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?


1 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.

  • $\begingroup$ 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 $Y$? $\endgroup$ Nov 1, 2019 at 23:23
  • $\begingroup$ 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. $\endgroup$
    – ralph
    Nov 1, 2019 at 23:28
  • $\begingroup$ Its alright. I'll look into it. Thanks a lot! $\endgroup$ Nov 1, 2019 at 23:32

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.