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Concrete mathematics : a foundation for computer science

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A ANSWERS TO EXERCISES 553<br />

8.19 (a) Gx,+x,(z) = Gx, (z)Gx>(z) = ervl+plrir ‘I. Hence the probability<br />

is eW1+p2 (~1 + pz)“/n!; the sum of independent Poisson variables is Poisson.<br />

(b) In general, if K,X denotes the mth cumulant of a random variable X, we<br />

have K,(aX, + bX2‘t = am(K,X1) + bm(K,Xz), when a, b 3 0. Hence the<br />

answer is 2mpl + Smu2.<br />

8.20 The general pgf will be G(z) = zm/F(z), where<br />

F(z) = zm - (1 -r)~aikl;A:kl=A,kl]Zm~~k,<br />

k=l<br />

F’(1) = m-f&,~[A(k’=A~k,],<br />

k=l<br />

F”(1) = m(m- 1) -2~(m-k)A:kI[Aik)=Alkl].<br />

k=l<br />

8.21 This is Ena0 qn, where q,, is the probability that the game between<br />

Alice and Bill is still incomplete after n flips. Let pn be the probability that<br />

the game ends at the nth flip; then p,, + q,, = q,-l. Hence the average time<br />

to play the game is Ena, wn=(q0-q41)+4ql -q2)+3(42-43)+...=<br />

qo + q1 + q2 +. . = N, since lim,,, nq, = 0.<br />

Another way to establish this answer is to replace H and T by :z.<br />

Then the derivative of the first equation in (8.78) tells us that N (1) + N’( 1) =<br />

N’(l)+S;(l)+Sf,(l).<br />

By the way, N = y.<br />

8.22 By definition we have V(XjY) = E(X’lY) - (E(XlY))’ and V(E(xlY)) =<br />

E(KW’l12) - (E(EIW’)))2; h ence E(V(XlY)) +V(E(XlY)) = E(E(X’lY)) -<br />

(E(E(X/Y)))‘. But E(E(XlY)) = EX and E(E(X’IY)) = E(X2), so the result is<br />

just VX.<br />

8.23 Let C& = { � , � }’ and fll =(m, � , � , H}‘; and let 02 be the<br />

other 16 elements of n. Then Prl1 (w) - Pro0 (cu) = $& &, & according<br />

as w E no, RI, 02. The events A must there<strong>for</strong>e be chosen with kj elements<br />

fromnj, where (ko,kl,kl) is one ofthe following: (O,O,O), (0,2,7), (0,4,14),<br />

(1,4,4), (1,6,11), (2,6,1), (2,&g), (3,&15), (3,10,5), (3,12,12), (4,12,2),<br />

(4,14,9), (4,16,16). Forexample, thereare ($(‘,“)(‘f) eventsoftype (2,6,1).<br />

The total number of such events is [z’] (1 + z”)~ (1 + z ‘)16( 1 + z2)16, which<br />

turns out to be 1304927002. If we restrict ourselves to events that depend<br />

on S only, we get 40 solutions S E’ A, where A = 0, { f2, ;b , t}, { f2, 5,9},<br />

(2,12 , PO, z, 5,9}, {2,4,6,8,10,12}, { f, ,7, z, 4, lo}, and the complements of<br />

these sets. (Here the notation ‘,:’ means either 2 or 12 but not both.)

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