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 \(\displaystyle{k}\in{N}_{{0}}\)