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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?

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  • $\begingroup$ type 1 and type are independent or not? $\endgroup$
    – user158565
    Jul 23, 2019 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, 2019 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, 2019 at 18:21
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    $\begingroup$ 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, 2019 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, 2019 at 15:23

1 Answer 1

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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.

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  • $\begingroup$ 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, 2019 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, 2019 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, 2019 at 20:04

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