This preview shows pages 1–3. Sign up to view the full content.
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 , , } is same as that of { Y 1 ,Y 2 , , } , it suf fices to show that the finite dimensional distributions of { X 1 ,X 2 , ,X n } and { Y 1 ,Y 2 , ,Y n } are same for all n N , and then appeal to the uniqueness of the Kolmogorovs 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 S 2 p r p r 1 r p w 1 w p b 1 b ( r + b + w ) ( r )( b )( w ) dp r dp w dp b = ( r + b + w ) ( r )( b )( w ) Z p r ,p w ,p b S 2 p r r p w 1 w p b 1 b dp r dp w dp b = ( r + b + w ) ( r )( b )( w ) ( r + 1)( b )( w ) ( r + b + w + 1) [ Since integrand is unnormalized D ( r + 1 ,b,w ) density ] = r r + w + b [ Using ( x + 1) = x ( 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 , ,e n +1 } { 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 , ,Y n = e n ) 1 = Z p r ,p w ,p b S 2 p r n r p w n w p b n b p r 1 r p w 1 w p b 1 b ( r + b + w ) ( r )( b )( w ) dp r dp w dp b = ( r + b + w ) ( r )( b )( w ) Z p r ,p w ,p b 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 = ( r + b + w ) ( r )( b )( w ) ( r + r n )( b + b n )( w + w n ) ( 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 , ,Y n +1 = e n +1 ) = ( r + b + w ) ( r )( b )( w ) ( r + r n + 1)( b + b n )( w + w n ) ( r + b + w + n + 1) Also note that by induction hypothesis, we have P ( X 1 = e 1 ,X 2 = e 2 , ,X n = e n ) = ( r + b + w ) ( r )( b )( w ) ( r + r n )( b + b n )( w + w n ) ( r + b + w + n ) Now consider the X process. Then P ( X 1 = e 1 ,X 2 = e 2 , ,X n +1 = e n +1 ) = P ( X n +1 = e n +1  X 1 = e 1 ,X 2 = e 2 ,...
View
Full
Document
This document was uploaded on 05/04/2010.
 Spring '09

Click to edit the document details