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Unformatted text preview: 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 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] otherwise F X ( x ) = 1 2 + 1 2 x, for x ∈ [ 1 , 1]. F Y ( y ) = P rob (exp( X ) ≤ y ) = P rob ( 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 2. X ∼ U [ 1 , 1] and Y = X 2 F Y ( y ) = P rob ( X 2 ≤ y ) = P rob ( √ 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 ) = P rob (1 X 2 ≤ y ) = P rob ( 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 )....
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This note was uploaded on 01/02/2010 for the course APSC 116 taught by Professor Tommazzuchi during the Fall '08 term at GWU.
 Fall '08
 TomMazzuchi

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