180asolns5 - 180A HW 5 Solutions 1) Suppose X has the...

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Unformatted text preview: 180A HW 5 Solutions 1) Suppose X has the uniform distribution on (0 , 1). Compute the probability density function and expected value of: ( a ) X , where > 0 ( b ) log( X ); ( c ) exp ( X ); ( d ) sin(2 X ) answer Let Y = X . Then, clearly: F Y ( y ) = P ( X y ) = P ( X y 1 / ) = y 1 / so f Y ( y ) = 1 y 1- for 0 y 1, 0 otherwise and E [ Y ] = integraldisplay 1 1 x 1 dx = 1 1 + x (1+ ) / vextendsingle vextendsingle vextendsingle vextendsingle 1 = 1 + 1 Let Y 2 = log( X ). Then, for y < 0, (as log x < 0 for x < 1), we have F Y 2 ( y ) = P (log( X ) < y ) = P ( X e y ) = e y Thus f Y 2 ( y ) = e y for y < 0, 0 otherwise. We note that via the change of variables x =- y and last weeks HW. E [ Y 2 ] = integraldisplay ye y dy =- integraldisplay xe x dx =- 1 Let Y 3 = exp( X ). Then for 1 < y < e (as if 0 < x < 1, we have 1 < e x < 1) F Y 3 ( y ) = P (exp( X ) < y ) = P ( X < log( y )) = log( y ) so f Y ( y ) = 1 y for 1 < y < e , and 0 otherwise. Then E [ Y 3 ] = integraldisplay e 1 y (1 /y ) dy = e- 1 . Let Y 4 = sin(2 X ). This is slightly trickier, as sin(2 x ) is not monotonic on (0 , 1). We can easily see that E [ Y 4 ] = integraltext 1 x sin(2 x ) dx = 0 (from our work last week). To get f Y 4 ( y ) we note that for 0 < y < 1 F Y 4 ( y ) = P (sin(2 X ) < y ) = P (0 2 X < arcsin( y ) or - arcsin( y ) < 2 X 2 ) = P (0 X < arcsin( y ) / (2 )) + P (1 /...
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180asolns5 - 180A HW 5 Solutions 1) Suppose X has the...

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