Convolution of Poisson with Binomial distribution?

Question

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.

Answer 1

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.

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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 $\theta$.

Answer 2

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.

related: Probability of compound Poisson process

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).

Answer 3

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.

Answer 4

Using the law of rare events

Apart from the mathematical demonstration, the process can be understood intuitively.

In the example in the question, the reasoning is that the laser has a very large number of excited particles (N). There is a tiny probability that a laser particle emits light ($p_{em}$). For this reason (Law of the rare events), the number of photons in the pulse follows a Poisson distribution with parameter $\mu_{em} = N \, p_{em}$.

Let's add the filter to the process. The probability that a laser particle emits a photon and that this photon is detected is, $p_{det} = p_ {em} * p_{trans}$. Here $p_{trans}$ is the probability that the filter transmits a photon.

Then, end-to-end, you have a process of many N laser particles, each with a small probability of emitting a photon that is detected $p_{det}$. This is again a Poisson process. The corresponding parameter representing the mean detected photons is:

$\mu_{det} = N \, p_{det} = N \, p_{em} \, p_{trans} = \mu_{em} \, p_{trans}$