Concrete mathematics : a foundation for computer science

Oh, now 1 unde&and
what
mathematicians
mean when they
say something is
“‘obvious,” “clear,”
or “trivial.”
8.5 HASHING 303
If X and Y are independent, the random variable Xly will be essentially
the same as X, regardless of the value of y, because Pr(X = x 1 Y = y ) is equal
to Pr(X =x) by (8.5); that’s what independence means. But if X and Y are
dependent, the random variables X/y and Xly’ need not resemble each other
inanywaywheny#y’.
If X takes only nonnegative integer values, we can decompose its pgf into
a sum of conditional pgf’s with respect to any other random variable Y:
Gx(z) = x WY=y)Gx,(z).
YEYIf~l
(8.91)
This holds because the coefficient of zx on the left side is Pr(X =x), for all
x E X(n), and on the right it is
x Pr(Y=y)Pr(x=xIY=y) = t Pr(X=x and Y=y)
YEytni YEYin)
= Pr(X=x).
For example, if X is the product of the spots on two fair dice and if Y is the
sum of the spots, the pgf for X16 is
Gx,6(z) = +z5 + $z8 + ;z9
because the conditional probabilities for Y = 6 consist of five equally probable
events { � m, � n, � m, � n, � m}. Equation (8.91) in this case
reduces to
Gx(z) = $x 2(z) + $x,3(z) + &Gx z,(z) + $x,5(z)
j$x,dz) + $&T(Z) + j$x,a(z) + &Gx~9(4
$%Io(z) + j$x,,, (~1 + &12(z),
a formula that is obvious once you understand it. (End of digression.)
In the case of hashing, (8.91) tells us how to write down the pgf for probes
in a successful search, if we let X = P and Y = K. For any fixed k between 1
and n, the random variable PI k is defined as a sum of independent random
variables X1 + . . . + Xk; this is (8.88). so it has the pgf
Gp,k(Z) = (m-;+Z)k-‘Z.