Dr. Hackney STA Solutions pg 27

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

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2-12 Solutions Manual for Statistical Inference b. 0 x r f 1 ( x ) sin(2 π log x ) dx = 0 x r 1 2 πx e - (log x ) 2 / 2 sin(2 π log x ) dx = -∞ e ( y + r ) r 1 2 π e - ( y + r ) 2 / 2 sin(2 πy + 2 πr ) dy (substitute y = log x, dy = (1 /x ) dx ) = -∞ 1 2 π e ( r 2 - y 2 ) / 2 sin(2 πy ) dy (sin( a + 2 πr ) = sin( a ) if r = 0 , 1 , 2 , . . . ) = 0 , because e ( r 2 - y 2 ) / 2 sin(2 πy ) = - e ( r 2 - ( - y ) 2 ) / 2 sin(2 π ( - y )); the integrand is an odd function so the negative integral cancels the positive one. 2.36 First, it can be shown that lim x →∞ e tx - (log x ) 2 = by using l’Hˆ opital’s rule to show lim x →∞ tx - (log x ) 2 tx = 1 , and, hence, lim x →∞ tx - (log x ) 2 = lim x →∞ tx = . Then for any k > 0, there is a constant c such that k 1 x e tx e ( log x ) 2 / 2 dx c k 1 x dx = c log x | k = . Hence M x ( t ) does not exist. 2.37 a. The graph looks very similar to Figure 2.3.2 except that f 1 is symmetric around 0 (since it is standard normal). b. The functions look like t 2 / 2 – it is impossible to see any difference. c. The mgf of
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