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# covind - = 0 = P Y = 0 | X =-1 P X =-1 = 0 Â 1 4 = 0 P X...

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Statistics 351 (Fall 2007) Covariance Zero Does Not Imply Independence The following exercise illustrates that two random variables can have covariance zero yet need not be independent. Exercise. Consider the random variable X deﬁned by P ( X = - 1) = 1 / 4, P ( X = 0) = 1 / 2, P ( X = 1) = 1 / 4. Let the random variable Y be deﬁned as Y := X 2 . Hence, P ( Y = 0 | X = 0) = 1, P ( Y = 1 | X = - 1) = 1, P ( Y = 1 | X = 1) = 1. Show that the density of Y is P ( Y = 0) = 1 / 2, P ( Y = 1) = 1 / 2. Find the joint density of ( X, Y ), and show that X and Y are not independent. Find the density of XY , compute E ( XY ), and show that X and Y are uncorrelated. Solution. We ﬁnd the density of Y simply using the law of total probability : P ( Y = 0) = P ( Y = 0 | X = 1) P ( X = 1) + P ( Y = 0 | X = 0) P ( X = 0) + P ( Y = 0 | X = - 1) P ( X = - 1) = 0 · 1 / 4 + 1 · 1 / 2 + 0 · 1 / 4 = 1 / 2 , P ( Y = 1) = P ( Y = 1 | X = 1) P ( X = 1) + P ( Y = 1 | X = 0) P ( X = 0) + P ( Y = 1 | X = - 1) P ( X = - 1) = 1 · 1 / 4 + 0 · 1 / 2 + 1 · 1 / 4 = 1 / 2 . The joint density of ( X, Y ) is given by P ( X = 0 , Y = 0) = P ( Y = 0 | X = 0) P ( X = 0) = 1 · 1 / 2 = 1 / 2; P ( X = 0 , Y = 1) = P ( Y = 1 | X = 0) P ( X = 0) = 0 · 1 / 2 = 0; P ( X = 1 , Y = 0) = P ( Y = 0 | X = 1) P ( X = 1) = 0 · 1 / 4 = 0; P ( X = 1 , Y = 1) = P ( Y = 1 | X = 1) P ( X = 1) = 1 · 1 / 4 = 1 / 4; P ( X = - 1 , Y

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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 / 4-1 / 4 = 0 . Since E ( X ) = 0 and E ( Y ) = 0, we see that Cov( X, Y ) = E ( XY )-E ( X ) E ( Y ) = 0-0 = 0; whence X and Y are uncorrelated....
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covind - = 0 = P Y = 0 | X =-1 P X =-1 = 0 Â 1 4 = 0 P X...

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