Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 5J (Statistical Modelling)

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- Solution 5J

Solution

An exponential dispersion family has densities of the form

f (y; θ, ϕ) = a (y, ϕ) exp {\frac{y θ - b ( θ )}{ϕ}},

where

θ

is the canonical parameter,

ϕ > 0

is the dispersion parameter, and the support does not depend on

θ

. Its mean and variance are

E Y = b^{'} (θ), var (Y) = ϕ b^{''} (θ) .

For the given Gamma distribution, set

ϕ = \frac{1}{α}, θ = - \frac{λ}{α} < 0, b (θ) = - lo g (- θ) .

Then

a (y, ϕ) = \frac{y ^{1/ ϕ - 1}}{Γ ( 1/ ϕ ) ϕ ^{1/ ϕ}}

gives

a (y, ϕ) exp {\frac{y θ - b ( θ )}{ϕ}} = \frac{λ ^{α}}{Γ ( α )} y^{α - 1} e^{- λ y} .

Thus the gamma laws form the Gamma exponential dispersion family.

Directly integrating, or using the cumulant-generating function of a gamma distribution,

M_{Y} (t) = (\frac{λ}{λ - t})^{α}, t < λ,

so

K_{Y} (t) = lo g M_{Y} (t) = - α lo g (1 - \frac{t}{λ}) .

Differentiation at zero yields

E Y = K_{Y}^{'} (0) = \frac{α}{λ}, var (Y) = K_{Y}^{''} (0) = \frac{α}{λ ^{2}} .

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