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### antithethic_Result

Course: STAT 340, Winter 2007
School: Waterloo
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Word Count: 139

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If Result: f() is a monotone function on (0, 1) then Cov ( f (U ), f (1 U ) ) &lt; 0 for U (0,1) Proof: Suppose U1 , U 2 iid U[0,1]. Since f() is monotone f (U1 ) f (U 2 ) f (1 U1 ) f (1 U 2 ) &lt; 0 . Thus f (U1 ) f (1 U1 ) + f (U 2 ) f (1 U 2 ) &lt; f (U1 ) f (1 U 2 ) + f (U 2 ) f (1 U1 ) Taking expected values and using the fact that E ( f (U1 ) f (1 U1 ) ) = E ( f (U 2 ) f U (1...

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If Result: f() is a monotone function on (0, 1) then Cov ( f (U ), f (1 U ) ) < 0 for U (0,1) Proof: Suppose U1 , U 2 iid U[0,1]. Since f() is monotone f (U1 ) f (U 2 ) f (1 U1 ) f (1 U 2 ) < 0 . Thus f (U1 ) f (1 U1 ) + f (U 2 ) f (1 U 2 ) < f (U1 ) f (1 U 2 ) + f (U 2 ) f (1 U1 ) Taking expected values and using the fact that E ( f (U1 ) f (1 U1 ) ) = E ( f (U 2 ) f U (1 2 ) ) and E ( f (U1 ) f (1 U 2 ) ) = E ( f (U1 ) ) E ( f (1 U 2 ) ) = E ( f (U 2 ) f (1 U1 ) ) = E ( f (U1 ) ) E ( f (1 U1 ) ) since U1 , U 2 are independent and 1-U is also U[0,1] we get E ( f (U1 ) f (1 U1 ) ) < E ( f (U1 ) ) E ( f (1 U1 ) ) Since Cov ( f (U1 ), f (1 U1 ) ) = E ( f (U1 ) f (1 U1 ) ) E ( f (U1 ) ) E ( f (1 U1 ) ) We have Cov ( f (U ), f (1 U ) ) < 0
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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Waterloo - STAT - 340
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