Is $(e^{i \lambda B_t + \frac{1}{2}\lambda^2t})_{t\geq 0}$ a martingale?

Question

Showing that $(e^{\lambda B_t - \frac{1}{2}\lambda^2t})_{t\geq 0}$ is a $\mathbb{R}$-valued martingale

Let $B$ be a standard $\mathbb{R}$-valued Brownian motion and let $\lambda\in\mathbb{R}$. From $B_t-B_s$ being independent of $\mathcal{F}_s$ and $B_t-B_s \sim \mathcal{N}\left(0, t-s\right)$ follows $\mathbb{E}\left[e^{\lambda\left(B_t - B_s\right)}\vert \mathcal{F}_s\right] = \mathbb{E}\left[e^{\lambda\left(B_t - B_s\right)}\right] = e^{\frac{1}{2}\lambda^2\left(t-s\right)}$. Because of $B_t = B_t - B_0 \sim \mathcal{N}\left(0, t\right)$ we also have $\mathbb{E}\left[e^{\lambda B_t}\right] = e^{\frac{1}{2}\lambda^2 t}$.

1. $(e^{\lambda B_t - \frac{1}{2} \lambda^2 t})_{t\geq 0}$ is an adapted process because $B$ is.
2. $e^{\lambda B_t - \frac{1}{2}\lambda^2t}$ is integrable: $$\mathbb{E}\left[\left\lvert e^{\lambda B_t - \frac{1}{2}\lambda^2t}\right\rvert\right] = \mathbb{E}\left[e^{\lambda B_t}\right]\cdot e^{-\frac{1}{2}\lambda^2t} = 1 < \infty\ \text{(a.s.)}$$
3. The fact that $B_s$ is $\mathcal{F}_s$-measurable yields the martingale condition:

\begin{align*} &\mathbb{E}\left[e^{\lambda B_t - \frac{1}{2}\lambda^2 t} \vert \mathcal{F}_s\right]\\ =\ &\mathbb{E}\left[e^{\lambda \left(B_t - B_s\right) + \lambda B_s - \frac{1}{2}\lambda^2 t} \vert \mathcal{F}_s\right]\\ =\ &\mathbb{E}\left[e^{\lambda \left(B_t - B_s\right)} \vert \mathcal{F}_s\right]\cdot e^{\lambda B_s - \frac{1}{2}\lambda^2 t}\\ =\ &e^{\frac{1}{2}\lambda^2\left(t-s\right) + \lambda B_s - \frac{1}{2}\lambda^2 t}\\ =\ &e^{\lambda B_s - \frac{1}{2}\lambda^2 s}\ \text{(a.s.)} \end{align*}

(Edited) Attempt at showing that $(e^{i \lambda B_t + \frac{1}{2}\lambda^2t})_{t\geq 0}$ is a $\mathbb{C}$-valued martingale

Again, let $B$ be a standard $\mathbb{R}$-valued Brownian motion and let $\lambda\in\mathbb{R}$.

1. $\left(e^{i \lambda B_t + \frac{1}{2}\lambda^2t}\right)_{t\geq 0}$ is an adapted process because $B$ is.
2. Recall that $\left\lvert e^{ix}\right\rvert = 1$ for all $x\in\mathbb{R}$. $e^{i \lambda B_t + \frac{1}{2}\lambda^2t}$ is integrable: $$\mathbb{E}\left[\left\lvert e^{i\lambda B_t + \frac{1}{2}\lambda^2t}\right\rvert\right] = \mathbb{E}\left[\left\lvert e^{i\lambda B_t}\right\rvert \cdot \left\lvert e^{\frac{1}{2}\lambda^2t}\right\rvert\right] = e^{\frac{1}{2}\lambda^2t} < \infty\ \text{(a.s.)}$$
3. Assuming that $\mathbb{E}\left[e^{c\left(B_t - B_s\right)}\right] = e^{\frac{1}{2}c^2\left(t-s\right)}$ also holds for $c\in\mathbb{C}$, keeping in mind that $(i\lambda)^2 = -\lambda^2$ and using the fact that $B_s$ is $\mathcal{F}_s$-measurable yields the martingale condition: \begin{align*} &\mathbb{E}\left[e^{i\lambda B_t + \frac{1}{2}\lambda^2 t} \vert \mathcal{F}_s\right]\\ =\ &\mathbb{E}\left[e^{i\lambda \left(B_t - B_s\right) + i\lambda B_s + \frac{1}{2}\lambda^2 t} \vert \mathcal{F}_s\right]\\ =\ &\mathbb{E}\left[e^{i\lambda \left(B_t - B_s\right)} \vert \mathcal{F}_s\right]\cdot e^{i\lambda B_s + \frac{1}{2}\lambda^2 t}\\ =\ &e^{-\frac{1}{2}\lambda^2\left(t-s\right) + i\lambda B_s + \frac{1}{2}\lambda^2 t}\\ =\ &e^{i\lambda B_s + \frac{1}{2}\lambda^2 s}\ \text{(a.s.)} \end{align*}

Questions

Does $\mathbb{E}\left[e^{c\left(B_t - B_s\right)}\right] = e^{\frac{1}{2}c^2\left(t-s\right)}$ also hold for $c\in\mathbb{C}$? If yes, is my work correct?

Edit

I corrected a missed sign change in my attempt.

Answer 1

$\mathbb{E}\left[e^{c\left(B_t - B_s\right)}\right] = e^{\frac{1}{2}c^2\left(t-s\right)}$ does hold for $c\in\mathbb{C}$. The proofs are correct.