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

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?

$$Z$$ 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.

• Thank you! But if I develop that I can find the $\phi$ for $Z$. Do you know how do I link it with $Y = Z/k$ and find the exponential form for $Y$? Nov 1, 2019 at 23:23
• Oh, sorry. I misread your question. $Y$ is a univariate transformation of $Z$. You need to determine the PDF of $Y$ first. $Y$ will also have a binomial distribution, but with 'scaled' parameters. See this question stats.stackexchange.com/questions/351452/…. If you're wondering how they got that answer, you should look into (univariate) transformations of random variables. Nov 1, 2019 at 23:28
• Its alright. I'll look into it. Thanks a lot! Nov 1, 2019 at 23:32