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solution8

# solution8 - EL630 Homework8 Solutions 1 In all these let Z...

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EL630: Homework8 Solutions 1) In all these, let Z be the resulting random variable whose density we want. (a) Since X and Y are independent we just convolve the two densities - - - - - - = - = = dx x z U e x U e dx x z f x f z f z f z f x z x Y X Y X Z ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( Since 0 ) ( = x U for 0 < x and 0 ) ( = - x z U for z x we have z z z z x z x Z ze dx e dx e e z f - - - - - = = = 0 0 ) ( ) ( for 0 z . (b) The random variable Y - has the density ) ( y f Y - (this is easy to show), so that the density of ) ( Y X Y X Z - + = - = is just - - - - - - = - = - = dx z x U e x U e dx z x f x f z f z f z f z x x Y X Y X Z ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( Since 0 ) ( = x U for 0 < x and 0 ) ( = - z x U for z x < we have: For 0 < z , 2 ) ( 0 2 0 ) ( z x z z x x Z e dx e e dx e e z f = = = - - - - . For 0 z , 2 2 ) ( 2 2 ) ( z z z z x z z z x x Z e e e dx e e dx e e z f - - - - - - = = = = . (c) To be posted later (d) From example 8.4 we have - = dy y yz f y z f XY Z ) , ( ) ( , and since X and Y are independent this becomes - - - - = = dy y U e yz U e y dy y f yz f y z f y z y Y X Z ) ( ) ( ) ( ) ( ) ( . Since ) ( y U is zero for 0 < y , we see that the integral is over positve values of y . In

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this case, ) ( zy U will be zero if 0 < z , so we can also restrict ourselves to positive values of z . Then, for 0 z we have 2 0 ) 1 ( 0 ) 1 ( 1 ) ( + = = = + - - - z dy e y dy e e y z f y z y z y Z . (e) From example 8.11 we have ) ( )) ( 1 ( ) ( )) ( 1 ( ) ( z f z F z f z F z f X Y Y X Z - + - = . Now, it is easy to calculate that ) ( ) 1 ( ) ( x U e x F x X - - = and ) ( ) 1 ( ) ( y U e x F y Y - - = . So, substituting in the above, we find ) ( 2 ) ( ) ( ) ( 2 z U e z U e e z U e e z f z z z z z Z - - - - - = + = .
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solution8 - EL630 Homework8 Solutions 1 In all these let Z...

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