Prove stochastic exponential to be a martingale

Question

I'm new in stochastic integral, and recently I met a problem as following:

Let $B$ be a Brownian motion, $\mu_t,\sigma_t$ be uniformly bounded progressive processes and $\sigma_t>\epsilon>0$ for every t. I want to show the local martingale

\begin{equation} X_t=\exp\left(-\frac{1}{2}\int_0^t\sigma_s^2ds+\int_0^t\sigma_sdB_s \right) \end{equation}

to be indeed a martingale.

I know the result that for a local martingale to be a martingale if it is of the class $DL$. I wanted to show that there exists a $p>1$ such that $\sup_{\tau\leq T}E[X_{\tau}^p]<\infty$ for every fixed $T>0$, but unfortunately I failed to prove it. The main difficulty I met is that I found it hard to describe all the stopping time $\tau$ to find the supreme value of these expectations.

So how can I prove this result?

Answer 1

A possible approach here would be to use some basic Itô calculus.

If we compute the stochastic differential of $X_t$ using Itô's chain rule , we get $$ d X_t = X_t \left( -\frac 12 \sigma_t^2 dt + \sigma_t dB_t \right) + \frac 12 X_t \sigma_t^2 dt = X_t \sigma_t dB_t. $$ Hence the differential of $X_t$ has no drift, implying that $X_t$ is an Itô integral, i.e. $$ X_t = \int_0^t X_s \sigma_s d B_s , $$ which is well-known to be a martingale.

Answer 2

Well...I worked it out by myself...

It is sufficient to prove the class of random variables $\mathcal{C}_T=\{X_\tau:\tau\leq T\}$ is uniformly integrable for any fixed $T>0$. In fact, since $\sigma_t$ is uniformly bounded, we can find an $M>0$ such that $0\leq\sigma_t\leq M$ for all $0\leq t\leq T$. Let's construct the process $W_t=\sup_{s\leq t}B_s$, by reflection principle we know $$ P[\sup_{s\leq T}B_s\geq\frac{K}{M}]=2P[B_T\geq\frac{K}{M}] $$ then it has probability density function $$ f(W_t)=\frac{2}{\sqrt{2\pi t}}e^{W_t^2/2t}I_{W\geq0} $$ and by definition it has the following property: \begin{align*} W_t\geq W_s\text{ whenever } t\geq s & \Rightarrow dW_t\geq 0;\\ W_t\geq B_t & \Rightarrow dW_t\geq dB_t \end{align*} Now to prove $\mathcal{C}_T$ is u.i., Denote $Y_t=\int_0^t\sigma_sdB_s$, it is clear to see that $X=\mathcal{E}(Y)$. For every $T<\infty$, consider \begin{align*} \sup_{\tau\leq T}E[\exp(\frac{1}{2}Y_{\tau})]&=\sup_{\tau\leq T}E[e^{\frac{1}{2}\int_0^{\tau}\sigma_sdB_s}]\\ &\leq \sup_{\tau\leq T}E[e^{\frac{1}{2}\int_0^{\tau}\sigma_sdW_s}]\\ &\leq\sup_{\tau\leq T}E[e^{\frac{1}{2}M\int_0^{\tau}dW_s}]\\ &\leq E[e^{\frac{1}{2}MW_{T}}]\\ &=\int_0^{\infty}e^{\frac{M}{2}x}\frac{2}{\sqrt{2\pi T}}e^{-\frac{x^2}{2T}}dx\\ &=2e^{\frac{M^2T}{8}}\int_0^{\infty}\frac{1}{\sqrt{2\pi T}}e^{-\frac{(x-MT/2)^2}{2T}}dx\\ &\leq 2e^{\frac{M^2T}{8}}P[B_T\geq-\frac{MT}{2}]<\infty \end{align*} By Novikov's theorem we know $\mathcal{C}_T$ is u.i.