Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 551
562 z 12.5. (This distribution has ~4 = 2974, which is rather large. Hence the
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standard deviation of this variance estimate when n = 10 is also rather large,
J2974/10 + 2(22)2/9 M 20.1 according to exercise 54. One cannot complain
that the students cheated.)
a.4 This follows from (8.38) and (88g), because F(z) = G(z)H(z). (A
similar formula holds for all the cumulants, even though F(z) and G(z) may
have negative coefficients.)
8.5 Replace H by p and T by q = 1 -p. If SA = Ss = i we have p2qN = i
and pq2N = iq + i; the solution is p = l/~$~, q = l/G.
8.6 In this case Xly has the same distribution as X, for all y, hence
E(XIY) = EX is constant and V(E(XlY)) = 0. Also V(XlY) is constant and
equal to its expected value.
8.7 We have 1 = (PI +pz+.+.+ps)’ 6 6(p~+p~+.~.+p~) by Chebyshev’s
summation inequality of Chapter 2.
8.8 Let p = Pr(wEAflB), q = Pr(~ueA), and r = Pr(w$B). Then
p+q+r=l,andtheidentitytobeprovedisp=(p+r)(p+q)-qr.
8.9 This is true (subject to the obvious proviso that F and G are defined
on the respective ranges of X and Y), because
Pr(F(X)=f and G(Y)=g) = x Pr(X=xandY=y)
Y&F-‘(~)
YEG-‘(9)
= x Pr(X=x) .Pr(Y=y)
c&F-‘(f)
YEG-‘(9)
= Pr(F(X) =f) . Pr(G(y) = g) .
8.10 Two. Let x1 < x2 be medians; then 1 < Pr(Xxz) <
1, hence equality holds. (Some discrete distributions have no median elements.
For example, let n be the set of all fractions of the form &l/n, with
Pr(+l/n) = Pr(-l/n) = $n2.)
8.11 For example, let K = k with probability 4/( k + 1) (k + 2) (k + 3)) for all
integers k 3 0. Then EK = 1, but E(K’) = 00. (Similarly we can construct
random variables with finite cumulants through K, but with K,+I = co.)
8.12 (a) Let pk = Pr(X = k). If 0 < x < 1, we have Pr(X 6 r) = tk