print2

# print2 - Transformations and Expectations of random...

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Transformations and Expectations of random variables X F X ( x ): a random variable X distributed with CDF F X . Any function Y = g ( X ) is also a random variable. If both X , and Y are continuous random variables, can we find a simple way to characterize F Y and f Y (the CDF and PDF of Y ), based on the CDF and PDF of X ? For the CDF: F Y ( y ) = P Y ( Y y ) = P Y ( g ( X ) y ) = P X ( x ∈ X : g ( X ) y ) ( X is sample space for X ) = Z { x ∈X : g ( X ) y } f X ( s ) ds. PDF: f Y ( y ) = F 0 y ( y ) Caution: need to consider support of y . Consider several examples: 1. X U [ - 1 , 1] and y = exp( x ) That is: f X ( x ) = 1 2 if x [ - 1 , 1] 0 otherwise F X ( x ) = 1 2 + 1 2 x, for x [ - 1 , 1]. F Y ( y ) = Prob (exp( X ) y ) = Prob ( X log y ) = F X (log y ) = 1 2 + 1 2 log y, for y [ 1 e , e ] . Be careful about the bounds of the support! f Y ( y ) = ∂y F Y ( y ) = f X (log y ) 1 y = 1 2 y , for y [ 1 e , e ]. 1

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2. X U [ - 1 , 1] and Y = X 2 F Y ( y ) = Prob ( X 2 y ) = Prob ( - y X y ) = F X ( y ) - F X ( - y ) = 2 F X ( y ) - 1 , by symmetry: F X ( - y ) = 1 - F X ( y ). f Y ( y ) = ∂y F Y ( y ) = 2 f X ( y ) 1 2 y = 1 2 y , for y [0 , 1]. 3. X U [ - 1 , 1] and Y = 1 - X 2 F Y ( y ) = Prob (1 - X 2 y ) = Prob ( X ≤ - p 1 - y X p 1 - y ) = 1 - F X ( p 1 - y ) + F X ( - p 1 - y ) = 2 - 2 F X ( p 1 - y ) , by symmetry: F X ( - p 1 - y ) = 1 - F X ( p 1 - y ). f Y ( y ) = ∂y F Y ( y ) = 2 f X ( 1 - y ) 2 1 - y = 1 2 1 - y , for y [0 , 1]. As the first example above showed, it’s easy to derive the CDF and PDF of Y when g ( · ) is a strictly monotonic function: Theorems 2.1.3, 2.1.5: When g ( · ) is a strictly increasing function, then F Y ( y ) = Z g - 1 ( y ) -∞ f X ( x ) dx = F X ( g - 1 ( y )) f Y ( y ) = f X ( g - 1 ( y )) ∂y g - 1 ( y ) using Leibniz’ rule .
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