# Poisson and exponential distribution for independent rvs

The number of times event $A$ occurs in a time $t$ has a Poisson distribution with parameter $a$ and the number of times event $B$ occurs in the same time $t$ has a Poisson distribution with parameter $b$. The events are independent.

What is the probability that event $A$ occurs first (i.e. before event $B$)?

What is the expected time interval until $A$ and $B$ have each occured at least once?

My question is this:

Let $X\sim \mathrm{Poisson}(a)$ and $X\sim \mathrm{Poisson}(b)$ represent the number of occurrences of these events.

For the first part, is the required probability simply $P(X=1|X+Y=1)$?

For the second part, is the expected time simply $\max(\frac{1}{a}, \frac{1}{b})$?

## 1 Answer

For the first part, is the required probability simply $P(X=1|X+Y=1)$?

Good try, but no. Consider, $\{X=2, Y=0\}$ still means that the first process has occured before the second.   Further note that the event you seek concerns the times until the first arrival of the processes, not the number of events in the given time.

You want $\mathsf P(T_{X=1}<T_{Y=1})$, the probabilty that the first process experiences some arrivals before the second process experiences any.   Here $T_{E}$ is the time until event $E$, and $T_{X=1}$ is the "time until the arrival of one event from process $A$", et cetera.

This can be calculate by a rather convoluted double integral.   It is not terribly difficult, but still a bit of work.

However a much simpler approach is to consider that the independence between the processes means that the events are arriving in a combined Poisson process.   That is that in time $t$ events from Process A are arriving at rate $a$ events in time $t$, those from Process $B$ at rate $b$ events in time $t$, and thus the combined rate of arivals is $a+b$ events in time $t$.   So if the average proportion of events arriving that are from Process A is $a/(a+b)$, what is the probability that the first such event to arrive is from Process A?

For the second part, is the expected time simply $\max(\frac{1}{a}, \frac{1}{b})$?

From the first part you can calculate the expected time of arrival for the first arrival, the probability that this is from one process or the other, and the expected time to arrival of the next event from the other Process when given whichever the first event's Process may be.

Then use the memoryless property of Poisson Processes, and the Law of Total Expectation.

${\mathsf E(\max\{T_{X=1},T_{Y=1}\}) ~}{=~ {\mathsf E(\min\{T_{X=1}, T_{Y=1}))}+{\mathsf P(T_{X=1}<T_{Y=1})~\mathsf E(T_{Y=1})}+{\mathsf P(T_{X=1}>T_{Y=1})~\mathsf E(T_{X=1})}}$