1. **State the problem:** We want to determine if the stochastic process $$S_t = S_0 e^{\mu t + \sigma W_t}$$ is a martingale under the risk-neutral measure, where $$W_t$$ is a standard Brownian motion. 2. **Recall the definition of a martingale:** A process $$M_t$$ adapted to a filtration $$\mathcal{F}_t$$ is a martingale if $$\mathbb{E}[M_t | \mathcal{F}_s] = M_s$$ for all $$t \geq s$$. 3. **Risk-neutral measure and drift adjustment:** Under the risk-neutral measure $$\mathbb{Q}$$, the drift term $$\mu$$ in the stock price process is replaced by the risk-free rate $$r$$ to reflect no-arbitrage pricing. Hence, the process becomes: $$S_t = S_0 e^{rt + \sigma W_t}$$ 4. **Express the expectation:** To test the martingale property, we compute $$\mathbb{E}^\mathbb{Q}[S_t | \mathcal{F}_s] = \mathbb{E}^\mathbb{Q}\left[S_0 e^{rt + \sigma W_t} | \mathcal{F}_s \right].$$ Write $$W_t = W_s + (W_t - W_s)$$ where $$W_t - W_s$$ is independent of $$\mathcal{F}_s$$ and normally distributed with mean zero and variance $$t - s$$. 5. **Calculate the conditional expectation:** $$\mathbb{E}^\mathbb{Q}[S_t | \mathcal{F}_s] = S_0 e^{rt + \sigma W_s} \mathbb{E}^\mathbb{Q}[e^{\sigma (W_t - W_s)}]$$ Since $$W_t - W_s \sim N(0, t-s)$$, $$\mathbb{E}^\mathbb{Q}[e^{\sigma (W_t - W_s)}] = e^{\frac{1}{2} \sigma^2 (t-s)}$$ 6. **Substitute back:** $$\mathbb{E}^\mathbb{Q}[S_t | \mathcal{F}_s] = S_0 e^{rt + \sigma W_s} e^{\frac{1}{2} \sigma^2 (t-s)} = S_0 e^{r t + \sigma W_s + \frac{1}{2} \sigma^2 (t-s)}$$ 7. **Check if it equals $$S_s$$:** The value of $$S_s$$ is $$S_s = S_0 e^{r s + \sigma W_s}$$ 8. **Compare and conclude:** For the conditional expectation to equal $$S_s$$, we would require $$S_0 e^{r t + \sigma W_s + \frac{1}{2} \sigma^2 (t-s)} = S_0 e^{r s + \sigma W_s}$$ Simplify by cancelling $$S_0 e^{\sigma W_s}$$: $$e^{r t + \frac{1}{2} \sigma^2 (t-s)} = e^{r s}$$ Taking logarithms: $$r t + \frac{1}{2} \sigma^2 (t-s) = r s$$ Rearranging: $$r(t - s) + \frac{1}{2} \sigma^2 (t-s) = 0$$ Which implies $$\left(r + \frac{1}{2} \sigma^2\right) (t-s) = 0$$ Since $$t > s$$, this requires $$r + \frac{1}{2} \sigma^2 = 0$$ which generally does **not** hold. 9. **Conclusion:** The process $$S_t = S_0 e^{rt + \sigma W_t}$$ is **not** a martingale because its conditional expectation is not equal to its current value unless a special condition on $$r$$ and $$\sigma$$ holds, which is unusual. 10. **Martingale adjustment (correct process):** The standard risk-neutral martingale process is $$S_t = S_0 e^{(r - \frac{1}{2} \sigma^2 )t + \sigma W_t}$$ which removes the extra term and has mean adjusted accordingly to satisfy $$\mathbb{E}^\mathbb{Q}[S_t | \mathcal{F}_s] = S_s$$. Thus, the initial process given is not a martingale but can be turned into one by adjusting the drift. **Final answer:** Under the risk-neutral measure, the given process is not a martingale as stated because it lacks the drift adjustment term $$- \frac{1}{2} \sigma^2$$ needed to correct for the exponential Brownian motion variance term.