I define an exponential dispersion family as any distribution whose PMF/PDF is
$$f(y \mid \boldsymbol\theta) = \exp\left\{\phi[y\theta – b(\theta)] + c(y, \phi) \right\}\text{, } y \in \Omega$$
where $\Omega$ is in the support of a random variable $Y$ in the family.
Suppose $Y_1, \dots, Y_m$ are independent and binomially distributed ($n$ trials, success probability $p_i$). I've already shown that the Binomial distribution satisfies the above, with
$$\begin{align}
\phi &= 1 \\
\theta_i &= \log\left(\dfrac{p_i}{1-p_i} \right) \\
b(\theta_i) &= n\log\left(\dfrac{1}{1-p_i}\right) \\
c(\phi, y_i) &= \log\binom{n}{y_i}\text{.}
\end{align}$$
After some work, I showed that, as a function of $\theta_i$,
$$b(\theta_i) = n\log(e^{\theta_i} + 1)$$
(this is consistent with what I found at http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/tutorials/xlghtmlnode38.html)
and I understand that
$$\mu_i = b^{\prime}(\theta_i) = n \cdot \dfrac{e^{\theta_i}}{e^{\theta_i}+1}\text{.}$$
I also understand that what we need to do is solve for $\theta_i$ in the above, and the canonical link function would be $g(\mu_i) = \theta_i$ according to above. But one thing bothers me: when I run the above through WolframAlpha, I obtain
$$g(\mu_i) = \theta_i = \log\left( \dfrac{\mu_i}{n-\mu_i}\right)\text{.}$$
Every source I've seen says that the $n$ above should be a $1$ for the binomial canonical link function. Did I do something wrong?
Best Answer
You're almost right, and it's such an easy fix:
$$\mu_i = p_i n$$ so $$\log(\frac{\mu_i}{n - \mu_i}) = \log(\frac{np_i}{n - np_i}) = ...$$
So, the $n$ can be a 1, as long as you swap out $\mu_i$ for $p_i$.