Concrete mathematics : a foundation for computer science

388 DISCRETE PROBABILITY
Repeating the process until n heads are obtained gives the pgf
P= n - = w& (n+;-yq,lk
( 1 -qz )
This, incidentally, is Z” times
(&)” = ; (ni-;-l)p.,q’z*.
(8.60)
the generating function for the negative binomial distribution.
The probability space in example (8.5g), where we flip a coin until
n heads have appeared, is different from the probability spaces we’ve seen
earlier in this chapter, because it contains infinitely many elements. Each element
is a finite sequence of heads and/or tails, containing precisely n heads
in all, and ending with heads; the probability of such a sequence is pnqkpn, Heads I win,
where k - n is the number of tails. Thus, for example, if n = 3 and if we tails you lose.
write H for heads and T for tails, the sequence THTTTHH is an element of the
probability space, and its probability is qpqqqpp = p3q4.
Let X be a random variable with the binomial distribution (8.57), and let
No? OK; tails you
lose, heads I win.
No? Well, then,
heads you ,ose
Y be a random variable with the negative binomial distribution (8.60). These
distributions depend on n and p. The mean of X is nH’(l) = np, since its
pgf is Hi; the variance is
tails I win.
’
n(H”(1)+H’(1)-H’(1)2) = n(O+p-p2) = npq. (8.61)
Thus the standard deviation is m: If we toss a coin n times, we expect
to get heads about np f fitpq times. The mean and variance of Y can be
found in a similar way: If we let
we have
G’(z) = (, T9sz,, ,
2pq2
G”(z) = (, _ qz13 ;
hence G’(1) = pq/p2 = q/p and G”(1) = 2pq2/p3 = 2q2/p2. It follows that
the mean of Y is nq/p and the variance is nq/p2.