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09_02ans - STAT 410 Examples for Fall 2009 Example 1...

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STAT 410 Examples for 09/02/2009 Fall 2009 Example 1 : Consider a continuous random variable X with p.d.f . f X ( x ) = < < - < < - otherwise 0 2 0 3 . 0 1 3 2 . 0 x x Find the probability distribution of Y = X 2 . y < 0 P ( X 2 y ) = 0 F Y ( y ) = 0. y 0 F Y ( y ) = P ( Y y ) = P ( X 2 y ) = P ( y X y ) Case 1: 0 y < 1 0 y < 1 F Y ( y ) = 0.3 y .
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Case 2: 1 y < 4 1 y < 2 F Y ( y ) = 0.2 ( 1 + y ) + 0.3 y . Case 3: 4 y < 9 2 y < 3 F Y ( y ) = 0.2 ( 1 + y ) + 0.6. Case 4: y 9 F Y ( y ) = 1 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Theorem 1.7.1 X continuous r.v. with p.d.f. f X ( x ) . Y = g ( X ) g ( x ) one-to-one, differentiable d x / d y = d [ g 1 ( y ) ] / d y f Y ( y ) = f X ( g 1 ( y ) ) y x d d _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
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Example 2 : Consider a continuous random variable X with p.d.f . f X ( x ) = < < o.w. 0 1 0 2 x x Find the probability distribution of Y = X . f X ( x ) = < < o.w. 0 1 0 2 x x F X ( x ) = < < 1 1 1 0 0 0 2 x x x x Y = X . y < 0 F Y ( y ) = P ( Y y ) = P ( X y ) = 0. y 0 F Y ( y ) = P ( Y y ) = P ( X y ) = P ( X y 2 ) = F X ( y 2 ) . 0 y < 1 F Y ( y ) = F X ( y 2 ) = y 4 .
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