Dr. Hackney STA Solutions pg 64

Dr. Hackney STA Solutions pg 64 - 4-20 Solutions Manual for...

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Unformatted text preview: 4-20 Solutions Manual for Statistical Inference 4.55 Let X1 , X2 , X3 be independent exponential() random variables, and let Y = max(X1 , X2 , X3 ), the lifetime of the system. Then P (Y y) = P (max(X1 , X2 , X3 ) y) = P (X1 y and X2 y and X3 y) = P (X1 y)P (X2 y)P (X3 y). y 1 -x/ e dx 0 by the independence of X1 , X2 and X3 . Now each probability is P (X1 y) = 1 - e-y/ , so P (Y y) = 1-e-y/ and the pdf is fY (y) = 4.57 a. A1 A-1 = = [ [ 1 1 x1 ] 1 n x=1 i = 3 , e 0 < y < , y>0 y 0. 3 1 - e-y/ 0 n 2 -y/ n = 1 xi , n x=1 1 1 n ( x1 the arithmetic mean. 1 , xn ) n 1 x-1 ]-1 = n x=1 i n n 1 + + the harmonic mean. r0 lim log Ar 1 1 = lim log[ xr ] r r0 n x=1 i 1 1 = lim log[ xr ] = r0 r n x=1 i = 1 n n r0 n lim 1 n n r-1 i=1 rxi n 1 r i=1 xi n = r0 lim 1 n n r i=1 xi log xi n 1 r i=1 xi n n i=1 log xi = i=1 n 1 1 log( xi ). n i=1 1 xi )) = ( i=1 xi ) n , the geometric mean. The term Thus A0 = limr0 Ar = exp( n log( r-1 r-1 d d r r rxi = xi log xi since rxi = dr xi = dr exp(r log xi ) = exp(r log xi ) log xi = xr log xi . i b. (i) if log Ar is nondecreasing then for r r log Ar log Ar , then elog Ar elog Ar . Therefore Ar Ar . Thus Ar is nondecreasing in r. (ii) (iii) d dr log Ar = -1 r2 1 log( n n x=1 xr ) + i 1 r 1 n 1 n n i=1 n rxr-1 i xr i=1 i = 1 r2 r n i=1 n xr log xi i xr i 1 - log( n n x=1 xr ) , i x=1 where we use the identity for rxr-1 showed in a). i r n r i=1 xi log xi n r x=1 xi - log( 1 xr ) n x=1 i n n = log(n) + r n n r i=1 xi log xi n r x=1 xi - log( x=1 xr ) i xr i n = log(n) + i=1 n xr i n i=1 xr i n i=1 xr i n i=1 xr i xr i r log xi - n i=1 xr i n log( x=1 xr ) i = = log(n) + i=1 n (r log xi - log( x=1 n x=1 xr i xr )) i n log(n) - i=1 xr i log( xr i ) = log(n) - i=1 ai log( 1 ). ai ...
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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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