4
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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 $p_i$.

$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – Clarinetist Oct 3 '18 at 18:11
  • $\begingroup$ Yes, exactly. If you find any sources that mix the $\mu$ and the $1$, they are probably assuming $n=1$. $\endgroup$ – eric_kernfeld Oct 3 '18 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.