Let $\{ W(t) \}_{t\in \mathbb [0,\infty)}$ be a Wiener-Process and define $$ X(t) := e^{\lambda t + \sigma W(t)} $$ for $\lambda, \sigma \in \mathbb R$. I want to know under what conditions on $\lambda$ and $\sigma$, the above process is a sub- or a supermartingale for the canonical filtration $\mathfrak F^W$ on $W(t)$?
Using the law of the unconscious statistician I find that $E[X] = \exp\left( \lambda t + \frac{\sigma^2 t}{2} \right)$. But I do not know how to compute the conditional expectation $E[X(t) \mid \mathfrak F^W(s)]$ in general for $s \le t$?
I know that $E[W(t) \mid \mathfrak F^W(s)] = E[W(t) - W(s) + W(s) \mid \mathfrak F^W(s)] = (t-s) + W(s)$ by independence. If $f : \mathbb R \to \mathbb R$ is convex, we have $f(E[W(t) \mid \mathfrak F^W(s)]) \le E(f(W(t)) \mid \mathfrak F^W(s))$ which gives by the above $e^{\lambda t + \sigma((t-s) + W(s))} \le E(e^{\lambda t + \sigma W(t)} \mid \mathfrak F(s))$, i.e. $$ e^{(\lambda + \sigma)(t - s)} e^{\lambda s + \sigma W(s)} \le E(e^{\lambda t + \sigma W(t)} \mid \mathfrak F(s)) $$ hence if $(e^{(\lambda + \sigma)})^{t-s} \ge 1$ this is a submartingale, which is fulfilled if $\lambda + \sigma \ge 0$. So a partial result is that if $\lambda + \sigma \ge 0$, the process is a submartingale. But I do not know what happens if $\lambda + \sigma < 0$, and when the process is a supermartingale?
Using the hint of @zhoraster, we have \begin{align*} E[X(t) \mid \mathfrak F(s)] & = \exp(\lambda t + \sigma W(s)) E[\exp(\sigma(W(t) - W(s)) \mid \mathfrak F(s) ] \\ & = \exp(\lambda t + \sigma W(s)) E[\exp(\sigma(W(t) - W(s))] \end{align*} by independence. In the following we use the fact that if $X \sim \mathcal N(0,1)$, then $$ E[e^{\sigma X}] = e^{\frac{\sigma^2}{2}}. $$ As by the law of the unconscious statistician (and completing the square) we have \begin{align*} E[e^{\sigma X}] & = \int_{-\infty}^{\infty} e^{\sigma z} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} ~ d z \\ & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{\sigma z - z^2/2} ~ d z \\ & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}\left(\left(z - \sigma\right)^2 - \sigma^2\right)} ~ dz \\ & = e^{\frac{\sigma^2}{2}}\cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}\left(z - \sigma\right)^2} ~ d z \\ & = e^{\frac{\sigma^2}{2}}. \end{align*} Using this we find $$ E[e^{\sigma(W(t) - W(s))}] = E\left[e^{\sigma \sqrt{t - s} \frac{W(t) - W(s)}{\sqrt{t-s}}}\right] = e^{\frac{\sigma^2(t-s)}{2}}. $$ Hence \begin{align*} E[X(t) \mid \mathfrak F(s)] & = e^{\lambda t + \sigma W(s)} e^{\frac{\sigma^2(t-s)}{2}} \\ & = e^{\lambda s + \sigma W(s)} e^{\lambda(t-s) e^{\frac{\sigma^2(t-s)}{2}}} \\ & = e^{\lambda s + \sigma W(s)} e^{(t-s)(\lambda + \sigma^2/2)} \end{align*} And so we have a supermartingale if $\lambda + \sigma^2/2 \le 0$, and a submartingale if $\lambda + \sigma^2/2 \ge 0$.