Concrete mathematics : a foundation for computer science

558 ANSWERS TO EXERCISES
8.37 The number of coin-toss sequences of length n is F,-‘, for all n > 0,
because of the relation between domino tilings and coin flips. Therefore the
probability that exactly n tosses are needed is F,-’ /2n, when the coin is fair.
Also q,, = Fn+‘/2n-‘, since xkbnFnzn = (Fnzn + F,-~z~+‘)/(l - z - zz).
(A systematic solution via generating functions is, of course, also possible.)
8.38 When k faces have been seen, the task of rolling a new one is equivalent
to flipping coins with success probability pk = (m - k)/m. Hence the pgf is
11
nki(,Pkd(l - qkz) = nk:zO( ’ ’ m - k)z/(m - kz). The mean is ~~~~pk’ =
M-b - H,-l); the variance is m’(H!’ - HE!,) - m(H, - H,-1); and
equation (7.47) provides a closed form for the requested probability, namely
mpnm!{;I,‘}/( m-L)!. (The problem discussed in this exercise is traditionally
called “coupon collecting!‘)
8.39 E(X) = P(-1); V(X) = P(-2) - P(-l)2; E(lnX) = -P’(O).
8.40 (a) We have K, =n(O!{';}p-l!{T}p2 +2!{y}p3 -...), by (7.49).
Incidentally, the third cumulant is npq(q-p) and the fourth is npq(l-6pq).
Theidentity q+pet = (p+qept)et shows that f,,,(p) = (-l)“f,(q)+[m=l];
hence we can write f,,,(p) = g,,,(pq)(q-p)'m Odd], where g,,, is a polynomial
of degree [m/2], whenever m > 1. (b) Let p = i and F(t) = ln( 5 + Set).
Then~,~,~,tm-‘/(m-1~t!=F’(t)=1-1/(et+1),andwecanuseexercise
6.23.
8.41 If G(z) is the pgf for a random variable X that assumes only positive
integer values, then s: G(z) dz/z = tk>, Pr(X=k)/k = E(X-‘). If X is the
distribution of the number of flips to obtain n + 1 heads, we have G(z) =
(PZ/(l - qz)y+’ by (8.5g), and the integral is
if we substitute w = pz/(l .- qz). When p = q the integrand can be written
(-l)n((l+w)~1-l+w-w2+~~~+(-l)nwn~’),sotheintegralis (-l)n(ln2l+~-~+...+(-l)n/n).
WehaveH~,-Hn=ln2-~n~‘+~n~Z+O(n~4)
by (g.28), and it follows tha.t E(X,:,) = :n-’ - $np2 + O(np4).
8.42 Let F,(z) and G,(z) be pgf’s for the number of employed evenings, if
the man is initially unemployed or employed, respectively. Let qh = 1 - ph
and qf := 1 -pf. Then Fo(z) = Go(z) = 1, and
F,,(z) = PhZGv’ (Z) + qhFm ’ (Z) ;
G,,(Z) = PfFn-1 (z) + qfZGn-1 (z).