# Convolution of Poisson with Binomial distribution?

Say I have a laser that emits pulses of light containing a random number of photons, which follow a Poisson distribution. So it has a mean number of photons per pulse.

These pulses go through a filter, which results in each photon having a probability of being either absorbed or transmitted.

I am trying to figure out what the distribution of the number of photons behind the filter will look like. How can this be described mathematically?

I understand that the resulting distribution should have something to do with a certain combination of the Poisson and a binomial distribution. Maybe some sort of convolution? Other than that I am pretty clueless, since I only have basic knowledge of statistics.

• "These pulses go through a filter, which results in each photon having a probability of being either absorbed or transmitted." -- what you're describing is not a convolution (as your title suggests). You appear to be describing a thinned Poisson process, which will just be Poisson with rate $\lambda p$. Mar 16 at 22:06

Let's start by looking at a single pulse and figure out the distribution of the number of photons in that pulse that get through the filter. To do this, let $$N$$ denote the initial number of photons in the pulse and let $$X$$ denote the number of photons that make it through the filter. Then you have the model:

\begin{align} N &\sim \text{Pois}(\lambda), \\[6pt] X|N &\sim \text{Bin}(N,\theta). \\[6pt] \end{align}

The marginal distribution of $$X$$ is obtained using the law of total probability, to wit:

\begin{align} p_X(x) \equiv \mathbb{P}(X=x) &= \sum_{n=0}^\infty \mathbb{P}(X=x|N=n) \cdot \mathbb{P}(N=n) \\[6pt] &= \sum_{n=0}^\infty \text{Bin}(x|n,\theta) \cdot \text{Pois}(n|\lambda) \\[6pt] &= \sum_{n=x}^\infty \frac{n!}{x! (n-x)!} \theta^x (1-\theta)^{n-x} \cdot \frac{\lambda^n}{n!} e^{-\lambda} \\[6pt] &= \frac{(\theta \lambda)^x}{x!} e^{-\theta \lambda} \sum_{n=x}^\infty \frac{((1-\theta)\lambda)^{n-x}}{(n-x)!} e^{-(1-\theta)\lambda} \\[6pt] &= \frac{(\theta \lambda)^x}{x!} e^{-\theta \lambda} \sum_{r=0}^\infty \frac{((1-\theta)\lambda)^r}{r!} e^{-(1-\theta)\lambda} \\[6pt] &= \text{Pois}(x| \theta \lambda) \sum_{r=0}^\infty \text{Pois}(r| (1-\theta)\lambda) \\[12pt] &= \text{Pois}(x| \theta \lambda). \\[6pt] \end{align}

This gives us the marginal distribution $$X \sim \text{Pois}(\theta \lambda)$$ for the number of photons that make it through the filter in a single pulse. This is called "thinning" the Poisson variable/process --- it leads to another Poisson variable/process but with the mean parameter reduced proportionately to the thinning. The result shown here can also be proved using the generating functions for the distribution; see e.g., here.

Now suppose we have $$k$$ independent pulses of the same type (i.e., with the same parameters) and let $$X_1,...,X_k \sim \text{Pois}(\theta \lambda)$$ denote the number of photons that go through the filter from each of these pulses. Then the total number of photons that make it through the filter is:

$$S_k = X_1 + \cdots + X_k.$$

The marginal distribution of $$S_k$$ is a $$k$$-fold convolution of the $$\text{Pois}(\theta \lambda)$$ distribution, which is:

$$S_k \sim \text{Pois}(k \theta \lambda).$$

This is the distribution of the number of photons that make it through the filter from $$k$$ pulses with mean-photons $$\lambda$$ and filter penetration probability $$\theta$$.

• I am not sure this models the situation I describe. Each puls contains a certain number of photons which is poisson distributed. One pulse hits the filter and each photon in this puls has a probability of getting absorbed or transmitted. Then the next puls comes in with a different number of photons and again, some might get absorbed others might not. Mar 16 at 21:47
• Okay, thanks for clarifying. I think I understand the problem now. I have updated the answer to reflect this information. Please let me know if this now answers your problem.
– Ben
Mar 17 at 0:17
• Splendid answer thank you! I almost get it. One last issue: Could you briefly explain how these sums come about or more specifically the start and end? Why from 0 to infinity and then suddenly from x to infinity? I am sure it is basic but I can't quite make sense of it. In my mind summation/integration is linked with the CDF but we are trying to find the PDF? I am quite confused as you might see. Mar 17 at 8:48
• The transition to the lower bound $x$ on the sum comes about because $\text{Bin}(x|n,\theta)=0$ for all $0 \leqslant n < x$ so those lower terms are effectively removed. For the other part, have a look at the link for the law of total probability.
– Ben
Mar 17 at 10:50

### Intuition 1

You can view it intuitively as following.

The Poisson distribution describes the number of counts for a Poisson process taking some time $$T$$ (like your pulse taking some time $$T$$ with photons being emitted randomly with a specific rate).

You could randomly designate each event/case/photon as $$X_i = 0$$ or $$X_i = 1$$ (in the image below this is shown as black/white circles on a line).

Effectively this is the same as generating two seperate independent Poisson processes (each taking times $$T_0, T_1$$ and with $$T_0+T_1 = T$$) and then mixing the points.

You can verify that this correct by the following thought: The sum of two Poisson variables is another Poisson variable, and each point will have $$T_i/T$$ probability of being $$i$$. So the number of cases $$X_i = 1$$ is another Poisson distributed variable.

### Intuition 2

We can also view it as two seperate 2D Poisson processes in a square, where we have split that square up in two strips. In one part of the square we observe cases A, in the other cases B. Then, when we consider the marginal distribution of that Poisson process, only the horizontal x-coordinate, then it is like the two seperate processes can be seen as a single Poisson process (and the color/case decided by a Bernoulli distribution). • I have a hard time understanding this. Can you relate your definitions to my specific case? I am really not used to the statistics lingo. What constitutes an event, a poisson process etc. in my example and how does the time come into play? Mar 16 at 21:52
• A poisson distributed variable relates to the number of counts in a Poisson process where events may occur at a certain rate for a certain amount of time (in your case the length of time that your pulse takes) or in a certain amount of space. Like a Geiger Muller apparatus counting ionised particles. With the filter you get the same type of distribution but just at a lower rate. Mar 16 at 21:58
• The same also works when you are compounding a binomial distribution with a binomial distribution, which gives just another binomial distribution. For that case there might be example as well. Mar 16 at 22:01

This sounds like a compound Poisson distribution. You have a Poisson distributed number $$N$$ of binomial trials $$X_i$$, each trial coming from one incoming photon. Each photon either passes through or not - as long as the absorption probability is constant, each such choice is a Bernoulli trial, i.e., $$P(X_i=1)=p$$. So in the end you have a binomial distribution for the total number of photons passing through, with the binomial parameter $$N$$ being Poisson distributed, or a Poisson-binomial compound. This may be helpful: Compound of Binomial and Poisson random variable.