I am analyzing daily data transaction data.

I am assuming that

  • The number of transactions in every day of length t has the Poisson distribution with mean λt
  • The number of transactions in evert collection of disjoint days are independent

Xi = {number of transactions on day i}
And thus, X~Poisson(λ) where λ is the mean number conversions per day.

I want to take a Bayesian approach - specifically, as I get more days of data, I want to update λ with the conjugate prior distribution to the Poisson, Gamma.

What are the parameters of this Gamma distribution and how do they relate to the Poisson process?


2 Answers 2


Your prior is $\lambda\sim\mathcal G(a, b)$, i.e. $$\pi(\lambda)\propto \lambda^{a-1}e^{-b\lambda}.$$

To get the posterior, multiply the prior by the likelihood:

$$\begin{eqnarray} \pi\left(\lambda|X_1,\ldots,X_n\right) &\propto& \lambda^{a-1}e^{-b\lambda} \prod_{i=1}^n e^{-\lambda}\frac{\lambda^{X_i}}{X_i!} \\ &\propto& \lambda^{a-1}e^{-b\lambda} e^{-n\lambda}\lambda^{\sum X_i}\\ & \propto&\lambda^{a+\sum X_i -1} e^{-\lambda(b+n)} \end{eqnarray}$$

and so the posterior is $\lambda|X_1,\ldots, X_n \sim \mathcal G\left(a+\sum X_i, b+n\right)$.

  • $\begingroup$ is there anyway, that $\mathcal G\left(a+\sum X_i, b+n\right)$ can be persented as this $\mathcal G\left(a+\sum X_i, b/(nb+1)\right)$, because I have been asked a question in an assignment about this, and I can only get the posterior you get. $\endgroup$
    – Bucephalus
    Sep 23, 2018 at 11:19
  • 3
    $\begingroup$ My answer uses the shape-rate parameterization for the gamma distribution. If you use the shape-scale parameterization, your formula is correct. $\endgroup$ Sep 23, 2018 at 17:10

The answer by @RobynRyder is great, +1. Just wanted to generalize it a bit and add some interpretation.

Imagine we observe $X_i$ events from the Poisson process in time $t_i$ for observation $i$. The equations in his answer get modified slightly:

$$\begin{eqnarray} \pi\left(\lambda|X_1,\ldots,X_n\right) &\propto& \lambda^{a-1}e^{-b\lambda} \prod_{i=1}^n e^{-\lambda t_i}\frac{(\lambda t_i)^{X_i}}{X_i!} \\ &\propto& \lambda^{a-1}e^{-b\lambda} e^{-\lambda \sum t_i}\lambda^{\sum X_i}\\ & \propto&\lambda^{a+\sum X_i -1} e^{-\lambda(b+\sum t_i)} \end{eqnarray}$$

and so the posterior is $\lambda|X_1,\ldots, X_n \sim \mathcal G\left(a+\sum X_i, b+\sum t_i\right)$.

This is equivalent to saying that we say $\sum X_i$ more events in a total observation period of $\sum t_i$. So, both the number of events, $a$ and total observation time, $b$ need to be updated.


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