1. ## canonical parameter and canonical link function

Given the exponential distribution: $\displaystyle f(x)= \theta^{-1}e^{-x/\theta$.

Converts it into exponential family form: $\displaystyle exp[\frac{(1/\theta)x-log(1/\theta)}{-1}]$

Isn't the canonical parameter and link function the same?

Which is $\displaystyle \frac{1}{\theta}$

Are there any differences between canonical parameter and canonical link function?

2. No, this isn't right. You need to move the $\displaystyle -1$ in the denominator upstairs (if you are thinking of this as an exponential dispersion family, you can't have the dispersion parameter be negative). The canonical parameter for this should be $\displaystyle \lambda = \frac {-1} \theta$. The canonical link turns out to be the same - except as a function of the mean instead of the canonical parameter. This isn't true in general, of course (unless your base parameterization is always in terms of the mean, I suppose).

The exponential dispersion form of the density should be: $\displaystyle \exp\left\{(-1 / \theta) x - \log \theta)\right\}$.

It's also worth mentioning that people apparently give the canonical link of the exponential as $\displaystyle 1 / \mu$ instead of $\displaystyle -1 / \mu$ because these two are equivalent in terms of actually fitting the model and the negative sign looks ugly. But the canonical parameter DOES have the negative in it.

3. Thanks for all ur reply to all my queries (Include the one on rats too).

I define my exponential family to be in this form:

$\displaystyle exp[\frac{Y\theta-a(\theta)}{\phi}+b(Y,\phi)]$ Will it make a difference to the conclusion?

To further verify it, it turns out that $\displaystyle b'(\theta)$ and $\displaystyle b''(\theta)\phi$ do fit well as $\displaystyle \theta$ and $\displaystyle \theta^2$

So do u mean $\displaystyle \theta$ is the parameter and $\displaystyle a(\theta)$ is the link function??

Sorry I very confused, it will help a lot if u can state the parameter and link function accordingly. Thanks!!

4. You can't have $\displaystyle \phi = -1$. In some sense $\displaystyle \phi$ is a variance parameter so you don't let that be negative. Every source I've ever seen takes $\displaystyle \phi > 0$. Usually this is called an exponential dispersion family, as opposed to an exponential family, but they are closely related.

You should have $\displaystyle \lambda = \frac {-1} \theta$ as the canonical parameter, and $\displaystyle g(\theta) = \frac {-1} \theta$ the canonical link.