Question

# Consider a Poisson process on [0, \infty) with parameter \lambda and let T be a random variable independent of the process. Assume T has an exponentia

Discrete math

Consider a Poisson process on $$[0, \infty)$$ with parameter $$\displaystyle\lambda$$ and let T be a random variable independent of the process. Assume T has an exponential distribution with parameter v. Let $$N_{T}$$ denote the number of particles in the interval $$[0, T]$$. Compute the discrete density of $$N_{T}$$.

2021-05-06

First of all, observe that $$N_T$$ is a counter of particle in a random interval $$[0,T].$$ Therefore, $$N_T$$ hs discrete support in $$N_0$$. For $$k \in N_0$$ we use law of the total probability to obtain that
$$\displaystyle{P}{\left({N}_{{T}}={k}\right)}={\int_{{{0}}}^{{\infty}}}{P}{\left({N}_{{T}}={k}{\mid}{T}={t}\right)}{{f}_{{T}}{\left({t}\right)}}{\left.{d}{t}\right.}={\int_{{{0}}}^{{\infty}}}{P}{\left({N}_{{t}}={k}{\mid}{T}={t}\right)}{{f}_{{T}}{\left({t}\right)}}{\left.{d}{t}\right.}$$
$$\displaystyle{\int_{{{0}}}^{{\infty}}}{P}{\left({N}_{{t}}={k}\right)}{{f}_{{T}}{\left({t}\right)}}{\left.{d}{t}\right.}={\int_{{{0}}}^{{\infty}}}{\frac{{{\left(\lambda{t}\right)}^{{{f}}}}}{{{k}!}}}{e}^{{-\lambda{t}}}\cdot{v}{e}^{{-{v}{t}}}{\left.{d}{t}\right.}={\frac{{\lambda^{{{k}}}\cdot{v}}}{{{k}!}}}{\int_{{{0}}}^{{\infty}}}{t}^{{{k}}}{e}^{{-{\left(\lambda+{v}\right)}^{{{t}}}}}{\left.{d}{t}\right.}$$
where in third equality we have used that T and considered Poisson process are independent. Now, let’s a calculate this integral. The key here is to use substitution $$\displaystyle{u}={\left(\lambda+{v}\right)}{t}$$ and use properties of gamma function. We have that
$$\displaystyle{\int_{{{0}}}^{{\infty}}}{t}^{{{k}}}{e}^{{-{\left(\lambda+{v}\right)}{t}}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{\left(\lambda+{v}\right)}^{{{k}+{1}}}}}}{\int_{{{0}}}^{{\infty}}}{u}^{{{k}}}{e}^{{-{u}}}{d}{u}={\frac{{Г{\left({k}+{1}\right)}}}{{{\left(\lambda+{v}\right)}^{{{k}+{1}}}}}}={\frac{{{k}!}}{{{\left(\lambda+{v}\right)}^{{{k}+{1}}}}}}.$$ Finally, we have obtained that
$$\displaystyle{P}{\left({N}_{{T}}={k}\right)}={\frac{{\lambda^{{{k}}}\cdot{v}}}{{{k}!}}}\cdot{\frac{{{k}!}}{{{\left(\lambda+{v}\right)}^{{{k}+{1}}}}}}={\frac{{\lambda^{{{k}}}\cdot{v}}}{{{\left(\lambda+{v}\right)}^{{{k}+{1}}}}}}$$
$$\displaystyle{P}{\left({N}_{{T}}={k}\right)}={\frac{{\lambda^{{{k}}}\cdot{v}}}{{{\left(\lambda+{v}\right)}^{{{k}+{1}}}}}}$$ for $$\displaystyle{k}\in{N}_{{0}}$$