Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 563
values of X in a sequence of independent trials will be a median or mode of
the random variable X.
8.53 We can disprove the statement, even in the special case that each
variableisOor1. LetpO=Pr(X=Y=Z=O),pl=Pr(X=Y=Z=O),...,
p7=Pr(X=Y=Z=O),whereX=l-X. Thenpo+pl+...+p7=1,and
the variables are independent in pairs if and only if we have
(p4+p5+p6+p7)(pL+p3+p6+p7) = p6+p7,
(p4 + p5 + p 6 -+ p7)(pl + p 3 + p5 + p7) = p5 + p7,
(p2 + p3 + p6 -t p7)tpl + p3 + p5 + p7) = p3 + p7.
But WX+Y=Z=O) # Pr(X+Y=O)Pr(Z=O) w p0 # (pO +p,)(pc +
pr + p4 + p61. One solution is
PO = P3 = Ps = P6 = l/4; p1 = p2 = p4 = p7 = 0.
This is equivalent to flipping two fair coins and letting X = (the first coin
is heads), Y = (the second coin is heads), Z = (the coins differ). Another
example, with all probabilities nonzero, is
PO = 4/64, PI = ~2 = ~4 = 5/64,
p3 = p5 = p6 = 10/64, p7 = 15/64.
For this reason we say that n variables XI, , X, are independent if
Pr(X1 =x1 and...and Xn=x,) = Pr(X, =xl)...Pr(X, = x,);
pairwise independence isn’t enough to guarantee this.
8.54 (See exercise 27 for notation.) We have
E(t:) = nll4 +n(n-1)~:;
E(LzLfI = np4 +2n(n-l)u3pl +n(n-1)~: +n(n-l)(n-2)p2&;
E(xy) = np4 +4n(n-l)p~u1 +3n(n-1)~:
it follows that V(\iX) = K4/n + 2K:/(n ~ 1).
+ 6n(n-l)(n-2)u2p: + n(n-l)(n-2)(np3)pT ;
8.55 There are A � = & .52! permutations with X = Y, and B = g .52!
permutations with X # Y. After the stated procedure, each permutation
with X = Y occurs with probability A/(( 1 - gp)A), because we return
to step Sl with probability $p. Similarly, each permutation with X # Y
occurs with probability g(l - p)/((l ~ sp)B). Choosing p = i makes
Pr (X = x and Y = 9) = & for all x and y (We could therefore make two flips
of a fair coin and go back to Sl if both come up heads.)