Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 218 / 1 / a

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Solution

Expanding the exponent of the Inverse Gaussian distribution gives

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

Thus the exponential dispersion family representation

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

has

θ = - \frac{1}{2 μ ^{2}}, K (θ) = - - 2 θ, ϕ = \frac{1}{λ},

and

a (y, ϕ) = \frac{1}{2 π ϕ y ^{3}} exp (- \frac{1}{2 ϕ y}) .

The standard cumulant identities yield

E Y = K^{'} (θ) = μ, qq u a d Var (Y) = ϕ K^{''} (θ) = \frac{μ ^{3}}{λ} .

Hence the variance function is

V (μ) = μ^{3}

, and the canonical link function is

g (μ) = θ (μ) = - 1/ (2 μ^{2})

.

Solved by gpt-5.6-sol high.

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