Advanced High-School Mathematics

328 CHAPTER 6 Inferential Statistics

E(X) =

∑∞

n=1

nP(X=n) =

∑∞

n=1

np(1−p)n−^1 = p

∑∞

n=1

n(1−p)n−^1.

Note that

∑∞

n=1

n(1−p)n−^1 =

d

dx

(1 +x+x^2 +···)

∣∣

∣∣

∣x=1−p

=

d

dx

(

1

1 −x

)∣∣

∣∣

∣x=1−p

=

1

(1−x)^2

∣∣

∣∣

∣x=1−p

=

1

p^2

,

which implies that the mean of the geometric random variable X is
given by

E(X) = p

∑∞

n=1

n(p−1)n−^1 =

1

p

.

Notice that the smallerpbecomes, the longer the game is expected to
last.

Next, we turn to the variance ofX. By Equation 6.4 we have

Var(X) = E(X^2 )−

1

p^2

=

∑∞

n=1

n^2 P(X=n)−

1

p^2

= p

∑∞

n=1

n^2 (1−p)n−^1 −

1

p^2

.