How to show that $X_t = (1+t)^{-1/2} \exp \biggl( \frac{B_t^2}{2(1+t)} \biggr)$ for a Brownian motion $B$

Question

Show that $X_t = (1+t)^{-1/2} \exp \biggl( \frac{B_t^2}{2(1+t)} \biggr)$, where $B$ is a Brownian motion, is a martingal.

I understand that we need to show that $\mathbb{E}(X_t \vert \mathcal{F}_s) = X_s$ for all $t \ge s \ge 0$, where $\mathcal{F}$ is the canonical filtration of $X$. However, I do not see how to do this computation as we have no density function here. Could you please explain this to me?

Answer 1

You can explicitly compute the filtered expectation value, given the distribution of the increments of the Brownian motion $W_{st}=B_t-B_s$ and their independence from the increment $W_{0s}=B_s$:

$$\mathbb{E}[X_t|B_{T}, 0\leq T\leq s]=\mathbb{E}\left[(1+t)^{-1/2}\exp\left(\frac{(B_s+W_{st})^2}{2(1+t)}\right)\Bigg|B_s\right]\\=\frac{1}{\sqrt{2\pi(1+t)(t-s)}}\int_{-\infty}^{\infty}dx \exp[(B_s+x)^2/2(1+t)]\exp[-x^2/2(t-s)]$$

This integral is Gaussian. Can you take it from here?

Answer 2

Let $f(t,x)=(1+t)^{-1/2}\exp(\frac{x^2}{2(1+t)})$. Using Itô's formula, $$ df(t,B_t)=\left[\frac{\partial f}{\partial t}(t,B_t)+\frac12\frac{\partial^2 f}{\partial x^2}(t,B_t)\right]dt+\frac{\partial f}{\partial x}(t,B_t)dB_t $$ To prove $f(t,B_t)$ is a Martingale, you just need to show the $dt$ part is zero, which only requires deterministic calculus.