Compound Poisson process property Let $X_t=\sum_{i=1}^{N_t}J_i$ be a compound Poisson Process, where $J_i$ are independent and equidistributed. 
I have to prove that for every $0<t_1<t_2 \leq t_3<t_4$ : $X_{t_4}-X_{t_3}$ is independent of $X_{t_3}-X_{t_2}$.
So if I demonstrate that, $$\mathbb{P}(X_{t_4}-X_{t_3}\leq s, X_{t_2}-X_{t_1}\leq t )= \mathbb{P}(X_{t_4}-X_{t_3} J_i\leq s)\mathbb{P}(X_{t_2}-X_{t_1} J_i\leq t)$$ it would be finished.
I have that:
$$\mathbb{P}(X_{t_4}-X_{t_3} J_i\leq s, X_{t_2}-X_{t_1}J_i \leq t )  = \mathbb{P}(\sum_{N_{t_3}+1}^{N_{t_4}} J_i\leq s, \sum_{N_{t_1}+1}^{N_{t_2}}J_i\leq t )$$
Since $N_t$ only takes integer values, the last expresion is the same as:
   $$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\mathbb{P}(\sum_{N_{t_3}+1}^{N_{t_4}}J_i \leq s, \sum_{N_{t_1}+1}^{N_{t_2}}J_i\leq t )|N_{t_4}-N_{t_3}=n,N_{t_2}-N_{t_1}=m)\mathbb{P}(N_{t_4}-N_{t_3}=n,N_{t_2}-N_{t_1}=m)$$ and since $N_t$ is independent of $J_i$ we have that:
In this context, I don´t know how to follow to complete my proof: I know that I have to use that $N_t$ is independent of $J_i$, $N_{t_4}-N_{t_3}$ is independent of $N_{t_2}-N_{t_1}$ and that $J_i$ are independent and equidistributed. 
Any suggestion? Thanks!
 A: I follow the same path as you:
\begin{eqnarray*}
&& P\left(X_{t_3}-X_{t_2}\leq s \;\bigcap\; X_{t_4}-X_{t_3}\leq t\right) \\
&&\qquad = P\left(\sum_{i=N_{t_2}+1}^{N_{t_3}} J_i\leq s \;\bigcap\; \sum_{i=N_{t_3}+1}^{N_{t_4}} J_i\leq t\right) \\
&& \\
&&\qquad = P\left(\sum_{i=1}^{N_{t_3}-N_{t_2}} J_i\leq s \;\bigcap\; \sum_{i=N_{t_3}-N_{t_2}+1}^{N_{t_4}-N_{t_2}} J_i\leq t\right) \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{re-index the $J_i$ from $1$ (OK since $J_i$ are i.i.d.)} \\
&& \\
&&\qquad = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} P\left(\sum_{i=1}^{N_{t_3} - N_{t_2}} J_i\leq s \;\bigcap\; \sum_{i=N_{t_3} - N_{t_2}+1}^{N_{t_4} - N_{t_2}} J_i\leq t \; \bigg| \; N_{t_3} - N_{t_2}=n\cap N_{t_4} - N_{t_3}=m\right) \\
&&\qquad\qquad \qquad \times P\left(N_{t_3} - N_{t_2}=n\right)P\left(N_{t_4} - N_{t_3}=m\right) \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{condition on the values of $N(t)$} \\
&& \\
&&\qquad = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} P\left(\sum_{i=1}^{n} J_i\leq s \;\bigcap\; \sum_{i=n+1}^{n+m} J_i\leq t \right) P\left(N_{t_3} - N_{t_2}=n\right)P\left(N_{t_4} - N_{t_3}=m\right) \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since $J_i$ and $N(t)$ are independent} \\
&& \\
&&\qquad = \left[ \sum_{n=1}^{\infty} P\left(\sum_{i=1}^{n} J_i\leq s \right) P\left(N_{t_3} - N_{t_2}=n\right) \right] \left[ \sum_{m=1}^{\infty} P\left(\sum_{i=1}^{m} J_i\leq t \right)P\left(N_{t_4} - N_{t_3}=m\right)\right] \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since the $J_i$ are independent of each other} \\
