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Unformatted text preview: = 0) = P ( Y = 0  X =1) P ( X =1) = 0 Â· 1 / 4 = 0; P ( X =1 , Y = 1) = P ( Y = 1  X =1) P ( X =1) = 1 Â· 1 / 4 = 1 / 4 . Since, for example, P ( X = 0 , Y = 0) = 1 / 2, but P ( X = 0) P ( Y = 0) = 1 / 2 Â· 1 / 2 = 1 / 4, we see that X and Y cannot be independent. â€¢ The possible values of XY are 0, 1,1. Hence, P ( XY = 0) = P ( X = 0 , Y = 0) = 1 / 2 and P ( XY = 1) = P ( X = 1 , Y = 1) = 1 / 4 using the computations above. By the law of total probability, P ( XY =1) = 1 / 4 . (Equivalently, P ( XY =1) = P ( X =1 , Y = 1) = 1 / 4.) Thus, E ( XY ) = 0 Â· P ( XY = 0) + 1 Â· P ( XY = 1) + (1) Â· P ( XY =1) = 0 + 1 / 41 / 4 = 0 . Since E ( X ) = 0 and E ( Y ) = 0, we see that Cov( X, Y ) = E ( XY )E ( X ) E ( Y ) = 00 = 0; whence X and Y are uncorrelated....
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 Fall '08
 MichaelKozdron
 Statistics, Covariance, Probability, Variance, Probability theory, xy

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