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180afinalsolns - 180 Final Solutions December 4 2006 1...

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180 Final Solutions December 4, 2006 1) Consider f ( x ) = 0 , x 0 2 9 , 0 < x a 1 x 2 , a < x What value(s) of a R make(s) f a probability density function? Answer Note that if f is a PDF 1 = Z a 0 2 9 dx + Z a 1 x 2 dx = 2 9 x a 0 + - 1 x a 0 = 2 9 a + 1 a Rearranging and multiplying through by 9 2 a we see that this occurs if 0 = a 2 - 9 2 a + 9 2 = ( a - 3 / 2)( a - 3) so a ∈ { 3 , 3 / 2 } . 2) Consider X with PDF f ( x ) = 5 2 x 4 , x ( - 1 , 1) , and f ( x ) = 0 otherwise 1. Compute the PDF of Y = X 2 + 1 . 2. Compute the PDF of Z = 1 / ( X 2 + 1) Answer We note that for y (1 , 2) F Y ( y ) = P ( Y y ) = P ( X 2 y - 1) = P ( - p y - 1 X p y - 1) = 2 Z y - 1 0 5 2 x 4 dx = ( y - 1) 5 / 2 1
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Thus f Y ( y ) = 5 2 ( y - 1) 3 / 2 with f Y ( y ) = 0 otherwise. For (2) we note that for z (1 / 2 , 1) F Z ( z ) = P (1 /Y z ) = P 1 x Y = Z 2 1 /z 5 2 ( y - 1) 3 / 2 dy = - Z 1 /z 2 5 2 ( y - 1) 3 / 2 dy Applying the FTC, we have f Z ( z ) = - 5 2 1 z - 1 3 / 2 - 1 z 2 = 5 2 z 2 1 z - 1 3 / 2 with f Z ( z ) = 0 otherwise. 3) Consider X, Y with joint PDF f ( x, y ) = C y x , x ( y, 1) , y (0 , 1) , and f ( x, y ) = 0 otherwise. 1. Compute C 2. Compute the PDF of X and the PDF of Y . For purposes of compute C it is slightly easier to change the order of integration. If f ( x, y ) is a joint PDF: 1 = C Z 1 x =0 Z x y =0 y x dydx = C Z 1 x =0 y 2 2 x x 0 dx = C Z 1 x =0 x 2 dx
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