are told that E (( X W )( X + W )) = 100; thus, E (( X W )( X + W )) = E ( X 2 W 2 ) = E ( X 2 ) E ( W 2 ) = E ( X 2 ) 107 = 100, so we know that E ( X 2 ) = 207. This tells us that var ( X ) = E ( X 2 ) [ E ( X )] 2 = 207 2 2 = 203. We now compute cov ( X 3 ,W 2 ) = E ( X 3 W 2 ) E ( X 3 ) E ( W 2 ). Now, by independence of X and W , we know that E ( g ( X ) h ( Y )) = E ( g ( X )) E ( h ( Y )). Thus, cov ( X 3 ,W 2 ) = E ( X 3 W 2 ) E ( X 3 ) E ( W 2 ) = E ( X 3 ) E ( W 2 ) E ( X 3 ) E ( W 2 ) = 0. (In fact, since X and W are independent, we see that cov ( g ( X ) ,h ( Y )) = 0 for any functions g and h .) (2). (15 points) The random variables X and Y have a joint density function given by f ( x,y ) = braceleftbigg 2 e 2 x /x if x < , y x otherwise Compute cov (2 X,Y + 3) ....
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 Spring '08
 Tucker,A

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