I have a variable that is Poisson distributed. Let's say I have a number of boxes each with a number of balls inside according to a Poisson distribution, with $\lambda=0.4$, (the average number of balls per box is 0.4):

$X\sim Poiss(\lambda=0.4)$

A random sample of 20 boxes would give us these values:

X=[0, 2, 1, 0, 0, 2, 1, 0, 2, 1, 1, 0, 0, 2, 0, 0, 1, 1, 0, 0]

An average of these values as an estimate of $\lambda$ gives us $0.7$, which is off due to the low sample count.

However, I can't actually see the number of balls in each box, just whether the box is empty or not:

X_1=[0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0]

I can estimate $\lambda$ from this via $\lambda=-ln(Empty Box\%)$ since the probability of a box having $k$ balls is:

$p(k) = \displaystyle\frac{\lambda^ke^{-\lambda}}{k!}$

and therefore:

$Empty Box\%=p(0) = \displaystyle\frac{\lambda^0e^{-\lambda}}{0!}=e^{-\lambda}$

In this case I get an estimate of $\lambda$ as $0.69$, again because of the low sample size for this demonstration. I'm aware that this approach works for sufficiently large sample size, and appropriate range of $\lambda$s.

In my case, I have several thousand boxes, and I am much closer to the true value when running simulations. But I am wondering if there is any Bayesian inference I can do in determining $\lambda$ for an unknown population. If I had the original counts per box instead of the binarized states, I'm aware that I could create a Gamma conjugate prior and perform Bayesian inference with my data to get a credible range of $\lambda$ predictions. Is there any way for me to do this with the data I have?


1 Answer 1


If $X \sim \mathcal{Pois}(\lambda)$ then $X^* = \mathcal{1}\{X>0\}$ have a Bernoulli distribution with parameter $p=1-e^{-\lambda}$. If you have observations for $n$ boxes then $\sum_i^n X_i^* \sim \mathcal{Bin}(n,p)$ and you can base inference on that.

The conjugate prior for the Binomial is the Beta distribution. An example is developed here.

  • 1
    $\begingroup$ Thanks @kjetil, in this case would it be $p=1 - e^{-\lambda}$? Is $X^* = \mathcal{1}\{X>0\}$ math notation for $X^*$ is 1 for all values greater than $0$, and $0$ elsewhere? $\endgroup$
    – Hector
    Commented May 19, 2019 at 19:10
  • $\begingroup$ Yes, and yes. I will edit. The $\mathcal{1}$ means the indicator function. $\endgroup$ Commented May 19, 2019 at 19:16
  • $\begingroup$ Great. I now have the following model: $\lambda \sim \mathcal{Gam}(\alpha, \beta)$; $p = 1 - e^{-\lambda}$; $\sum_i^n X_i^* \sim \mathcal{Bin}(n,p)$; In my Gamma prior for $\lambda$, I'm using $\alpha = \lambda_{ideal} \cdot 10000$, $\beta = 10000$. Where I would expect to measure $\lambda_{ideal}$ after observing $10000$ boxes. A less confident prior would use a smaller number. $\endgroup$
    – Hector
    Commented May 19, 2019 at 20:09

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