elemprob-fall2010-page39

# elemprob-fall2010-page39 - E XY = E X E Y Proof By the...

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15 Expectations As in the one variable case, we have E g ( X,Y ) = XX g ( x,y ) p ( x,y ) in the discrete case and E g ( X,Y ) = Z Z g ( x,y ) f ( x,y ) dxdy in the continuous case. If we set g ( x,y ) = x + y , then E ( X + Y ) = Z Z ( x + y ) f ( x,y ) dxdy = Z Z xf ( x,y ) dxdy + Z Z yf ( x,y ) dxdy. If we now set g ( x,y ) = x , we see the ﬁrst integral on the right is E X , and similarly the second is E Y . Therefore E ( X + Y ) = E X + E Y. Proposition 15.1 If X and Y are independent, then E [ h ( X ) k ( Y )] = E h ( X ) E k ( Y ) . In particular,
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Unformatted text preview: E ( XY ) = ( E X )( E Y ) . Proof. By the above with g ( x,y ) = h ( x ) k ( y ), E [ h ( X ) k ( Y )] = Z Z h ( x ) k ( y ) f ( x,y ) dxdy = Z Z h ( x ) k ( y ) f X ( x ) f Y ( y ) dxdy = Z h ( x ) f X ( x ) Z k ( y ) f Y ( y ) dy dx = Z h ( x ) f X ( x )( E k ( Y )) dx = E h ( X ) E k ( Y ) . 39...
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## This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.

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