PROBABILITY
an expression for the PGF ΞS(t)ofSN:
ΞS(t)=
∑∞
k=0
ξktk=
∑∞
k=0
tk
∑∞
n=0
hn×coefficient oftkin [ΦX(t)]n
=
∑∞
n=0
hn
∑∞
k=0
tk×coefficient oftkin [ΦX(t)]n
=
∑∞
n=0
hn[ΦX(t)]n
=χN(ΦX(t)). (30.84)
In words, the PGF of the sumSNis given by the compound functionχN(ΦX(t))
obtained by substituting ΦX(t)fortin the PGF for the number of termsNin
the sum. We illustrate this with the following example.
The probability distribution for the number of eggs in a clutch is Poisson distributed with
meanλ, and the probability that each egg will hatch isp(and is independent of the size of
the clutch). Use the results stated in (30.72) and (30.73) to show that the PGF (and hence
the probability distribution) for the number of chicks that hatch corresponds to a Poisson
distribution having meanλp.
The number of chicks that hatch is given by a sum of the form (30.83) in whichXi=1if
theith chick hatches andXi= 0 if it does not. As given by (30.72), ΦX(t) is thus (1−p)+pt.
The value ofNis given by a Poisson distribution with meanλ; thus, from (30.73), in the
terminology of our previous discussion,
χN(t)=e−λeλt.
We now substitute these forms into (30.84) to obtain
ΞS(t)=exp(−λ)exp[λΦX(t)]
=exp(−λ)exp{λ[(1−p)+pt]}
=exp(−λp)exp(λpt).
But this is exactly the PGF of a Poisson distribution with meanλp.
That this implies that the probability is Poisson distributed is intuitively obvious since,
in the expansion of the PGF as a power series int, every coefficient will be precisely
that implied by such a distribution. A solution of the same problem by direct calculation
appears in the answer to exercise 30.29.
30.7.2 Moment generating functions
As we saw in section 30.5 a probability function is often expressed in terms of
its moments. This leads naturally to the second type of generating function, a
moment generating function. For a random variableX, and a real numbert,the
moment generating function (MGF) is defined by
MX(t)=E
[
etX
]
=
{∑
ie
txif(x
i) for a discrete distribution,
∫
etxf(x)dx for a continuous distribution. (30.85)