1 theorem 44 if a continuous time markov chain xt is

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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 ....
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