Solved – Deriving the canonical link for a binomial distribution

I define an exponential dispersion family as any distribution whose PMF/PDF is
$$f(y mid boldsymboltheta) = expleft{phi[ytheta – b(theta)] + c(y, phi) right}text{, } y in Omega$$
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
phi &= 1 \
theta_i &= logleft(dfrac{p_i}{1-p_i} right) \
b(theta_i) &= nlogleft(dfrac{1}{1-p_i}right) \
c(phi, y_i) &= logbinom{n}{y_i}text{.}

After some work, I showed that, as a function of $theta_i$,
$$b(theta_i) = nlog(e^{theta_i} + 1)$$
(this is consistent with what I found at
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
$$g(mu_i) = theta_i = logleft( dfrac{mu_i}{n-mu_i}right)text{.}$$
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?

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 $p_i$.

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