Concrete mathematics : a foundation for computer science

390 DISCRETE PROBABILITY
There’s a curious relatio:n between S and the sum of domino tilings
in equation (7.1). Indeed, we obtain S from T if we replace each 0 by T and
each E by HT, then tack on an HH at the end. This correspondence is easy to
prove because each element of n has the form (T + HT)"HH for some n 3 0,
and each term of T has the form (0 + E)n. Therefore by (7.4) we have
s = (I-T-HT)-'HH,
and the probability generatin.g function for our problem is
G(z) = (1 -w- (P~W-‘(PZ)Z
p*2*
= 1 - qz-pqz* .
(8.64)
Our experience with the negative binomial distribution gives us a clue
that we can most easily calcmate the mean and variance of (8.64) by writing
where
F(z) =
1 - qz-pqz*
P2 ’
and by calculating the “mean” and “variance” of this pseudo-pgf F(z). (Once
again we’ve introduced a function with F( 1) = 1.) We have
F’(1) = (-q-2pq)/p* = 2-p-l -P-*;
F”(1) = -2pq/p* = 2 - 2pP’ .
Therefore, since z* = F(z)G(z), Mean = 2, and Var(z2) = 0, the mean
and variance of distribution G(z) are
Mean(G) = 2 - Mean(F) = pp2 + p-l ; (8.65)
Var(G) = -Va.r(F) = pP4 � t&-3 -2~-*-~-1. (8.66)
When p = 5 the mean and variance are 6 and 22, respectively. (Exercise 4
discusses the calculation of means and variances by subtraction.)