This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: STAT 300C HW2 solution 1 Credit goes to SUMIT MUKHERJEE. a. To show that the joint distribution of { X 1 ,X 2 , Â·Â·Â· , } is same as that of { Y 1 ,Y 2 , Â·Â·Â· , } , it suf fices to show that the finite dimensional distributions of { X 1 ,X 2 , Â·Â·Â· ,X n } and { Y 1 ,Y 2 , Â·Â·Â· ,Y n } are same for all n âˆˆ N , and then appeal to the uniqueness of the Kolmogorovâ€™s extension theo rem. To show that the finite dimensional distributions are the same, we use induction on n . For n = 1 , the distribution of X 1 is given by X 1 = R w.p. r r + w + b = W w.p. w r + w + b = B w.p. b r + w + b Also, the distribution of Y 1 is given by P ( Y 1 = R ) = Z p r ,p w ,p b âˆˆ S 2 p r Ã— p r 1 r p w 1 w p b 1 b Î“( r + b + w ) Î“( r )Î“( b )Î“( w ) dp r dp w dp b = Î“( r + b + w ) Î“( r )Î“( b )Î“( w ) Z p r ,p w ,p b âˆˆ S 2 p r r p w 1 w p b 1 b dp r dp w dp b = Î“( r + b + w ) Î“( r )Î“( b )Î“( w ) Ã— Î“( r + 1)Î“( b )Î“( w ) Î“( r + b + w + 1) [ Since integrand is unnormalized D ( r + 1 ,b,w ) density ] = r r + w + b [ Using Î“( x + 1) = x Î“( x ) ] By symmetry, it follows that P ( Y 1 = W ) = w r + w + b ,P ( Y 1 = B ) = b r + w + b . Thus the distribu tion of X 1 and Y 1 are the same, and so the result holds for n = 1 . Now suppose the result holds for n . Need to show that the result holds for n + 1 . Let { e 1 ,e 2 , Â·Â·Â· ,e n +1 } âˆˆ { R,W,B } n +1 be any string of lenght n + 1 with R,W,B as the characters in the string. Also, let r n ,w n ,b n be the number of times that a ball colored R,B,W respectively, was drawn from the urn upto time n . Then r n + w n + b n = n . Also, number of R colored balls at time n is r + r n , number of W colored balls is w + w n , number of B colored balls is b + b n . Also, total number of balls at time n is r + b + w + n . For the purpose of definiteness, let us assume e n +1 = R . First consider the Y process. From the process specification, we have P ( Y 1 = e 1 ,Y 2 = e 2 , Â·Â·Â· ,Y n = e n ) 1 = Z p r ,p w ,p b âˆˆ S 2 p r n r p w n w p b n b Ã— p r 1 r p w 1 w p b 1 b Î“( r + b + w ) Î“( r )Î“( b )Î“( w ) dp r dp w dp b = Î“( r + b + w ) Î“( r )Î“( b )Î“( w ) Z p r ,p w ,p b âˆˆ S 2 p r + r n 1 r p w + w n 1 w p b + b n 1 b dp r dp w dp b = Î“( r + b + w ) Î“( r )Î“( b )Î“( w ) Ã— Î“( r + r n )Î“( b + b n )Î“( w + w n ) Î“( r + b + w + n ) [ Since integrand is unnormalized D ( r + r n ,b + b n ,w + w n ) density ] Doing similar computations for n + 1 , and noting that e n +1 = R by assumption, we have P ( Y 1 = e 1 ,Y 2 = e 2 , Â·Â·Â· ,Y n +1 = e n +1 ) = Î“( r + b + w ) Î“( r )Î“( b )Î“( w ) Ã— Î“( r + r n + 1)Î“( b + b n )Î“( w + w n ) Î“( r + b + w + n + 1) Also note that by induction hypothesis, we have P ( X 1 = e 1 ,X 2 = e 2 , Â·Â·Â· ,X n = e n ) = Î“( r + b + w ) Î“( r )Î“( b )Î“( w ) Ã— Î“( r + r n )Î“( b + b n )Î“( w + w n ) Î“( r + b + w + n ) Now consider the X process. Then P ( X 1 = e 1 ,X 2 = e 2 , Â·Â·Â· ,X n +1 = e n +1 ) = P ( X n +1 = e n +1  X 1 = e 1 ,X 2 = e 2 , Â·Â·Â·...
View
Full Document
 Spring '09
 Trigraph, yj, rSx p2

Click to edit the document details