HW9_sol_v2

# HW9_sol_v2 - ISyE 2027 Spring 2010 HW 9 Solution 1 Let Y =...

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ISyE 2027, Spring 2010 HW 9 Solution 1. Let Y = 1 + μX , or X = Y - 1 μ . F Y ( y ) = Pr ( Y y ) = Pr ( X y 1 μ ) = F X ( y 1 μ ) . f Y ( y ) = dF X ( y - 1 μ ) dy = dF X ( x ) dx vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle x = y - 1 μ dx dy = f X ( x ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle x = y - 1 μ dx dy . The pdf of Y , f Y ( y ), becomes f Y ( y ) = f X ( x ) dx dy = 2 e - 2 x 1 μ , x 0 , = 2 μ e - 2( y - 1) μ , y 1 , and zero otherwise. 2. Let Y = 1 + 2 Z , or Z = ( Y - 1) 2 4 . Y is one of 1 , 3 , 1 + 2 2 , 1 + 2 3 , and 5, since Z is one of 0 , 1 , 2 , 3 , and 4. The pmf of Y , p Y ( y ), becomes p Y ( y ) = Pr ( Y = y ) = Pr (1 + 2 Z = y ) = Pr ( Z = ( y 1) 2 4 ) = ( 4 ( y - 1) 2 4 ) ( 1 4 ) ( y - 1) 2 4 ( 3 4 ) 4 - ( y - 1) 2 4 , y = 1 , 3 , 1 + 2 2 , 1 + 2 3 , 5 , 0 , otherwise. 3. Let Y = sin ( U ). The distribution function of Y , F Y ( y ), becomes F Y ( y ) = Pr ( Y y ) = Pr (sin( U ) y ) = Pr ( U a 1 ) + Pr ( U a 2 ) , where a 1 and a 2 are such that sin a 1 = y, 0 < a 1 π 2 and sin a 2 = y, π 2 a 2 < π . Therefore, F Y ( y ) = 1 π a 1 + 1 1 π a 2 = 1 π sin - 1 ( y ) + 1 1 π ( π sin - 1 ( y ) ) = 2 π sin - 1 ( y ) , 0 < y < 1 , 1, y 1, and 0 otherwise.

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