I am trying to incorporate a prior into a model I am working on. From available data, I have found that the likelihood follows a Poisson distribution with $\lambda = 1.5$.

I have then used R to generate random Poisson values with $\lambda = 1.5$.

However, I suspect that the $\lambda$ value is more likely to be $\lambda=2$. How would I turn this into a prior I could use and what would the posterior be?

Thanks in advance, Neil


1 Answer 1


Common choice of bayesian prior for the Poisson distribution is the Gamma distribution. Which ultimately has a Gamma distribution for the posterior.

If you have a large set of data which you believe comes from a Poisson distribution. The assumption is that $Y_i \text{ iid}\sim \text{Poisson}(\lambda)$. $$f(y|\lambda) = \frac{\lambda^ye^{-\lambda}}{y!}$$

The idea behind the Bayesian approach of estimation / regression is that there is uncertainty in the parameter $\lambda$ for each $Y_i$. And that for each observation, there may be a natural variation of $\lambda$s such that they have their own distribution $g(\lambda| \nu)$ with hyper-parameters $\nu$. As mentioned, a convenient choice of prior for the Poisson distribution is the gamma distribution because with hyper-parameters $\lambda \sim \Gamma(v, r)$: $$f(\lambda|y) \propto f(y|\lambda)\cdot g(\lambda|\nu)$$ $$\propto \big(\frac{\lambda^ye^{-\lambda}}{y!}\big)\cdot g(\lambda|\nu)$$ $$\propto \big(\frac{\lambda^ye^{-\lambda}}{y!}\big)\cdot\big(\frac{v^r}{\Gamma(r)}\lambda^{r-1}e^{-v\lambda}\big)$$ $$\propto \frac{v^r}{\Gamma(r)}\lambda^{r+y-1}e^{-(v+1)\lambda}\sim\Gamma(r' = r + y, v' = v + 1)$$

Now you use the fact that $E[\lambda] = \frac{r'}{v'},\ \text{Var}[\lambda] = \frac{r'}{v'^2}$ you can solve for the regressed maximum likelihood estimate of $\lambda$, where $r$, and $v$ come from a historical set of data to inform the prior, and $y$ is the next observation or $\sum y_i$ the 'next' set of information.

  • $\begingroup$ Yes. I've been trying a Gamma Distribution but couldn't figure out how to choose the shape and rate parameters that would resemble my prior $\endgroup$
    – Neil
    Apr 6, 2015 at 22:58
  • $\begingroup$ That's a common difficulty--with various solutions. Do you have a mean in mind for your prior, and maybe an interval that should contain most of the probability? Or do you want a noninformative prior? Tell me what info you have and I'll try to show you how to pick a prior. (Put @ followed by my name in your comment when you reply and I'll be notified to look at it.) $\endgroup$
    – BruceET
    Apr 7, 2015 at 1:38

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.