# How to write the log likelihood for the sum of independent poisson events

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?

• type 1 and type are independent or not? Jul 23, 2019 at 17:58
• ... and are you only interested in the distribution of $n_1+n_2$, or do you want the joint distribution of $(n_1, n_2)$? Jul 23, 2019 at 18:11
• Type 1 and Type 2 events are independent and I want the likelihood for $k_{total}$ across multiple events of each type Jul 23, 2019 at 18:21
• 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 $n$. Jul 23, 2019 at 21:07
• So the following would work? $L = (\lambda_1 + \lambda_2)^k / k! * e^{-(n_1 + n_2) * (\lambda_1 + \lambda_2)}$ Jul 29, 2019 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.

• 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) Aug 5, 2019 at 19:35
• 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. Aug 5, 2019 at 19:48
• 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. Aug 5, 2019 at 20:04