If I have a Poisson process, $N_t$ with intensity $\lambda$, then we let $X$ be a random variable that is exponentially distributed and independent of the poisson process. I am trying to find the distribution of $N_X$.
I thought the best way to go about this would be to compute $P(N_X \geq k)$ and then write $$ P(N_X = k)=P(N_X \geq k)-P(N_X \geq k+1) $$
Do you guys think this is the best way to go about it? If so I am having a bit of trouble in computing $P(N_X \geq k)$.
Thankyou for any help!
I am assuming that $X$ is exponentially distributed with mean $1$. Then for any integer $k$, $$ \mathbb P(N_X=k\mid X)=e^{-\lambda X}\frac{(\lambda X)^k}{k!}, $$ since $N_X$ is Poisson-distributed with mean $\lambda X$. Therefore since the density of $X$ is $e^{-x}\ dx$, $$ \mathbb P(N_X=k)=\mathbb E\bigl[\mathbb P(N_X=k\mid X)\bigr]=\int_0^{\infty}e^{-\lambda x}\frac{(\lambda x)^k}{k!}\ e^{-x}dx.\qquad (\star) $$ To compute the latter integral, we use the fact that $$ \int_0^{\infty}y^ke^{-y}\ dy=k!, $$ which is easily verified by integration by parts and induction (or using the Gamma function).
Letting $y=ax$ in the latter integral yields that $$ \int_0^{\infty}x^ke^{-ax}\ dx=k!\ a^{-k-1}. $$ Finally, we are ready to compute $(\star)$. Taking $a=\lambda +1$ above yields $$ \int_0^{\infty}e^{-\lambda x}\frac{(\lambda x)^k}{k!}\ e^{-x}dx=\frac{\lambda^k}{k!}\int_0^{\infty}x^ke^{-(\lambda+1)x}\ dx=\frac{\lambda^k}{(\lambda+1)^{k+1}}. $$ Thus $N_X$ is a geometric random variable with mean $\lambda$, since $$ \sum_{k=0}^{\infty}\frac{k\ \lambda^k}{(\lambda+1)^{k+1}}=\frac{1}{\lambda+1}\frac{\lambda/(\lambda+1)}{(1-\lambda/(\lambda+1))^2}=\lambda. $$