From wikipedia:

In Bayesian inference, the conjugate prior for the rate parameter $λ$ of the Poisson distribution is the gamma distribution. Let

$$\lambda \sim \mathrm{Gamma}(\alpha, \beta)$$

denote that $λ$ is distributed according to the gamma probability density function $g$ parameterized in terms of a shape parameter $α$ and an inverse scale parameter $β$:

$$ g(\lambda \mid \alpha,\beta) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \; \lambda^{\alpha-1} \; e^{-\beta\,\lambda} \qquad \text{ for } \lambda>0 \,$$

Then, given the same sample of $n$ measured values $k_i$, and a prior of Gamma($α$, $β$), the posterior distribution is

$$\lambda \sim \mathrm{Gamma}\left(\alpha + \sum_{i=1}^n k_i, \beta + n\right). $$

The last equation shows how to update the Gamma prior when given $n$ observations of counts $k_i$ measured from $n$ experiments.

Given my prior belief that $\lambda \sim \mathrm{Gamma}(\alpha, \beta)$, if I run another experiment to get counts of $k_1$, then I know how to update my prior to give the posterior, using $n = 1$ in the equation from wikipedia.

However, I am now told that instead of running another experiment, the count from the next experiment is a random variate from a random variable with distribution $\mathrm{Pois}(\kappa)$. Knowing this information, I wish to produce the posterior.

  • $\begingroup$ is $\kappa$ non-random? Does it have anything to do with the random $\lambda$? Do you really mean "process?" Are you updating in this way after you have already updated once with the data? This doesn't seem clear to me. $\endgroup$
    – Taylor
    Commented Jul 22, 2016 at 5:47
  • $\begingroup$ Can you clarify? For the new experiment, do you not solimply have a point prior at $\kappa$ and the outcome of the experiment does not matter? $\endgroup$
    – Björn
    Commented Jul 22, 2016 at 5:50
  • $\begingroup$ Sorry, let me try and make the situation clearer by editing my original pots. $\endgroup$
    – Alex
    Commented Jul 22, 2016 at 6:07
  • $\begingroup$ I think the answer should be $\lambda \sim \mathrm{Gamma}\left(\alpha + \kappa, \beta + 1\right)$. $\endgroup$
    – Alex
    Commented Jul 22, 2016 at 6:10

1 Answer 1


I am assuming $\kappa$ here is a known non-random quantity. So here you have two data points, $k_1 \sim Pois(\lambda)$ and $k_2 \sim Pois(\kappa)$. $\kappa$ is known and $\lambda$ has a Gamma$(\alpha, \beta)$ prior. You need to find the posterior of $\lambda$.

Intuitively, the posterior of $\lambda$ should not depend on $\kappa$ since $k_2$ carries no information about $\lambda$, so $k_2$ will not impact the posterior of $\lambda$. If you do the math

\begin{align*} g(\lambda|\alpha, \beta, k_1, k_2) & \propto g(\lambda| \alpha, \beta) f(k_1 | \lambda)f(k_2 | \kappa)\\ & \propto g(\lambda| \alpha, \beta) f(k_1 | \lambda),\\ \end{align*}

And that would just give you the posterior Gamma$(\alpha + k_1, \beta + 1)$.

  • $\begingroup$ Thank you for your answer, I am afraid I do not understand how you go from the left to the right in the equation. It would be much appreciated if you could point me to somewhere where I can read and understand your notation. $\endgroup$
    – Alex
    Commented Jul 25, 2016 at 2:54
  • $\begingroup$ Also, I don't understand the following: "the posterior of $\lambda$ should not depend on $\kappa$ since $k_2$ carries no information about $\lambda$". If we knew that the counts in the next experiment are drawn from Pois($\kappa$) wouldn't that affect our belief in $\lambda$? $\endgroup$
    – Alex
    Commented Jul 25, 2016 at 2:57
  • $\begingroup$ Why does 𝑘2 carries no information about 𝜆? $\endgroup$
    – Yu Chen
    Commented May 21, 2019 at 21:20

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