Distribution of successes of poisson process followed by a binomial distribution

Question

I have been stuck with a problem for a couple of days regarding the distribution of outcomes from a two-stage process. Specifically, what is the distribution of the number of successes in a poisson process followed by a series of bernoulli trials, where the number of trials is determined by the result of the poisson process. If this is not clear, I’ll explain by way of example.

Suppose we have a hunter who is setting out to catch rabbits. To do this he sets up a trap. Rabbits pass over the trap at a rate of $r$ times per day (poisson distributed). However, the trap is not very good. If a rabbit walks over the trap, there is only a $p$ (probability) chance that it activates.

If the trap has no limit on the amount of rabbits it can catch, what does the distribution of rabbits caught per day look like?

My intuition is that it should follow a poisson distribution with mean $r.p$. However, I have been unable to prove this analytically. I figure that the probability of a given number of ‘successes’ should be calculable through the poisson distribution and the binomial distribution. Something like:

$ P(X=x) = \sum_{i=x}^{Inf} poisson_{pmf}(i | r) . binomial_{pmf}(x | i, p) $

We can then substitute in the respective mass functions into the above equation. However, beyond that I am getting stuck. Any help or a point in the right direction would be great appreciated.

Answer 1

Partition the event $E_k:$ "$k$ rabbits were caught" into disjoint events of the form $E_{k}(j):$ "$k+j$ rabbits entered traps and $k$ were caught," for $j=0, 1, 2, \ldots.$ The chance of $E_k(j)$ is the product of the Poisson chance of $k+j$ entering traps and the Binomial chance that $k$ of them were caught, whence (using $\binom{j+k}{k} = (j+k)!/(j!k!)$)

$$\eqalign{ \Pr(E_k) &= \sum_{j=0}^\infty \Pr(E_k(j)) = \sum_{j=0}^\infty e^{-r} \frac{r^{k+j}}{(j+k)!}\quad \binom{j+k}{k} p^k(1-p)^j \\ &= e^{-r} \frac{r^k p^k}{k!}\sum_{j=0}^\infty \frac{(r(1-p))^j}{j!}. }$$

The last sum is the power series for $e^{r(1-p)}$, which when combined with $e^{-r}$ gives

$$\Pr(E_k) = e^{-pr}\frac{(pr)^k}{k!},$$

precisely the Poisson probability of $k$ for parameter $pr$.

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A statistical way of viewing this problem is to consider a Poisson process on the line with rate $r$. "Thin" the process by randomly removing each event independently with chance $p$. Clearly the resulting process still enjoys all the defining properties of the Poisson process: homogeneous rate, independence of events, and no chance of simultaneous events. Moreover, the rate obviously is $pr$, QED.

Answer 2

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 there many rabbits (N). Then there is a small probability $p_1$ that a rabbit pass over the trap. These are the premises of the law of rare events, so the number of rabbits passing the trap follows a Poisson distribution with a parameter $r = N \, p_1$.

Note that the problem implicitly assumes large N and small $p_1$. If not the case, the number of the passing rabbits would be a binomial variable instead of a Poisson one.

Let us add the trap efficiency. The probability that a single rabbit pass over the trap and it is caught is the product of the probabilities of these two independent events, $p' = p_1 \, p$. Then, end-to-end, there N rabbits each with a small probability $p'$ of being caught. This is again a Poisson process. The corresponding parameter representing the mean caught rabbits is:

$\mu = N \, p_1 \, p = r \, p $

I adapted this answer from another I made about a convolution of a Poisson with a binomial variable that uses another example.