$M$ Poisson r.v. with parameter $\lambda T(F(b)-F(a))$ I have problems with the following statement:
Let $\tau_1,\tau_2,\ldots$ i.i.d. r.v. with distribution $\exp(\lambda)$, and $N_{t}$ the Poisson associated process. Let $X_1,X_2, \ldots$ continuos r.v. i.i.d. with distribution function $F : \mathbb{R} \to [0,1]$; also independent of the r.v. $\tau_k$'s. Difine the random variable
$$M=\#\left\{n:\left(\sum_{k=1}^n \tau_k,X_n\right) \in [0,T] \times [a,b]\right\}.$$
Prove that $M$ is a Poisson random variable with parameter $\lambda T(F(b)-F(a))$.
I need to prove that
$$P[M=m]=\frac{(\lambda T(F(b)-F(a)))^m}{m!} e^{-\lambda T(F(b)-F(a))},$$
but I cannot come with that solution. A hint was given in the exercise. Hint: Calcule $P(M = m)$ using the total probability law.
With this I think I need to make the following:
$$\sum_{i=m}^{\infty} P[M=m,N_T=i],$$
but the problem came when I have to manipulate the random variable $M$. I cannot came with anything. I guess I have to write down an equivalent way of the random variable $M$. Also I think I have to use in some step that $\sum_{k=1}^n \tau_k \sim \Gamma(n,\lambda)$.
Any help? Thanks in advance.
 A: Take $N_T=\max\Big\{n:\sum_{j=1}^n\tau_j\leq T\Big\}\sim \text{Poisson}(\lambda T)$.
In the context of your problem, $N_T$ counts the numbers of points from $\Big\{\left(\sum_{j=1}^n\tau_j,X_n\right)\Big\}_{n=1}^{\infty}$ that reside in the vertical strip $[0,T]\times \mathbb{R}$.
Let us suppose $N_T=k$ i.e. exactly $k$ points from $\Big\{\left(\sum_{j=1}^n\tau_j,X_n\right)\Big\}_{n=1}^{\infty}$ reside in the vertical strip $[0,T]\times \mathbb{R}$. The probability that any one of these $k$ points resides in $[0,T]\times [a,b]$ is $F(b)-F(a)$ which happens independently from any other point. So, $M|N_T=k\sim \text{Binomial}\left(k,F(b)-F(a)\right)$ which we'll express succinctly as $$M|N_T\sim \text{Binomial}\left(N_T,F(b)-F(a)\right)$$
From the total law, $$\begin{eqnarray*}\mathbb{P}(M=m) &=& \sum_{k=m}^{\infty}\mathbb{P}\left(M=m|N_T=k\right)\times \mathbb{P}\left(N_T=k\right) \\ &=&\sum_{k=m}^{\infty}{ k \choose m}\left(F(b)-F(a)\right)^m\left(1-F(b)+F(a)\right)^{k-m}\times \frac{e^{-\lambda T}(\lambda T)^k}{k!} \\ &=& \frac{\left(F(b)-F(a)\right)^m e^{-\lambda T}}{m!\left(1-F(b)+F(a)\right)^m} \sum_{k=m}^{\infty}\frac{\left(\lambda T(1-F(b)+F(a))\right)^k}{(k-m)!} \\ &=& e^{-\lambda T(F(b)-F(a))}\times \frac{(F(b)-F(a))^m}{m!}\end{eqnarray*}$$ The result follows.
