To have j
toy types after sampling i
toys, we either have j−1
toy types after sampling i−1
toys, and the i-th
toy is of a previously unseen type, or, we have j
toy types after sampling i−1
toys, and the i-th
toy has an already seen type.
Thus,
pi,j=pi−1,j−1n−j+1n+pi−1,jjn
(b)
Note that p1,0=0,p1,1=1
and pi,j=0
for j>i.
Using strong induction, a proof of the recursion in part a
follows.