Dr. Hackney STA Solutions pg 27

Dr. Hackney STA Solutions pg 27 - 2-12 Solutions Manual for...

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Unformatted text preview: 2-12 Solutions Manual for Statistical Inference b. xr f1 (x) sin(2log x)dx = 0 = = = 2 2 2 2 1 e-(log x) /2 sin(2 log x)dx 2x 0 2 1 e(y+r)r e-(y+r) /2 sin(2y + 2r)dy 2 - (substitute y = log x, dy = (1/x)dx) 2 2 1 e(r -y )/2 sin(2y)dy 2 - (sin(a + 2r) = sin(a) if r = 0, 1, 2, . . .) 0, xr because e(r -y )/2 sin(2y) = -e(r -(-y) )/2 sin(2(-y)); the integrand is an odd function so the negative integral cancels the positive one. 2.36 First, it can be shown that 2 lim etx-(log x) = x 2 by using l'H^pital's rule to show o tx - (log x)2 = 1, x tx lim and, hence, x k lim tx - (log x)2 = lim tx = . x k Then for any k > 0, there is a constant c such that 1 tx ( log x)2 /2 e e dx c x 1 dx = c log x|k = . x Hence Mx (t) does not exist. 2.37 a. The graph looks very similar to Figure 2.3.2 except that f1 is symmetric around 0 (since it is standard normal). b. The functions look like t2 /2 it is impossible to see any difference. c. The mgf of f1 is eK1 (t) . The mgf of f2 is eK2 (t) . d. Make the transformation y = ex to get the densities in Example 2.3.10. x d 2.39 a. dx 0 e-t dt = e-x . Verify d dx b. d d -t e dt 0 x e-t dt = 0 d -t e dt 0 d d 1 - e-t dx 0 x = 0 d dx 1 1 - e-x + = e-x . = = d d 1 -te-t dt = - (2) = - 2 . Verify 2 e-t dt = 0 d 1 1 = - 2. d c. d dt 1 1 dx t x2 = - t1 . 2 Verify d dt 1 t 1 d dx = 2 x dt 1 (x-t)2 - 1 1 x 1 = t d dt -1 + 1 t =- 1 . t2 d. d dt 1 dx 1 (x-t)2 = d dt 1 dx = 2(x - t)-3 dx = -(x - t) -2 = 1 1 . (1-t)2 Verify d dt (x - t)-2 dx = 1 d -1 -(x - t) dt = 1 d 1 1 = 2. dt 1 - t (1 - t) ...
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