Concrete mathematics : a foundation for computer science

554 ANSWERS TO EXERCISES
8.24 (a) Any one of the dice ends up in J’s possession with probability
p, hence p = A. Let q = A. Then the pgf for J’s total holdings
; I,“+‘;$+;
with mean (2n + l)p and variance (2n + l)pq, by (8.61).
(b) (;)p3q2 + (;)p;'4 + (;)p' = w = .585.
8.25 The pgf for the current stake after n rolls is G,(z), where
Go(z) = z*;
G,(z) = I;=, G, ,(~"~~')9/6, for n > 0.
This problem can
(The noninteger exponents cause no trouble.) It follows that Mean(G,,) =
Mean(G, I ), and Var(G,) -t Mean( = f$(Var(G,, 1) + Mean(G, I 1’).
perhaps be so’ved
more easily without
geprat;ngfunct;ons
So the mean is always A, but the variance grows to ((g)” ~ l)A2.
than with them.
8.26 The pgf FL,,(Z) satisfies F,',,,(z) = FL,, L(z)/L; hence Mean(Ft,,) =
FL,, (1) = [n 3 L]/l and F,lln (1) = [n 3 21]/L2; the variance is easily computed.
(In fact, we have
which approaches a Poisson distribution with mean l/1 as n + co.)
8.27 (n2L3 - 3nZ21-1 + Zt:)/n(n - l)(n - 2) has the desired mean, where
xk = XF + + Xk. This follows from the identities
E(x211) = np3 +n(n- 1)p2p1 ;
E(x:) = w3+3n(n-1)p2pl +n(n-l)(n-2)p:.
Incidentally, the third cumulant is ~3 = E( (X-EX)3), but the fourth cumulant
does not have such a simple expression; we have ~~ = E (( X - EX)4) - 3( VX)'.
8.28 (The exercise implicitly calls for p = q =: i, but the general answer is
given here for completeness.) Replace H by pz and T by qz, getting S*(z) =
p2qz3/(1 -pz)(l - qz)(l -pqz’) and SB(Z) = pq2z3/(1 ~ qz)(l -pqz2). The
pgf for the conditional probability that Alice wins at the nth flip, given that
she wins the game, is
S*(Z) _ 3 4 P 1-Pq
SAtI)
l-pz l-qz l-pqz2
This is a product of pseudo-pgf’s, whose mean is 3+p/q+q/p+2pq/( 1 -pq).
The formulas for Bill are the same but without the factor q/( 1 -pz), so Bill’s
mean is 3 + q/p + 2pq/(l -pq). When p = q = i, the answer in case (a) is