Update gamma prior with new rate parameter instead of observations

Question

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.

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$(\alpha + k_1, \beta + 1)$.