1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your reasoning for the first question is correct. The probability that a given jump is from the process 1 is $p$. $\endgroup$ – Harto Saarinen Sep 22 '19 at 19:30
  • $\begingroup$ @HartoSaarinen - I'm looking for a concrete proof as this will also help with the second part (re Compound Poisson processes). $\endgroup$ – Rohit Pandey Sep 22 '19 at 19:32
  • $\begingroup$ But that is the concrete proof! $\endgroup$ – Harto Saarinen Sep 22 '19 at 19:35
  • $\begingroup$ @HartoSaarinen - the problem is that I believed quite strongly at first that the logic should apply quite readily to the Compound Poisson process as well. Until I thought deeply about the orange curve referenced in the question. If the reasoning is solid, why doesn't it extend to the Compound Poisson? $\endgroup$ – Rohit Pandey Sep 22 '19 at 19:36
  • 2
    $\begingroup$ Compute $\mathbb{P}(X=j | X+Y=n)$ and show it is in the form of the PMF of a binomial distributed RV $\mathcal{B}(n,p)$ where $p=\lambda_X/(\lambda_X+\lambda_Y)$, where $X\sim Pois(\lambda_X)$ and $Y\sim Pois(\lambda_Y)$ and $X+Y$ is (by independence) $Pois(\lambda_X+\lambda_Y)$. $\endgroup$ – Nap D. Lover Sep 22 '19 at 22:35
1
$\begingroup$

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 $\lambda$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Nice thought experiment, +1. I have good reason to think that dividing by the average value of the compounding r.v. would make it binomial. Any thoughts? $\endgroup$ – Rohit Pandey Sep 23 '19 at 21:24
  • 1
    $\begingroup$ The average value of the compounding r.v. -- do you mean in this case the average of $\{1,2,3,4,5,6\}$, which is $3.5$? If $X_1$ above is bimodal, then ${X_1 \over 3.5}$ is still bimodal. Dividing by any constant doesn't affect the shape. $\endgroup$ – antkam Sep 23 '19 at 21:37
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.