Page 162 - Elements of Distribution Theory

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148 Parametric Families of Distributions

and let denote the range of λ. Then

E[g(X); θ] = E{E[g(x)|λ; θ]; θ}= g(x)p(x|λ; θ) dx dF λ (λ; θ).
R d
By Fubini’s Theorem,

g(x)p(x|λ; θ) dx dF λ (λ; θ) = g(x) p(x|λ; θ) dF λ (λ; θ) dx
R d R d
so that X has an absolutely continuous distribution with density p X ,asgiven above.
Now suppose that the conditional distribution of X given λ is discrete with frequency
function p(x|λ; θ) and support X. Then

E{p(x|λ; θ); θ}, if x ∈ X
Pr(X = x; θ) = ,
0 if x /∈ X
proving the result.

Example 5.17 (Negative binomial distribution). Let X denote a random variable with
a Poisson distribution with mean λ and suppose that λ has a gamma distribution with
parameters α and β. Then the marginal distribution of X is discrete with frequency function
β α ∞ x α−1
p(x; α, β) = λ λ exp{−λ} dλ
(α)x! 0
β α (x + α)
=
(α)x! (β + 1) x+α
x + α − 1 β
α
= , x = 0, 1,... ;
α − 1 (β + 1) x+α
here α> 0 and β> 0. This distribution is called the negative binomial distribution with
parameters α and β.
This distribution has moment-generating function

α
M(t) = β (1 + β − exp{t}) −α , t < log(1 + β).
It is straightforward to show that E(X; θ) = α/β and
α β + 1
Var(X; θ) = .
β β
The geometric distribution is a special case of the negative binomial distribution corre-
sponding to α = 1.
The Poisson distribution may be obtained as a limiting case of the negative binomial
distribution. Suppose X has a negative binomial distribution with parameters α and β where
α = λβ, λ> 0. Then
λβ
x + λβ − 1 β
Pr(X = x; λ, β) = , x = 0, 1,...
λβ − 1 (β + 1) x+λβ
and
x
λ exp{−λ}
lim Pr(X = x; λ, β) = , x = 0, 1,....
x!
β→∞