# Poisson arrival with exponential time service

The number $$N$$ of custoners that arrive at time $$t$$ is a Poisson random variable with parameter $$\beta t$$. The time $$T$$ required to service each customer has exponential distribution. Find the pmf for $$N$$ that arrive during the service time of a specific customer.

The solution is $$f_N(k)=\Pr(N=k)=\int_0^{\infty} \Pr(N=k|T=t)f_T(t)dt$$ The book simply used $$\Pr(N=k|T=t)=\frac{(\beta t)^k}{k}e^{-\beta t}$$ which is indeed, the PMF of the Poisson random variable.

My question is, how do the condition $$T=t$$ affected the PMF because it seems to me that the condition itself is not needed at all, that is they are independent which is intuitively wrong.

In fact, if $$N$$ and $$T$$ are independent then the formula becomes:

$$f_N(k)=\Pr(N=k)=\int_0^{\infty} \Pr(N=k)f_T(t)dt$$

Probability that $$k$$ customers arrive in in the time interval $$[0.t]$$ is $$e^{-\beta t} \frac {(\beta t)^{k}} {k!}$$. We are asked to find the probability that $$k$$ customers arrive in in the time interval $$[0,T]$$ where the service time $$T$$ is random. If we condition on $$T=t$$ then we get the probability as $$e^{-\beta t} \frac {(\beta t)^{k}} {k!}$$.
• @qcpz Yes, that is correct. $\beta$ is the parameter of the the Poisson process but the number of customers arriving by time $t$ is $Pois(\beta t)$. Commented Feb 6, 2022 at 8:44