Solution to HW 7 - Solution to EE351K homework assignment 7 2 E FX x = FX x f X x)dx = F X 1 1 x)dFX x = FX x]2 | = 2 2 3 The PDF of X is a

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2. 2 1 1 [ ( )] ( ) ( ) ( ) ( ) [ ( )] | 2 2 X X X X X X E F x F x f x dx F x dF x F x +∞ +∞ +∞ -∞ -∞ -∞ = = = = 3. The PDF of X is a piecewise function ( ) ( ) ( 2 ) 2 0 X h x a a x b b a h f x x b a b x b a b a otherwise - ≤ ≤ - = - - + ≤ ≤ - - a. By definition of a PDF, we have ( ) 1 X f x dx +∞ -∞ = Put in the piecewise function, and we get sum of two integrals equals to 1. 2 ( ) ( 2 ) 1 b b a a b h h x a dx x b a dx b a b a - - + - - + = - - In order to simplify computation, we can have y=x-b, hence 0 0 ( ) ( ) 1 b a a b h h y b a dy y b a dy b a b a - - + - + - - + = - - Do another substitution u= -y for the first integral, hence 0 0 ( )( ) ( ) 1 b a b a h h u b a du y b a dy b a b a - - - + - - + - - + = - - Now the first integral is the same as the second, hence 0 1 ( ) 2 b a h y b a dy b a - - - + = - 0 1 ( ) 2 b a a b y b a dy h - - - + = 2 0 1 1 [ ( ) ]| 2 2 b a a b y b a y h - - - - = 2
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This note was uploaded on 04/21/2008 for the course EE 351k taught by Professor Bard during the Fall '07 term at University of Texas at Austin.

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Solution to HW 7 - Solution to EE351K homework assignment 7 2 E FX x = FX x f X x)dx = F X 1 1 x)dFX x = FX x]2 | = 2 2 3 The PDF of X is a

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