Mathematics for Computer Science

“mcs” — 2017/3/3 — 11:21 — page 824 — #832
824
Chapter 19
Random Variables
his own hat back is 1=n. There are many different probability distributions of hat
permutations with this property, so we don’t know enough about the distribution of
G to calculate its expectation directly using equation (19.2) or (19.3). But linearity
of expectation lets us sidestep this issue.
We’ll use a standard, useful trick to apply linearity, namely, we’ll express G as
a sum of indicator variables. In particular, let G i be an indicator for the event that
the ith man gets his own hat. That is, G i D 1 if the ith man gets his own hat, and
G i D 0 otherwise. The number of men that get their own hat is then the sum of
these indicator random variables:
G D G 1 C G 2 C C G n : (19.9)
These indicator variables are not mutually independent. For example, if n 1 men
all get their own hats, then the last man is certain to receive his own hat. But again,
we don’t need to worry about this dependence, since linearity holds regardless.
Since G i is an indicator random variable, we know from Lemma 19.4.2 that
ExŒG i D PrŒG i D 1 D 1=n: (19.10)
By Linearity of Expectation and equation (19.9), this means that
ExŒG D ExŒG 1 C G 2 C C G n
D ExŒG 1 C ExŒG 2 C C ExŒG n
n
‚ …„ ƒ
1
D
n C 1 n C C 1 n
D 1:
So even though we don’t know much about how hats are scrambled, we’ve figured
out that on average, just one man gets his own hat back, regardless of the number
of men with hats!
More generally, Linearity of Expectation provides a very good method for computing
the expected number of events that will happen.
Theorem 19.5.4. Given any collection of events A 1 ; A 2 ; : : : ; A n , the expected
number of events that will occur is
nX
PrŒA i :
iD1
For example, A i could be the event that the ith man gets the right hat back. But
in general, it could be any subset of the sample space, and we are asking for the
expected number of events that will contain a random sample point.