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In my lecture notes I have that the distribution of a random variable $Y$ is said to be in the exponential family if it can be written as $f(y;\theta)=exp(a(y)b(\theta)+c(\theta)+d(y))$, where $a,b,c$ and $d$ are fixed functions. If we have (in the exponential family form) that $a(y)=y$, then the distribution is said to be in canonical form and then $b(\theta)$ is the called the natural parameter and $b$ is called the natural link function. Then we have shown that for a random variable $Y$ in exponential family form we have $\mathbb{E}(a(Y))=\frac{-c'(\theta)}{b'(\theta)}$ and $var(a(Y))=\frac{b''(\theta)c'(\theta)-c''(\theta)b'(\theta)}{[b'(\theta)]^{3}}$.

I am now working on the following problem: the gamma pdf of r.v. $Y$ is given by

$\begin{equation} f(y) = (s^{a}\Gamma (a))^{-1}y^{a-1}e^{y/s} \end{equation}$,

where $y \geq 0$, $s$ is the scale parameter, $a$ the shape parameter. The first question asks me to reparameterise this pdf by setting $a=1/\phi$ and $s=\mu\phi$, and hence show that it is a member of the exponential family. So after introducing $a=1/\phi$ and $s=\mu\phi$ into the pdf and rearranging I get

$ f(y) = exp((\frac{1}{\phi}-1)log(y)-\frac{y}{\mu\phi}-\frac{1}{\phi}log(\mu\phi)-log(\Gamma(1/\phi))=exp(a(y)b(\theta)+c(\theta)+d(y))$,

where $a(y)=y$,$b(\mu)=-\frac{1}{\mu\phi}$,$c(\mu)=-\frac{1}{\phi}log(\mu\phi)$ and $d(y)=(\frac{1}{\phi}-1)log(y) - log(\Gamma(\frac{1}{\phi}))$, where we treat the (dispersion) parameter $\phi$ as a nuisance parameter. Is that correct?

The next question says: deduce that the canonical link for the gamma is $\theta=\frac{1}{\mu}=\eta=\textbf{X}\beta$. So I'm thinking since $f$ is in a canonical form, the canonical parameter is $b(\mu)=-\frac{1}{\mu\phi}$ and so ignoring all the constants of proportionality we have that the canonical link, in its simplest form, is $\frac{1}{\mu}$, as required. Does that make sense? I still don't have a good enough understanding of link/canonical link functions, I'm afraid.

Then the next question asks me to deduce further that the variance function is $b''(\theta)=-1/\theta^{2}=-\mu^{2}$. I don't really know how to do this. Why is this the variance function? What is $\theta$ here? A canonical link? That clearly doesn't make sense. Or is just the dummy variable for our parameter of interest ( $\mu$ in our case )? I've tried to differentiate $b(\mu)$ w.r.t. to $\mu$ but it doesn't work. I'd really appreciate some help.

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