Let $B_t$ be a standard Brownian motion and let $T_1,T_2,...$ be the successive arrival times of a Poisson process with intensity $\lambda$. Let $X_n = B_{T_n}$ be the value of BM at time $T_n$. Moreover, we assume that $T_n$ and $(B_t)$ are independent.
Show that $X_n$ may be written as $X_n = (Y_1+Y_2+ ... + Y_n) - (Z_1 + Z_2 + ... Z_n)$ where $Y$ and $Z$ are i.i.d exponential RVs with parameter $\sqrt{2\lambda}$.
My attempt:
First find the moment generating function of $X_n$. We have $E[e^{rB_{T_n}}]$ \begin{align} & = E[E[e^{rB_{T_n}}|T_n]] \\ & = E[e^{T_nr^2/2}] \quad (\text{since }B_t \text{ is normal}) \\ & = \frac{\lambda}{\lambda - r^2/2} \quad (\text{arrival times are i.i.d exponential}) \\ & = \frac{2\lambda}{2\lambda -r^2} \end{align}
However, now that I have the mgf I'm not quite sure how to proceed. The only thing I could think of was that maybe this process is a Levy process (since both Brownian motion and Poisson processes are) and then by definition it can be represented as a sum of i.i.d random variables. But we looking for the difference of two sums not just one sum.