# Sum of two Poisson random variables

I am stuck on a problem which actually I have the answer to. First, let me give you another example question then I'll try to explain why I am stuck on this other question.

The emails I get can be modeled by a Poisson distribution with an 0.2 emails per minute. Then what is the probability that I get no emails in an interval of 5 minutes?

The solution is as follows:

Since I get 0.2 emails per minute, $$\lambda = 0.2 \times 5 =1$$. Then $$P(X=0)=\frac{e^{-\lambda} \lambda^k}{k!} = \frac{1}{e}$$

Okay now let's get to the question I got stuck.

Let $$X\sim \text{Poisson}(\alpha)$$ and $$Y\sim \text{Poisson}(\beta)$$ be two independent random variables. Define a new random variable as $$Z = X+Y$$. Find the PMF of $$Z$$.

The problem is solved as follows:

$$P_Z(k) = P(X+Y =k)$$ $$= \sum_{i = 0}^k P(X+Y=k\mid X=i)P(X=i)$$ $$= \sum_{i = 0}^k P(Y=k-i)P(X=i)$$ then from the Binomial theorem it ends up being: $$= \frac{e^{-(\alpha + \beta)}}{k!}(\alpha + \beta)^k$$

I have no problems following the solution and understanding why the first statement ends to the last one. What I don't understand is how can we just write $$P_Z(k) = P(X+Y =k)$$. In the first example to calculate $$\lambda$$, we multipled 0.2 with 5 since we receive 0.2 emails per minute and we are asked about a 5 minute interval. What if this $$\alpha$$ and $$\beta$$ parameters belong to different time framed distributions? If $$\alpha$$ is for per minute and $$\beta$$ is for per hour don't we have to do something like $$P_Z(k) = P(60X+Y =k)$$?

Thank you

• this makes sense to me -- they have to be in the same units to be reasonably combined :) – gt6989b Aug 21 '20 at 14:40

The formula $$\frac{e^{-\lambda}\lambda^k}{k!}$$ requires $$\lambda$$ to be dimensionless. The number of times a specific type of event occurs in a time period $$t$$ is $$\operatorname{Poisson}(\omega t)$$-distributed, for a constant $$\omega$$ associated with that event type. Since $$\omega=\lambda/t$$ has the dimension of inverse time (also called frequency), adding $$\omega$$s requires us to express them in the same units. For example, $$m$$ per minute plus $$h$$ per hour is $$60m+h$$ per hour, or $$m+h/60$$ per minute. In particular, independent$$X\sim\operatorname{Poisson}(\omega_Xt),\,Y\sim\operatorname{Poisson}(\omega_Yt)$$satisfy$$Z:=X+Y\sim\operatorname{Poisson}(\omega_Xt+\omega_Yt)=\operatorname{Poisson}((\omega_X+\omega_Y)t),$$so frequencies add as per the above rules.
• Thank you so much for your input. So for my question we have $\alpha$ and $\beta$ as $\lambda$s and the solution just adds them up. Does this mean that the solution assumed $\alpha = w_1t$ and $\beta = w_2t$ where $t$s are the same? So they just added them up? – diane Aug 22 '20 at 16:31
• @diane You asked about two problems, one with an explicit discussion of time, and one just asking about parameters $\alpha$ and $\beta$. But you can relate these two problem types as you've understood. – J.G. Aug 22 '20 at 16:38