= Solution
For $y=0,1,2,\ldots$ and $\lambda>0$, the <Poisson distribution> has mass
$$
\Pr(Y=y)=\frac{e^{-\lambda}\lambda^y}{y!}
=\exp\{y\log\lambda-\lambda-\log(y!)\}.
$$
Match this to the <exponential dispersion family> with
$$
\boxed{\theta=\log\lambda,\quad b(\theta)=e^\theta,\quad
\phi=1,\quad c(y,1)=-\log(y!).}
$$
Then $\mu=b'(\theta)=e^\theta=\lambda$ and $V(\mu)=b''(\theta)=\mu$. The <canonical link function> expresses the <natural parameter> in terms of the mean, so the <Poisson canonical link> is \b[$g(\mu)=\log\mu$]. Its conditional <variance> equals its conditional mean.
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