Dr. Hackney STA Solutions pg 21

Dr. Hackney STA Solutions pg 21 - 2-6 m set Solutions...

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Unformatted text preview: 2-6 m set Solutions Manual for Statistical Inference 1/3 2.17 a. 0 3x2 dx = m3 = 1 m = 1 = .794. 2 2 b. The function is symmetric about zero, therefore m = 0 as long as the integral is finite. 1 - 1 1 dx = tan-1 (x) 1+x2 a - = - 1 + 2 2 = 1. This is the Cauchy pdf. 2.18 E|X - a| = - |x - a|f (x)dx = -(x - a)f (x)dx + a (x a - a)f (x)dx. Then, set d E|X - a| = da f (x)dx - - a f (x)dx = 0. The solution to this equation is a = median. This is a minimum since d2 /da2 E|X -a| = 2f (a) > 0. 2.19 d E(X - a)2 da = = - d da (x - a)2 fX (x)dx = - - d (x - a)2 fX (x)dx da -2(x - a)fX (x)dx = -2 - xfX (x)dx - a - fX (x)dx = -2[EX - a]. d Therefore if da E(X - a)2 = 0 then -2[EX - a] = 0 which implies that EX = a. If EX = a then d 2 2 2 2 da E(X - a) = -2[EX - a] = -2[a - a] = 0. EX = a is a minimum since d /da E(X - a) = 2 > 0. The assumptions that are needed are the ones listed in Theorem 2.4.3. 2.20 From Example 1.5.4, if X = number of children until the first daughter, then P (X = k) = (1 - p)k-1 p, where p = probability of a daughter. Thus X is a geometric random variable, and EX = k=1 k(1 - p)k-1 p = p - k=1 d (1 - p)k dp = -p d dp (1 - p) -1 k=0 k = -p d 1 -1 dp p = 1 . p Therefore, if p = 1 ,the expected number of children is two. 2 2.21 Since g(x) is monotone Eg(X) = - g(x)fX (x)dx = - yfX (g -1 (y)) d -1 g (y)dy = dy yfY (y)dy = EY, - where the second equality follows from the change of variable y = g(x), x = g -1 (y) and d dx = dy g -1 (y)dy. 2.22 a. Using integration by parts with u = x and dv = xe-x 0 2 / 2 we obtain that 2 x2 e-x 2 / 2 dx2 = 2 2 0 e-x / 2 dx. The integral can be evaluated using the argument on pages 104-105 (see 3.3.14) or by trans 2 2 forming to a gamma kernel (use y = -2 / 2 ). Therefore, 0 e-x / dx = /2 and hence the function integrates to 1. ...
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