HW27Answers(10.2)

HW27Answers(10.2) - Math331, Spring 2008 Instructor: David...

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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 10.2 Homework Answers Homework: pgs. 424 - 425, #'s 4, 5, 7, 9. 4. Let X and Y denote the numbers of sheep and goats stolen, respectively. We want Cov(X, Y ) = E[XY ] - E[X]E[Y ]. I will first calculate E[X] and E[Y ]. To see these, note that X and Y are both hypergeometric random variables. Therefore, calculating E[X] directly and E[Y ] using the formula for the hypergeometric RV (just to show the two ways to do this) gives 4 E[X] = x=0 x 7 1 7 x 13 20 / 4-x 4 7 20 13 +3 / 3 4 2 7 20 13 +4 / 4 4 1 20 13 / 4 0 = = 7 20 13 +2 / 2 4 3 7 = 1.4. 5 48 8 E[Y ] = = = 1.6. 20 5 Now we need E[XY ]. This is E[XY ] = x+y4 xy 7 x 8 y 5 20 / . 4-x-y 4 The only nonzero terms of the above are (x, y) {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}. Therefore, E[XY ] = x+y4 xy 20 4 7 2 7 x 8 y 11 8 1 5 20 / . 4-x-y 4 7 8 5 +12 1 1 2 5 7 8 +22 1 2 2 7 8 5 +13 1 2 1 5 7 8 +31 0 3 1 7 1 5 0 8 3 5 0 = 1/ +2 1 = 168 95 = 1.76842. Therefore, Cov(X, Y ) = E[XY ] - E[X]E[Y ] = 1.76842 - 1.4 1.6 = -.47157. 5. Note that Y = n - X. Therefore, E[XY ] = E[X(n - X)] = nE[X] - E[X 2 ]. 1 However, X is a binomial(n, p) RV and so E[X] = np and E[X 2 ] = n2 p2 - np2 + np. Y is binomial(n, 1 - p) and so E[Y ] = n(1 - p). Therefore, E[XY ] = nnp - n2 p2 + np2 - np = n2 p - n2 p2 + np2 - np and Cov(X, Y ) = E[XY ] - E[X]E[Y ] = n2 p - n2 p2 + np2 - np - npn(1 - p) = n2 p - n2 p2 + np2 - np - n2 p + n2 p2 = np2 - np = -np(1 - p). 7. We have that X, Y are independent. Therefore, Cov(X, Y + Z) = E [(X - E[X])(Y + Z - (E[Y ] + E[Z]))] = E [(X - E[X])((Y - E[Y ]) + (Z - E[Z]))] = E [(X - E[X])(Y - E[Y ])] + E [(X - E[X])(Z - E[Z])] = E [(X - E[X])] E [(Y - E[Y ])] + Cov(X, Z) = Cov(X, Z), where the fourth equality follows by the independence of X and Y . 9. Simply calculating using definitions gives V ar(X - Y ) = E[((X - Y ) - (E[X] - E[Y ]))2 ] = E[((X - E[X]) - (Y - E[Y ]))2 ] = E[(X - E[X])2 - 2(X - E[X])(Y - E[Y ]) + (Y - E[Y ])2 ] = V ar(X) + V ar(Y ) - 2Cov(X, Y ). 2 ...
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This note was uploaded on 06/24/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at Wisconsin.

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