Two Poisson processes - conditional on total events, number of events from first.

Question

There are two Poisson processes that run for an equal amount of time. The rates are $\lambda_1$ and $\lambda_2$. Conditional on the total number of events (from both processes) seen being $n$, the number of events that came from the first process is distributed Binomial with $p=\frac{\lambda_1}{\lambda_1+\lambda_2}$ and $n$ number of trials. This is what the uniformly most powerful test for Poisson process rates is based on. How do I prove this? I have some vague intuition of each event belonging to the first process being a Bernoulli with $p$ as specified above and so, the sum of those Bernoulli's across all events being Binomial. But in terms of concrete proof's, I don't know where to start. Normally, we try and construct the CDF. But there is no nice closed form for the binomial CDF.

Now for the second part of my question and what I'm really after: replace the two Poisson processes with compound poisson processes. In other words, each time a Poisson arrivals happens, toss a fair die (with first six numbers) and those represent the number of point events for that arrival. Now, given that we observe $n$ events from both of them, what is the conditional distribution of the number of events from the first one? I have good reason to believe it isn't binomial. If it were, the orange line in the second graph here: https://stats.stackexchange.com/questions/425425/hypothesis-test-for-poisson-process-failure-rates-stays-just-as-powerful-for-com would have been a straight line.

Answer 1

There is really no reason to think the compound case will still be binomial, right? E.g.

• $\lambda_1 = \lambda_2 = 10^{-9}$, i.e. super-low-rate Poisson

• You observe total number of points $= 6$

Let $X_1$ be the number of points from the $\lambda_1$ process. Intuitively one would think $X_1$ is basically bimodal, i.e. it is either $0$ or $6$, right? I.e. all $6$ events come from the same "burst" (rolling $6$ on the die), and the single burst could be from either process -- this case being much more likely than having two or more bursts (because of the tiny $\lambda$).

Answer 2

Using the approach proposed by @Nap D. Lover in the comments, we require:

$$P(X=j | X+Y=n) = \frac{P(X=j \cap X+Y=n)}{P(X+Y=n)}$$

$$=\frac{P(X=j)P(Y=n-j)}{P(X+Y=n)}$$ $$=\frac{\left(\frac{e^{-\lambda_1} \lambda_1^j}{j!}\right)\left(\frac{e^{-\lambda_2} \lambda_2^{n-j}}{(n-j)!}\right)}{\left(\frac{e^{-\lambda_1+\lambda_2} (\lambda_1+\lambda_2)^n}{ n!}\right)}$$ $$ ={n \choose j} \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^j \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^{n-j}$$

And this is the binomial PMF as expected.