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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:

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

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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 $e^{-\beta t} \frac {(\beta t)^{k}} {k!}$.

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    $\begingroup$ @qcpz Yes, that is correct. $\beta$ is the parameter of the the Poisson process but the number of customers arriving by time $t$ is $Pois(\beta t)$. $\endgroup$ Commented Feb 6, 2022 at 8:44

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