Let $X_t = \sum _{i=1}^{N_t} Y_i$ and $N_t$ be a Poisson process with intensity $\lambda >0$. Suppose $Y_i$ are i.i.d. random variables independent of $N_t$, having $N(m,\sigma^2)$. Find $\mu>0$ such that $X_t - \mu t$ is a martingale.
My attempt:
First note that the filtration generated by the Poisson counting process only depends on $N_s$. Therefore, $E[X_t - \mu t \mid F_s] = E[X_t\mid N_s] - \mu t$.
Compute: $E[X_t\mid N_s] = E[\sum _{i=1}^{N_t} Y_i\mid N_s] = E[\sum _{i=1}^{N_s} Y_i+ \sum _{i=N_{s+1}}^{N_t} Y_i\mid N_s]=N_s E[Y_1]+E[\sum _{i=N_{s+1}}^{N_t} Y_i\mid N_s]$
With the memoryless property, the second term is the expression above can be written as: $$E[\sum _{i=N_{s+1}}^{N_t} Y_i\mid N_s] = E[\sum _{i=1}^{N_{t-s}} Y_i]=E[E\sum _{i=1}^{N_{t-s}} Y_i\mid N_{t-s}]]=E[N_{t-s} m]=\lambda (t-s) m$$
Therefore, $E[X_t-\mu t|N_s] = N_s m + \lambda(t-s)m - \mu t = N_s m + E[X_t] - E[X_s] - \mu t$.
How can I relate the above expression to $X_s$ so that I can choose $\mu$ to ensure the martingale condition? Are there any mistakes in the above computation?