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. (independent of $N_t$) with normal distribution $N(m,\sigma^2)$. Determine $\mu>0$ such that $\exp(X_t-\mu t)$ is a martingale.
My attempt:
I compute: $E[e^{u X_t}] = e^{(\lambda t) \exp(u m + 0.5 u \sigma ^2)}$.
Then, $$\begin{align*} E[\exp{(X_t)} \mid F_s] &= E[\exp{(X_t - X_s + X_s)} \mid F_s] = E[\exp{(X_{t-s})}]e^{X_s} \\ &= e^{(\lambda (t-s)) \exp(u m + 0.5 u \sigma ^2)}e^{X_s}\end{align*}$$
So, $E[\exp{(X_t - \mu t)} \mid F_s] = e^{(\lambda (t-s)) \exp(u m + 0.5 u \sigma ^2)}e^{X_s}e^{-\mu t}$
So martingale condition holds if and only if
$e^{(\lambda (t-s)) \exp(u m + 0.5 u \sigma ^2)}e^{X_s}e^{-\mu t} = e^{X_s - \mu s}$
But I cannot find $\mu$ from this equation.
Note that $u=1$ in your calculation (since you need $\mathbb{E}\exp(X_{t-s})$ in your calculation), i.e.
$$\mathbb{E}(\exp(X_t) \mid \mathcal{F}_s) = e^{\lambda(t-s) \cdot \exp(m+ \frac{1}{2} \sigma^2)} e^{X_s}$$
Therefore the martingale condition boils down to
$$e^{\lambda(t-s) \cdot \exp(m+ \frac{1}{2} \sigma^2)} e^{X_s} e^{-\mu t} = e^{X_s} e^{-\mu s},$$
i.e.
$$e^{\lambda(t-s) \cdot \exp(m+ \frac{1}{2} \sigma^2)} = e^{\mu(t-s)}$$
Taking logarithm yields
$$\mu = \lambda \exp \left( m+ \frac{1}{2} \sigma^2 \right).$$