Show that Y from a exponential family has $\phi = 1/k$ if $Y=Z/k$ where $Z \sim B(k, p)$

Question

Suppose Y from an exponential family

$$ f(y;\theta,\phi) = \exp\left\{\frac{y\theta - b(\theta)}{a(\phi)} + c(y, \phi)\right\} $$

I'm struggling to show that if $Y = Z / k$ and $Z \sim B(k, p)$ (binomial distribution) then $Y$ has a $\phi = 1 / k$. Is $Y$ even a binomial too in this case?

Answer 1

$Z$ is binomial. It has parameters $k$ (number of trails) and $p$ (probability of success). Since we know that $Z$ is a sample from a binomial distribution we can say

$$Pr(Z = r)={k\choose r}p^{r}(1-p)^{k-r} $$

Getting it in exponential form requires a trick. Consider

$$ p^{r}(1-p)^{k-r} = \exp\Bigg\{ \log \big(p^{r}(1-p)^{k-r} \big) + \log{k\choose r} \Bigg\}$$

I'll let you take it from here.