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This is a marked homework question, so please try not to write complete solutions here:

The number of customers that arrive at a service station during a time t is a Poisson random variable with parameter $\beta t$. The time required to service each customer is an exponential random variable with parameter $\alpha$. Identify the distribution of the number of customers $N$ that arrive during the service time $T$ of a specific customer by (a) finding the PMF of $N$ directly (b) finding the MGF of $N$.

My understanding is that we have $P(N=n|T=t)\tilde{}Poisson(\beta t)$ i.e. a Poisson process with rate $\beta$, $P(T=t)\tilde{}exp(\alpha)$, and the question is asking for $P(N=n)$. I tried to find this using marginal distribution, but one random variable is discrete and the other is continuous and I don't know how to take marginal distribution in that case.

Another idea was to find $P(N=n)=P(Y_n<T)=\int_0^\infty\int_0^tf_{Y_n}(y)f_T(t) dydt$ where $Y_n$ denotes time to $n^{th}$ arrival which is $\tilde{}Erlang(n)$. But that looks like overcomplicating it and it certainly doesn't involve MGFs.

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Got something from an answer below? –  Did Jul 18 at 9:29

2 Answers 2

$Poi(\beta t)$ is a non-homogenous poisson process. Don't forget that the Poisson process describes counts in a given interval. Therefore, the number of counts within a given instant will be zero. That is what you were doing in the first formulation.

For a nonhomogenous poisson process, the number of counts between time $t_1$ and $t_2$ (again, we need an interval), is distributed $Poisson(\Lambda (t_1,t_2))$ where $\Lambda(t_1,t_2) = \int\limits_{t_1}^{t_2} \beta t\;dt$

Does that help?

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Thanks. But I never made that assumption. In the first formulation P(N=n|T=t) can be looked at as either number of arrivals until time t or number of arrivals in an interval of length t –  user798275 Oct 19 '13 at 4:20
    
I see. My misunderstanding. I think my answer is still applicable to part (a), but you would need to integrate over all values of T. I.e. $P(N=n)=\int\limits_{0}^{\infty}f_{N|T}(N=n|T=t)f_{T}(T=t) dT = \int\limits_{0}^{\infty} \alpha e^{-aT}poi(n,\frac{\beta T^{2}}{2}) dT$ –  Eupraxis1981 Oct 20 '13 at 16:48

For every $n\geqslant0$, $$ P(N=n)=\int_\mathbb RP(N=n\mid T=t)\,f_T(t)\,\mathrm dt=\int_0^\infty\mathrm e^{-\beta t}\,\frac{(\beta t)^n}{n!}\,\alpha\,\mathrm e^{-\alpha t}\,\mathrm dt. $$ The change of variable $s=(\alpha+\beta)t$ yields $$ P(N=n)=\frac{\beta^n\alpha}{(\alpha+\beta)^{n+1}}\int_0^\infty\frac{s^n}{n!}\,\mathrm e^{-s}\,\mathrm ds=(1-p)p^n, $$ where the parameter $p$ in $(0,1)$ is defined by $$ p=\frac{\beta}{\alpha+\beta}. $$ Thus, the distribution of $N$ is geometric with parameter $p$ (of course, this result has a pretty neat "physical" interpretation based on the conditioning properties of the exponential distributions at the basis of Poisson processes).

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