# Deriving the canonical link for a binomial distribution

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?

$$\mu_i = p_i n$$ 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 $$p_i$$.
• So if I'm understanding what you're saying correctly, it's this: if I wanted to write the canonical link in terms of $p_i$, it would be the logit, but if I wanted to write it in terms of $\mu_i$, it would be what I have above (and in both cases, they are identical for when $n = 1$ , i.e., the Bernoulli)? – Clarinetist Oct 3 '18 at 18:11
• Yes, exactly. If you find any sources that mix the $\mu$ and the $1$, they are probably assuming $n=1$. – eric_kernfeld Oct 3 '18 at 21:33