lecture8 - ISYE 2028 A and B Lecture 8 Kobi Abayomi March...

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Unformatted text preview: ISYE 2028 A and B Lecture 8 Kobi Abayomi March 25, 2009 1 Independent Random Variables Two random variables are independent if p X,Y ( x,y ) = p X ( x ) p Y ( y ) (1) or f X,Y ( x,y ) = f X ( x ) f Y ( y ) (2) This is directly analogous to the general probability rules. The conditional probability mass and density functions, are then just: p X | Y ( X | Y ) = p X,Y ( x,y ) p Y ( y ) = p X ( x ) p Y ( y ) p Y ( y ) = p X ( x ) f X | Y ( X | Y ) = f X,Y ( x,y ) f Y ( y ) = f X ( x ) f Y ( y ) f Y ( y ) = f X ( x ) Dependence is any violation of this condition. 1.1 Example Let f X 1 ,X 2 ( x 1 ,x 2 ) = x 1 + x 2 1 { <x 1 < 1 , <x 2 < 1 } Are x 1 and x 2 independent? 1 Well...: f 1 ( x 1 ) = Z f ( x 1 ,x 2 ) dx 2 = Z 1 x 1 + x 2 dx 2 = x + 1 / 2 1 { x 1 < 1 } f 2 ( x 2 ) = Z f ( x 1 ,x 2 ) dx 1 = Z 1 x 1 + x 2 dx 1 = 1 / 2 + x 2 1 { x 2 < 1 } But: x 1 + x 2 6 = ( x 1 + 1 / 2)( x 2 + 1 / 2). The answer is no. 1.2 Independence is factorization of the pdf In general, for X 1 ,X 2 f ( x 1 ,x 2 ), if f ( x 1 ,x 2 ) = g ( x 1 ) h ( x 2 ) this implies that X 1 is indepen- dent of X 2 This is not a formal proof 1 : We know that we can always write a joint pdf as product of a conditional and marginal... f X 1 ,X 2 ( x 1 ,x 2 ) = f X 2 | X 1 ( x 2 | x 1 ) f X 1 ( x 1 ) If the functional form of f X 2 | X 1 ( x 2 | x 1 ) does not include (depend) x 1 , say f X 2 | X 1 ( x 2 | x 1 ) = h ( x 2 ) then - integrate both sides over x 1 ... Z f X 1 ,X 2 ( x 1 ,x 2 ) dx 1 = Z f X 2 | X 1 ( x 2 | x 1 ) f X 1 ( x 1 ) dx 1 which yields f X 2 ( x 2 ) = h ( x 2 ) Z f X 1 ( x 1 ) dx 1 and of course f X 2 ( x 2 ) = h ( x 2 ) = f X 2 | X 1 ( x 2 | x 1 ) 1 But then, what is these days...? 2 Thus, independence of X 2 and X 1 is equivalent to the factorability of the joint distribution. The definitions for independence are straightforward. Often we can exploit the strong as- sumption of independence to simplify modelling, and get interesting results. 1.3 Example In n + m independent trials, each with probability p of success: let X the successes in the 1 st the first n trials; let Y the successes in the final m trials. Then P ( X = x,Y = y ) = C n x p x (1- p ) n- x C m y p y (1- p ) n- y and it is apparent that X and Y are independent. What about Z = X + Y ? Are Z and X independent? P ( X = x,Z = z ) = P ( X = x,Y = z- x ) = C n x p x (1- p ) n- x C m z- x p z- x (1- p ) m- ( z- x ) which implies that X and Z are not independent. 2 Mutual Independence Heres an example of mutual independence Let f ( x,y,z ) = e- ( x + y + z ) 1 { <x,y,z< } The cumulative distribution, then, is F ( x,y,z ) = Z z Z y Z x e- ( u + v + w ) dudvdw = = (1- e- x )(1- e- y )(1- e- z ) The joint distribution is completely factorable X , Y and Z are mutually independent....
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This note was uploaded on 11/08/2009 for the course ISYE 2028 taught by Professor Shim during the Spring '07 term at Georgia Institute of Technology.

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lecture8 - ISYE 2028 A and B Lecture 8 Kobi Abayomi March...

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