Advanced High-School Mathematics

SECTION 6.1 Discrete Random Variables 333

In order to appreciate the method em-
ployed, let’s again consider the geo-
metric distribution. That is, assume
that the probability of flipping a head
(H) is p, and that X measures the
number of trials before observing the
first head. We may writeX=B+(1−
B)(1 +Y), where B is the Bernoulli
random variable withP(B = 1) = p
and P(B = 0) = 1−p, and where
Y andX have the same distribution.
(See the tree diagram to the right.)

• HH

HHH

T

(B= 0 and we start

the experiment over

again with one trial

already having been

performed.)





H (is over)B= 1 and the game

1 −p

p

It follows, therefore that

E(X) = E(B+ (1−B)(1 +Y))

= E(B) +E(1−B)E(1 +Y) (sinceBandY are independent)

= p+ (1−p)(1 +E(X))

= 1 + (1−p)E(X).

Solving forE(X) quickly yields the correct result, viz.,E(X) = 1/p.

The above method quickly generalizes to sequences. Let’s consider
tossing a coin withP(heads) =p, stopping after two consecutive heads
are obtained. LettingX be the number of trials,Y have the same dis-
tribution asXand lettingB 1 andB 2 be independent Bernoulli random
variables, we may set

X=B 1 +B 2 +B 1 (1−B 2 )(2 +Y) + (1−B 1 )(1 +Y). (6.6)