Concrete mathematics : a foundation for computer science

8.4 FLIPPING COINS 389
A simpler way to derive the mean and variance of Y is to use the reciprocal
generating function
F(z) =
and to write
l-q2 1 q
- = ---2,
P P P
(8.62)
G(z)” = F(z)-“. (8.63)
This polynomial F(z) is not a probability generating function, because it has
a negative coefficient. But it does satisfy the crucial condition F(1) = 1.
Thus F(z) is formally a binomial that corresponds to a coin for which we
The probability is get heads with “probability” equal to -q/p; and G(z) is formally equivalent
negative that I’m to flipping such a coin -1 times(!). The negative binomial distribution
getting younger.
with parameters (n,p) can therefore be regarded as the ordinary binomial
Oh? Then it’s > 1 distribution with parameters (n’, p’) = (-n, -q/p). Proceeding formally,
that you’re getting
older, or staying the mean must be n’p’ = (-n)(-q/p) = nq/p, and the variance must be
the same. n’p’q’ = (-n)(-q/P)(l + 4/p) = w/p2. This formal derivation involving
negative probabilities is valid, because our derivation for ordinary binomials
was based on identities between formal power series in which the assumption
0 6 p 6 1 was never used.
Let’s move on to another example: How many times do we have to flip
a coin until we get heads twice in a row? The probability space now consists
of all sequences of H’s and T's that end with HH but have no consecutive H’s
until the final position:
n = {HH,THH,TTHH,HTHH,TTTHH,THTHH,HTTHH,. . .}.
The probability of any given sequence is obtained by replacing H by p and T
by q; for example, the sequence THTHH will occur with probability
Pr(THTHH) = qpqpp = p3q2.
We can now play with generating functions as we did at the beginning
of Chapter 7, letting S be the infinite sum
S = HH + THH + TTHH + HTHH + TTTHH + THTHH + HTTHH + . . .
of all the elements of fI. If we replace each H by pz and each T by qz, we get
the probability generating function for the number of flips needed until two
consecutive heads turn up.