Lect12_2700_s09

# Lect12_2700_s09 - ENGRD 2700 Basic Engineering Probability...

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Review: Independence 2 Conditional Expectation Covariance Title Page JJ II J I Page 1 of 32 Go Back Full Screen Close Quit ENGRD 2700 Basic Engineering Probability and Statistics Lecture 12: Multivariate Distributions David S. Matteson School of Operations Research and Information Engineering Rhodes Hall, Cornell University Ithaca NY 14853 USA [email protected] March 3, 2009

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Review: Independence 2 Conditional Expectation Covariance Title Page JJ II J I Page 2 of 32 Go Back Full Screen Close Quit 1. Review:
Review: Independence 2 Conditional Expectation Covariance Title Page JJ II J I Page 3 of 32 Go Back Full Screen Close Quit Recall p ( x, y ) satisfies 1. p ( x, y ) 0 , 2. { ( x,y ) possible values } p ( x, y ) = 1 . Get the marginal mass function by summing over the other variable: p X ( x ) = y p ( x, y ); p Y ( y ) = x P ( x, y ) Reason: [ X = x ] = { s S : X ( s ) = x } ∩ S = { s S : X ( s ) = x } ∩ [ y { s : Y ( s ) = y } = [ y { s S : X ( s ) = x, Y ( s ) = y } = [ y [ X = x, Y = y ] . Take probabilities and use the fact that the union is a disjoint union so the probabiity of a union is the sum of the probabilities.

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Review: Independence 2 Conditional Expectation Covariance Title Page JJ II J I Page 4 of 32 Go Back Full Screen Close Quit Review: continuous case
Review: Independence 2 Conditional Expectation Covariance Title Page JJ II J I Page 5 of 32 Go Back Full Screen Close Quit Properties of joint density f ( x, y ): 1. f ( x, y ) 0 , for all ( x, y ) R 2 ; 2. R -∞ R -∞ f ( x, y ) dxdy = 1 . Getting marginal densities: Integrate out the other variable: f X ( x ) = Z y = -∞ f ( x, y ) dy, f Y ( y ) = Z x = -∞ f ( x, y ) dx. Reason: F X ( x ) = P [ X x ] = P [ X x, Y R ] = Z x -∞ Z -∞ f ( u, v ) dudv.

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Review: Independence 2 Conditional Expectation Covariance Title Page JJ II J I Page 6 of 32 Go Back Full Screen Close Quit Differentiate with respect to x on both sides to get f X ( x ) = Z v = -∞ f ( x, v ) dv.
Review: Independence 2 Conditional Expectation Covariance Title Page JJ II J I Page 7 of 32 Go Back Full Screen Close Quit Example. Suppose F ( x, y ) = ( k ( x 2 + y 2 ) , if 0 x 2 , 1 y 4 , 0 , otherwise. Good habit: Instead of specifying the function by cases, use indicator functions 1 A ( x ) = ( 1 , if x A, 0 , if x / A. So f ( x, y ) = k ( x 2 + y 2 )1 [0 , 2] ( x )1 [1 , 4] ( y ) or f ( x, y ) = k ( x 2 + y 2 )1 { ( u,v ):0 u 2 , 1 v 4 } ( x, y ) .

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Review: Independence 2 Conditional Expectation Covariance Title Page JJ II J I Page 8 of 32 Go Back Full Screen Close Quit What is k ?
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