From this expression it is easy to see that the mean

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Unformatted text preview: is different. The joint pmf of Xi and Yi for any i is: pXi ,Yi (1, 0) = 1 6 pXi ,Yi (0, 1) = 1 6 4 pXi ,Yi (0, 0) = . 6 In particular, Xi Yi is always equal to zero, because it is not possible to for both a one and a two to show on a single roll of the die. Thus, E [Xi Yi ] = 0. Therefore, Cov(Xi , Yi ) = E [Xi Yi ]−E [Xi ]E [Yi ] = 1 1 0 − 1 6 = − 36 . Not surprisingly, Xi and Yi are negatively correlated. 6 (d) Using the answer to part (c) yields n n Cov(X, Y ) = Cov Xi , i=1 n Yj j =1 n = Cov(Xi , Yj ) i=1 j =1 n = Cov(Xi , Yi ) + i=1 n i,j :i=j − = i=1 Cov(Xi , Yj ) 1 36 + 0=− n . 36 (e) Using the definition of correlation coefficient and answers to (b) and (d) yields: ρX,Y = n − 36 5n 5n 36 36 1 =− . 5 Since Cov(X, Y ) < 0, or, equivalently, ρX,Y < 0, X and Y are negatively correlated. This makes sense; if X is larger than average, it means that there were more one’s showing than average, which would imply that there should be somewhat fewer two’s showing than average. Example 4.8.6 Suppose X1 , . . . , Xn are independent and i...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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