&& \\
&&\qquad = P\left(\sum_{i=N_{t_2}+1}^{N_{t_3}} J_i\leq s \right) P\left(\sum_{i=N_{t_3}+1}^{N_{t_4}} J_i\leq t \right) \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{undo the conditioning on the values of $N(t)$} \\
&& \\
&&\qquad = P\left(X_{t_3}-X_{t_2}\leq s\right) P\left(X_{t_4}-X_{t_3}\leq t\right). \\
\end{eqnarray*}
A: Thanks a lot! So according to my  case (for any $t_1<t_2 \leq t_3 <t_4$) it would be:
\begin{eqnarray*}
&& P\left(X_{t_4}-X_{t_3}\leq s \;\bigcap\; X_{t_2}-X_{t_1}\leq t\right) \\
&&\qquad = P\left(\sum_{i=N_{t_3}+1}^{N_{t_4}} J_i\leq s \;\bigcap\; \sum_{i=N_{t_1}+1}^{N_{t_2}} J_i\leq t\right) \\
&& \\
&&\qquad = P\left(\sum_{i=1}^{N_{t_4}-N_{t_3}} J_i\leq s \;\bigcap\; \sum_{i=1}^{N_{t_2}-N_{t_1}} J_i\leq t\right) \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{re-index the $J_i$ from $1$ (OK since $J_i$ are i.i.d.)} \\
&& \\
&&\qquad = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} P\left(\sum_{i=1}^{N_{t_4} - N_{t_3}} J_i\leq s \;\bigcap\; \sum_{i=1}^{N_{t_2} - N_{t_1}} J_i\leq t \; \bigg| \; N_{t_4} - N_{t_3}=n\cap N_{t_2} - N_{t_1}=m\right) \\
&&\qquad\qquad \qquad \times P\left(N_{t_4} - N_{t_3}=n,N_{t_2} - N_{t_1}=m\right) \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{condition on the values of $N(t)$} \\
&&\qquad = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} P\left(\sum_{i=1}^{N_{t_4} - N_{t_3}} J_i\leq s \;\bigcap\; \sum_{i=1}^{N_{t_2} - N_{t_1}} J_i\leq t \; \bigg| \; N_{t_4} - N_{t_3}=n\cap N_{t_2} - N_{t_1}=m\right) \\
&&\qquad\qquad \qquad \times P\left(N_{t_3} - N_{t_2}=n\right) P\left(N_{t_4} - N_{t_3}=m\right) \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{$N(t)$ has independent increments} \\
&& \\
&&\qquad = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} P\left(\sum_{i=1}^{n} J_i\leq s \;\bigcap\; \sum_{i=1}^{m} J_i\leq t \right) P\left(N_{t_4} - N_{t_3}=n\right)P\left(N_{t_2} - N_{t_1}=m\right) \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since $J_i$ and $N(t)$ are independent} \\
&& \\
&&\qquad = \left[ \sum_{n=1}^{\infty} P\left(\sum_{i=1}^{n} J_i\leq s \right) P\left(N_{t_4} - N_{t_3}=n\right) \right] \left[ \sum_{m=1}^{\infty} P\left(\sum_{i=1}^{m} J_i\leq t \right)P\left(N_{t_2} - N_{t_1}=m\right)\right] \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since the $J_i$ are independent of each other} \\
&& \\
&&\qquad = P\left(\sum_{i=N_{t_3}+1}^{N_{t_4}} J_i\leq s \right) P\left(\sum_{i=N_{t_1}+1}^{N_{t_2}} J_i\leq t \right) \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{undo the conditioning on the values of $N(t)$} \\
&& \\
&&\qquad = P\left(X_{t_4}-X_{t_3}\leq s\right) P\left(X_{t_2}-X_{t_1}\leq t\right). \\
\end{eqnarray*}
