Dr. Hackney STA Solutions pg 62

# Dr. Hackney STA Solutions pg 62 - 4-18 Solutions Manual for...

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Unformatted text preview: 4-18 Solutions Manual for Statistical Inference 4.49 a. fX (x) = (af1 (x)g1 (y) + (1 - a)f2 (x)g2 (y))dy g1 (y)dy + (1 - a)f2 (x) g2 (y)dy = af1 (x) = af1 (x) + (1 - a)f2 (x). fY (y) = (af1 (x)g1 (y) + (1 - a)f2 (x)g2 (y))dx f1 (x)dx + (1 - a)g2 (y) f2 (x)dx = ag1 (y) = ag1 (y) + (1 - a)g2 (y). b. () If X and Y are independent then f (x, y) = fX (x)fY (y). Then, f (x, y) - fX (x)fY (y) = af1 (x)g1 (y) + (1 - a)f2 (x)g2 (y) - [af1 (x) + (1 - a)f2 (x)][ag1 (y) + (1 - a)g2 (y)] = a(1 - a)[f1 (x)g1 (y) - f1 (x)g2 (y) - f2 (x)g1 (y) + f2 (x)g2 (y)] = a(1 - a)[f1 (x) - f2 (x)][g1 (y) - g2 (y)] = 0. Thus [f1 (x) - f2 (x)][g1 (y) - g2 (y)] = 0 since 0 < a < 1. () if [f1 (x) - f2 (x)][g1 (y) - g2 (y)] = 0 then f1 (x)g1 (y) + f2 (x)g2 (y) = f1 (x)g2 (y) + f2 (x)g1 (y). Therefore fX (x)fY (y) = a2 f1 (x)g1 (y) + a(1 - a)f1 (x)g2 (y) + a(1 - a)f2 (x)g1 (y) + (1 - a)2 f2 (x)g2 (y) = a2 f1 (x)g1 (y) + a(1 - a)[f1 (x)g2 (y) + f2 (x)g1 (y)] + (1 - a)2 f2 (x)g2 (y) = a2 f1 (x)g1 (y) + a(1 - a)[f1 (x)g1 (y) + f2 (x)g2 (y)] + (1 - a)2 f2 (x)g2 (y) = af1 (x)g1 (y) + (1 - a)f2 (x)g2 (y) = f (x, y). Thus X and Y are independent. c. Cov(X, Y ) = a1 1 + (1 - a)2 2 - [a1 + (1 - a)2 ][a1 + (1 - a)2 ] = a(1 - a)[1 1 - 1 2 - 2 1 + 2 2 ] = a(1 - a)[1 - 2 ][1 - 2 ]. To construct dependent uncorrelated random variables let (X, Y ) af1 (x)g1 (y) + (1 - a)f2 (x)g2 (y) where f1 , f2 , g1 , g2 are such that f1 - f2 = 0 and g1 - g2 = 0 with 1 = 2 or 1 = 2 . d. (i) f1 binomial(n, p), f2 binomial(n, p), g1 binomial(n, p), g2 binomial(n, 1 - p). (ii) f1 binomial(n, p1 ), f2 binomial(n, p2 ), g1 binomial(n, p1 ), g2 binomial(n, p2 ). p p (iii) f1 binomial(n1 , n1 ), f2 binomial(n2 , n2 ), g1 binomial(n1 , p), g2 binomial(n2 , p). ...
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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