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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 Spring '10
 DURRETT
 The Land

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