Stochastic

# 1 theorem 44 if a continuous time markov chain xt is

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) tj 1 e Adding the ﬁrst two terms then comparing with (4.14) shows that the above is = j pt (1, j ) + (j 1)pt (1, j 1) Having worked to ﬁnd pt (1, j ), it is fortunately easy to ﬁnd pt (i, j ). The chain starting with i individuals is the sum of i copies of the chain starting from 1 individual. Using this one can easily compute that ✓ ◆ j1 pt (i, j ) = (e t )i (1 e t )j i (4.15) i1 In words, the sum of i geometrics has a negative binomial distribution. Proof. To begin we note that if N1 , . . . Ni have the distribution given in (4.13) and n1 + · · · + ni = j , then P (N1 = n1 , . . . , Ni = ni ) = i Y e t (1 e ) t nk 1 = (e ) (1 ti e ) tj i k=1 To count the number of possible (n1 , . . . , ni ) with nk 1 and sum j , we think of putting j balls in a row. To divide the j balls into i groups of size n1 , . . . , ni , we will insert cards in the slots between the balls and let nk be the number of balls in the k th group. Having made this transformation it is clear that the number of (n1 , . . . , ni ) is the number of ways of picking i 1 of the j 1 slot to put 1 the cards or j 1 ....
View Full Document

Ask a homework question - tutors are online