Find the probability density of $N$, the customer of customers arriving at $T$, the service time of a specific customer.

Question

The number of clients arriving at a service at time $t$ is distributed Poisson with mean $\beta t$. The service time $T$ is distributed exponentially with parameter $\alpha$, that is, $E[T] =\frac{1}{\alpha}$ Find the probability density of $N$, the customer of customers arriving at $T$, the service time of a specific customer.

My try:

We know that $$f_N(n)= \int_{0}^{\infty} f(n\mid t)\cdot f_T(t)\,dt$$ $$= \int_{0}^{\infty} \alpha e^{-\alpha t}\frac{e^{-\beta t}(\beta t)^n}{n!}dt$$

I´m stuck with the algebra here. I know that I must add constants to the integral so we can form a "nucleus" of a know probability density. Any suggestions would be great!

Answer 1

Rewriting your integral (I did a little edit because it is $e^{-\beta t}$) you get

$$\frac{\alpha \beta^n}{n!(\alpha+\beta)^{n+1}}\int_0^{+\infty}[(\alpha+\beta)t]^ne^{-(\alpha+\beta)t}d[(\alpha+\beta)t]=\frac{\alpha \beta^n}{n!(\alpha+\beta)^{n+1}}n!=\frac{\alpha \beta^n}{(\alpha+\beta)^{n+1}}$$

this because now the integral is in the form

$$\int_0^{+\infty}x^{\alpha}e^{-x}dx=\Gamma(\alpha+1)=\alpha!$$

Concluding: the requested density is

$$f_N(n)=\frac{\alpha \beta^n}{(\alpha+\beta)^{n+1}}$$

$n=0,1,2,...$

Answer 2

Two minor issues - first, $N$ does not have a density, as it is a discrete random variable. Second, your idea is correct, though the $e^{\beta t}$ should be $e^{-\beta t}$. In that case we find that $$ \mathbb P(N=n) = \int_0^\infty \alpha e^{-(\alpha+\beta)t}\frac{(\beta t)^n}{n!}\ \mathsf dt =\alpha\beta^n(\alpha+\beta)^{-(n+1)},\ n=0,1,2,\ldots. $$ Summing over $n$ we find that $$ \sum_{n=0}^\infty \mathbb P(N=n) = \sum_{n=0}^\infty \alpha\beta^n(\alpha+\beta)^{-(n+1)} = 1, $$ so indeed this is a valid probability distribution - no "constants" needed to be added.