Expanding the exponent of the <Inverse Gaussian distribution> gives
$$
-\frac{\lambda(y-\mu)^2}{2\mu^2y}
=\lambda\left(-\frac{y}{2\mu^2}+\frac1\mu-\frac1{2y}\right).
$$
Thus the <exponential dispersion family> representation
$$
f(y;\theta,\phi)=a(y,\phi)
\exp\left\{\frac{y\theta-K(\theta)}\phi\right\}
$$
has
$$
\theta=-\frac1{2\mu^2},\qquad
K(\theta)=-\sqrt{-2\theta},\qquad
\phi=\frac1\lambda,
$$
and
$$
a(y,\phi)=\frac1{\sqrt{2\pi\phi y^3}}
\exp\left(-\frac1{2\phi y}\right).
$$
The standard cumulant identities yield
$$
\mathbb EY=K'(\theta)=\mu,qquad
\operatorname{Var}(Y)=\phi K''(\theta)=\frac{\mu^3}{\lambda}.
$$
Hence the <variance function> is $V(\mu)=\mu^3$, and the <canonical link function> is $g(\mu)=\theta(\mu)=-1/(2\mu^2)$.

Solved by gpt-5.6-sol high.

Solution

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

{title2=$V(\mu)$}

The variance function writes an exponential dispersion family's variance as $\operatorname{Var}(Y)=\phi V(\mu)/w$.


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