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

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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 α , where α > 0 ( b ) log( X ); ( c ) exp ( X ); ( d ) sin(2 πX ) answer Let Y = X α . Then, clearly: F Y ( y ) = P ( X α y ) = P ( X y 1 ) = y 1 so f Y ( y ) = 1 α y 1 - α α for 0 y 1, 0 otherwise and E [ Y ] = integraldisplay 1 0 1 α x 1 α dx = 1 α · α 1 + α x (1+ α ) vextendsingle vextendsingle vextendsingle vextendsingle 1 0 = 1 α + 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 0 −∞ ye y dy = - integraldisplay 0 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 πX ). This is slightly trickier, as sin(2 πx ) is not monotonic on (0 , 1). We can easily see that E [ Y 4 ] = integraltext 1 0 x sin(2 πx ) 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 πX ) < y ) = P (0 2 πX < arcsin( y ) or π - arcsin( y ) < 2 πX 2 π ) = P (0 X < arcsin( y ) / (2 π )) + P (1 / 2 - arcsin( y ) / (2 π ) < X 1) = arcsin( y ) / (2 π ) + 1 - (1 / 2 - arcsin( y ) / (2 π )) = 1 / 2 + arcsin( y ) 1

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Another way to see that, is to use the fact that sin(2 πX ) and sin(2 πX -
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