Say that the per each event of TYPE 1, the average number of occurrences is $\lambda_1$. Then the likelihood for the number of occurrences in a single event, $k_1$, is $\lambda_1^k / k_1! * e^{-\lambda_1}$. Now let's say I also have events of TYPE 2 for which the average number of occurrences is $\lambda_2$.

I want to write a likelihood for the total number of occurrences across multiple events of each type (i.e., across $n_1$ events of TYPE 1 and $n_2$ events of TYPE 2)? The main unknown in this case would be the total number of occurrences, $k$. The "occurrences" in these two types of events are the same (e.g., number of traffic stops on weekdays [TYPE 1] and weekends [TYPE 2])

As an example, let's say $\lambda_1 = 1.2$ and $\lambda_2 = 1.8$ and there are $n_1 = 10$ events of TYPE 1 and $n_2 = 20$ events of TYPE 2. The likelihood should maximize right around $k = \lambda_1*n_1 + \lambda_2*n*2 = 48$.

How would I write out such a likelihood?

  • $\begingroup$ type 1 and type are independent or not? $\endgroup$ – user158565 Jul 23 '19 at 17:58
  • $\begingroup$ ... and are you only interested in the distribution of $n_1+n_2$, or do you want the joint distribution of $(n_1, n_2)$? $\endgroup$ – jbowman Jul 23 '19 at 18:11
  • $\begingroup$ Type 1 and Type 2 events are independent and I want the likelihood for $k_{total}$ across multiple events of each type $\endgroup$ – ZombiePlan37 Jul 23 '19 at 18:21
  • 3
    $\begingroup$ Given $n_1$ and $n_2$ are independent and follow Poisson distribution, then $n_1+n_2 = n$ also follows Poisson distribution with parameter $\lambda_1+\lambda_2$. Then you should be able to write the likelihood for $n$. $\endgroup$ – user158565 Jul 23 '19 at 21:07
  • $\begingroup$ So the following would work? $L = (\lambda_1 + \lambda_2)^k / k! * e^{-(n_1 + n_2) * (\lambda_1 + \lambda_2)}$ $\endgroup$ – ZombiePlan37 Jul 29 '19 at 15:23

You are interested in the probability distribution associated with the distribution of the sum of $n_1$ and $n_2$ independent Poisson random variates with means $\lambda_1$ and $\lambda_2$ respectively. The sum $k$ of two (or more) independent Poisson variates is distributed Poisson with mean equal to the sum of the individual means:

$$P(k|n_1, \lambda_1, n_2, \lambda_2) = {(n_1\lambda_1+n_2\lambda_2)^k\text{e}^{-(n_1\lambda_1+n_2\lambda_2)} \over k!}$$

This distribution has mean $\lambda = n_1\lambda_1+n_2\lambda_2$.

The mode of a Poisson distribution with non-integer mean is the mean rounded down, $\lfloor \lambda \rfloor $. If $\lambda$ is an integer, then there are two modes: $\lambda$ and $\lambda-1$. These would be the values of $k$ that maximize the probability distribution.

| cite | improve this answer | |
  • $\begingroup$ The goal was to find the value of $k$ occurrences that would maximize the corresponding likelihood given that we know the values of $n_1$, $n_2$, $\lambda_1$, and $\lambda_2$ , but I don't see that parameter in the equation - am I missing something? Also, just to clarify, I treat $n_1$, $n_2$, $\lambda_1$, and $\lambda_2$ as known parameters in this example, but this will be just one component of a larger joint likelihood that will simultaneously estimate multiple parameters - so I would want to include all parameters in the equation (even if they would otherwise be constant) $\endgroup$ – ZombiePlan37 Aug 5 '19 at 19:35
  • $\begingroup$ Ah, I think I misread your question slightly... will revise my answer. However, you seem to have confused "probability" with "likelihood"; you are trying to find $k$ that maximizes the probability. $\endgroup$ – jbowman Aug 5 '19 at 19:48
  • $\begingroup$ Yes, likelihood is maximized w.r.t to parameters and NOT number of observations. You can estimate the required sample size etc. but that is not what the question explicitly asks. $\endgroup$ – asifzuba Aug 5 '19 at 20:04

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.