Concrete mathematics : a foundation for computer science

552 ANSWERS TO EXERCISES
8.13 (Solution by Boris Pittel.) Let us set Y = (XI + . . . + X,)/n and
Z = (X,+1 + . + X2,)/n. Then
= Pr(JZ-cxJ 6 IY-al) 3 t.
The last inequality is, in fact, ‘>’ in any discrete probability distribution,
because Pr(Y = Z) > 0.
8.14 Mean(H) = pMean(F) + qMean(G); Var(H) = pVar(F) + qVar(G) +
pq(Mean(F)-Mean(G))‘. (A mixture is actually a special case of conditional
probabilities: Let Y be the coin, let XIH be generated by F(z), and let XIT
be generated by G(z). Then VX = EV(XIY) + VE(XlY), where EV(XlY) =
pV(XIH) + qV(XIT) and VE(XlY) is the variance of pzMeanCF) + qzMeanfG).)
8.15 By the chain rule, H’(z) = G’(z)F’(G(z)); H”(z) = G”(z)F’(G(z)) +
G’(z)‘F”(G(z)). Hence
Mean(H) = Mean(F) Mean(G) ;
Var(H) = Var(F) Mean(G)’ + Mean(F) Var(G)
(The random variable corresponding to probability distribution H can be understood
as follows: Determine a nonnegative integer n by distribution F;
then add the values of n independent random variables that have distribution
G. The identity for variance in this exercise is a special case of (8.105),
when X has distribution H and Y has distribution F.)
8.16 ewizmm’)/(l -w).
8.17 Pr(Y,,, 6 m) = Pr(Y,,, + n 6 m + n) = probability that we need <
n + n tosses to obtain n heads = probability that m + n tosses yield 3 n
heads = Pr(X m+n,p 3 n). Thus
n+k-1 pnqk = t (m;n)p*qm+n~k
k >
k>n
= x (m;n)pm+n-kqk;
k