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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 5J / Solution

Expand the exponent in the Inverse Gaussian distribution with unit shape parameter:

- \frac{( x - μ ) ^{2}}{2 μ ^{2} x} = - \frac{x}{2 μ ^{2}} + \frac{1}{μ} - \frac{1}{2 x} .

Consequently its probability density function can be written as

f (x; μ) = h (x) exp {η x - A (η)},

where

h (x) = \frac{1}{2 π x ^{3}} exp (- \frac{1}{2 x}), η = - \frac{1}{2 μ ^{2}} < 0, A (η) = - - 2 η .

Indeed,

- A (η) = - 2 η = 1/ μ

. This is the canonical form of a one-parameter exponential family, and its natural parameter is therefore

η = - \frac{1}{2 μ ^{2}} .

Here the natural statistic is

T (X) = X

. The exponential-family derivative identities give

E_{η} X = A^{'} (η) = (- 2 η)^{- 1/2} = μ

and

var_{η} X = A^{''} (η) = (- 2 η)^{- 3/2} = μ^{3} .

Thus

E X = μ, var X = μ^{3} .

Solved by gpt-5.6-sol high.

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