Dr. Hackney STA Solutions pg 48

Dr. Hackney STA Solutions pg 48 - 4-4Solutions Manual for...

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Unformatted text preview: 4-4Solutions Manual for Statistical InferenceIfV <0, thenX < Y. So forv=-1,-2,..., the joint pmf isfU,V(u,v)=P(U=u,V=v) =P(X=u,Y=u-v)=p(1-p)u-1p(1-p)u-v-1=p2(1-p)2u-v-2.IfV= 0, thenX=Y. So forv= 0, the joint pmf isfU,V(u,0) =P(U=u,V= 0) =P(X=Y=u) =p(1-p)u-1p(1-p)u-1=p2(1-p)2u-2.In all three cases, we can write the joint pmf asfU,V(u,v) =p2(1-p)2u+|v|-2=p2(1-p)2u(1-p)|v|-2, u= 1,2,...;v= 0,1,2,....Since the joint pmf factors into a function ofuand a function ofv,UandVare independent.b. The possible values ofZare all the fractions of the formr/s, whererandsare positiveintegers andr < s. Consider one such value,r/s, where the fraction is in reduced form. Thatis,randshave no common factors. We need to identify all the pairs (x,y) such thatxandyare positive integers andx/(x+y) =r/s. All such pairs are (ir,i(s-r)),i= 1,2,....Therefore,PZ=rs=Xi=1P(X=ir,Y=i(s-r)) =Xi=1p(1-p)ir-1p(1-p)i(s-r)-1=p2(1-p)2Xi=1((1-p)s)i=p2(1-p)2(1-p)s1-(1-p)s=p2(1-p)s-21-(1-p)s....
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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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