Mathematical Methods for Physics and Engineering : A Comprehensive Guide

30.7 GENERATING FUNCTIONS

and differentiating once more we obtain

d^2 ΦX(t)

dt^2

=

∑∞

n=0

n(n−1)fntn−^2 ⇒ Φ′′X(1) =

∑∞

n=0

n(n−1)fn=E[X(X−1)].

(30.75)

Equation (30.74) shows that Φ′X(1) gives the mean ofX. Using both (30.75) and

(30.51) allows us to write

Φ′′X(1) + Φ′X(1)−

[

Φ′X(1)

] 2

=E[X(X−1)] +E[X]−(E[X])^2

=E

[

X^2

]

−E[X]+E[X]−(E[X])^2

=E

[

X^2

]

−(E[X])^2

=V[X], (30.76)

and so express the variance ofXin terms of the derivatives of its probability

generating function.

A random variableXis given by the number of trials needed to obtain a first success

when the chance of success at each trial is constant and equal top. Find the probability

generating function forXand use it to determine the mean and variance ofX.

Clearly, at least one trial is needed, and sof 0 =0.Ifn(≥1) trials are needed for the first
success, the firstn−1 trials must have resulted in failure. Thus

Pr(X=n)=qn−^1 p, n≥ 1 , (30.77)

whereq=1−pis the probability of failure in each individual trial.
The corresponding probability generating function is thus

ΦX(t)=

∑∞

n=0

fntn=

∑∞

n=1

(qn−^1 p)tn

=

p

q

∑∞

n=1

(qt)n=

p

q

×

qt

1 −qt

=

pt

1 −qt

, (30.78)

where we have used the result for the sumof a geometric series, given in chapter 4, to
obtain a closed-form expression for ΦX(t). Again, as must be the case, ΦX(1) = 1.
To find the mean and variance ofXwe need to evaluate Φ′X(1) and Φ′′X(1). Differentiating
(30.78) gives

Φ′X(t)=

p

(1−qt)^2

⇒ Φ′X(1) =

p

p^2

=

1

p

,

Φ′′X(t)=

2 pq

(1−qt)^3

⇒ Φ′′X(1) =

2 pq

p^3

=

2 q

p^2

.

Thus, using (30.74) and (30.76),

E[X]=Φ′X(1) =

1

p

,

V[X]=Φ′′X(1) + Φ′X(1)−[Φ′X(1)]^2

=

2 q

p^2

+

1

p

−

1

p^2

=

q

p^2

.

A distribution with probabilities of the general form (30.77) is known as ageometric
distributionand is discussed in subsection 30.8.2. This form of distribution is common in
‘waiting time’ problems (subsection 30.9.3).