Dr. Hackney STA Solutions pg 9

Dr. Hackney STA Solutions pg 9 - 1-6Solutions Manual for...

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Unformatted text preview: 1-6Solutions Manual for Statistical Inferencec.nk=1(-1)k+1k(nk)=nk=1(-1)k+1(n-1k-1)=nn-1j=0(-1)j(n-1j)= 0 from part a).1.28 The average of the two integrals is[(nlogn-n) + ((n+ 1) log (n+ 1)-n)]/2=[nlogn+ (n+ 1) log (n+ 1)]/2-n(n+ 1/2) logn-n.Letdn= logn!-[(n+ 1/2) logn-n], and we want to show that limnmdn=c, a constant.This would complete the problem, since the desired limit is the exponential of this one. Thisis accomplished in an indirect way, by working with differences, which avoids dealing with thefactorial. Note thatdn-dn+1=n+12log1 +1n-1.Differentiation will show that ((n+12)) log((1 +1n)) is increasing inn, and has minimumvalue (3/2) log 2 = 1.04 atn= 1. Thusdn-dn+1>0. Next recall the Taylor expansion oflog(1 +x) =x-x2/2 +x3/3-x4/4 +. The first three terms provide an upper bound onlog(1 +x), as the remaining adjacent pairs are negative. Hence< dndn+1<n+121n12n2+13n3-1 =112n2+16n3....
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