= Solution
A one-parameter <exponential family> with natural statistic $Y$ has density or probability mass function
$$
f_\theta(y)=h(y)\exp\{\theta y-K(\theta)\},
$$
where $\theta$ is the <natural parameter of an exponential family> and $K$ is the <cumulant function of an exponential family>. Its <mean parameter of an exponential family>[mean parameter] is
$$
\mu(\theta)=K'(\theta).
$$
The <exponential-family deviance> from $\theta_1$ to $\theta_2$ is twice the <Kullback-Leibler divergence>:
$$
D(\theta_1,\theta_2)
=2\mathbb E_{\theta_1}
\left[\log\frac{f_{\theta_1}(Y)}{f_{\theta_2}(Y)}\right].
$$
The carrier $h(Y)$ cancels from the likelihood ratio, so, with $\mu_1=\mathbb E_{\theta_1}Y$,
$$
\boxed{
D(\theta_1,\theta_2)
=2\{(\theta_1-\theta_2)\mu_1-K(\theta_1)+K(\theta_2)\}.}
$$
For the <Poisson distribution> with mean $\mu$,
$$
\theta=\log\mu,
\qquad
K(\theta)=e^\theta=\mu.
$$
Therefore
$$
\boxed{
D(\mu_1,\mu_2)
=2\left\{\mu_1\log\frac{\mu_1}{\mu_2}-\mu_1+\mu_2\right\}.}
$$
Writing $\mu_2=\mu_1+\delta$ and applying the <Taylor series> of $\log(1+\delta/\mu_1)$ gives
$$
D(\mu_1,\mu_1+\delta)
=\frac{\delta^2}{\mu_1}
+O(\delta^3),
$$
so the second-order approximation is
$$
\boxed{D(\mu_1,\mu_2)\simeq
\frac{(\mu_2-\mu_1)^2}{\mu_1}.}
$$
Back to article page
