# Conditional distribution of compounded Poisson process

Consider a Poison a process $N = (N_t )_{t\geq 0}$ of intensity $\lambda >0$ whose instants of jumps are $(T_n)_{n\geq0}$ $(T_0 =0)$ and a process $\tilde{N} =(\tilde N_t )_{t\geq 0}$ defined as follows

$$\tilde {N_t} := \left( \sum _{k=1}^{N_t} \mathbf {1}_{A_k} \right) \ \mathbf {1}_{\{ N_t \geq 1\}}$$

where $\{A_k\}_{k\geq 0}$ are independent events of identical probability $p \ (0<p<1)$, also independent of the $(T_{n+1}- T_n )_{n \geq0}$.

I am interested in calculate $\mathbb P \{ \tilde N_t =m \ |\ N_t =n \}$ but I'm stucked . Obviously it's $0$ if $n<m$, otherwise we must have

$$\mathbb P \{ \tilde N_t =m \ |\ N_t =n \}= \mathbb P \left\{ \left( \sum _{k=1}^{n} \mathbf {1}_{A_k} \right) \ \mathbf {1}_{\{ n \geq 1\}} =m \ , \ N_t =n \right\}/ \mathbb P\{ N_t =n\} = \mathbb P \left\{ \left( \sum _{k=1}^{n} \mathbf {1}_{A_k} \right) \ \mathbf {1}_{\{ n \geq 1\}} =m \ , T_n \leq t <T_{n+1} \right\} \frac{n!}{(\lambda t)^n} \exp(\lambda t)$$

I know $\mathbb P \{ \left( \sum _{k=1}^{n} \mathbf {1}_{A_k} \right) \ \mathbf {1}_{\{ n \geq 1\}} =m \}=C_{n, m} p^m (1-p)^{m-n}$ if $n\geq n$ and $0$ otherwise. As you may see I am stuked without knowing how to manage the independence of $(A_k)$ and $(T_{n+1}- T_n)$ in order to finish the calculation.