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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 1

An <exponential dispersion family> has densities of the form
$$
f(y;\theta,\phi)
=a(y,\phi)\exp\left\{\frac{y\theta-b(\theta)}{\phi}\right\},
$$
where $\theta$ is the <natural parameter of an exponential family>[canonical parameter], $\phi>0$ is the <dispersion parameter>, and the support does not depend on $\theta$. Its mean and variance are
$$
\mathbb E Y=b'(\theta),
\qquad
\operatorname{var}(Y)=\phi b''(\theta).
$$

For the given <Gamma distribution>, set
$$
\phi=\frac1\alpha,
\qquad
\theta=-\frac\lambda\alpha<0,
\qquad
b(\theta)=-\log(-\theta).
$$
Then
$$
a(y,\phi)
=\frac{y^{1/\phi-1}}{\Gamma(1/\phi)\phi^{1/\phi}}
$$
gives
$$
a(y,\phi)
\exp\left\{\frac{y\theta-b(\theta)}\phi\right\}
=\frac{\lambda^\alpha}{\Gamma(\alpha)}y^{\alpha-1}e^{-\lambda y}.
$$
Thus the gamma laws form the <Gamma exponential dispersion family>.

Directly integrating, or using the <cumulant-generating function of a gamma distribution>,
$$
M_Y(t)=\left(\frac\lambda{\lambda-t}\right)^\alpha,
\qquad t<\lambda,
$$
so
$$
K_Y(t)=\log M_Y(t)
=-\alpha\log\left(1-\frac t\lambda\right).
$$
Differentiation at zero yields
$$
\boxed{\mathbb E Y=K_Y'(0)=\frac\alpha\lambda},
\qquad
\boxed{\operatorname{var}(Y)=K_Y''(0)=\frac\alpha{\lambda^2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 5J / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 5J / Solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 5J / Solution

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