Sum of two Poisson random variables

Question

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

Answer 1

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.