Conditional expectation equation containing martingales

Question

Say $M_t$ is a process such that $e^{i\lambda M_t+\frac12\lambda^2t}$ is a martingale with respect to the filtration $(\mathcal{F}_t)_{t\ge 0}$. By definition $E[e^{i\lambda M_t+\frac12\lambda^2t}|\mathcal{F}_s]=e^{i\lambda M_s+\frac12\lambda^2t}$. Apparently it is trivial to show from this that $E[e^{i\lambda (M_t-M_s)}|\mathcal{F}_s]=e^{-\frac12\lambda^2(t-s)}$ but I don't see it since $E\left[e^{i\lambda (M_t-M_s)}|\mathcal{F}_s\right]=E\left[e^{i\lambda M_t}\cdot e^{-i\lambda M_s}|\mathcal{F}_s\right]$ but then we cannot use linearity.

Answer 1

Actually, the definition of martingale gives $$\tag{*} \mathbb E\left[e^{i\lambda M_t+\frac12\lambda^2t}|\mathcal{F}_s\right]=e^{i\lambda M_s+\frac12\lambda^2s}. $$ We have $$ \mathbb E\left[e^{i\lambda (M_t-M_s)}|\mathcal{F}_s\right]=\mathbb E\left[e^{i\lambda M_t}e^{-i\lambda M_s}|\mathcal{F}_s\right]=e^{-i\lambda M_s}\mathbb E\left[e^{i\lambda M_t}|\mathcal{F}_s\right] $$ and $$ \mathbb E\left[e^{i\lambda M_t}|\mathcal{F}_s\right]=\mathbb E\left[e^{i\lambda M_t+\frac12\lambda^2t}|\mathcal{F}_s\right]e^{-\frac12\lambda^2t} $$ and (*) gives the wanted result.