# An exponential martingale [closed]

Let $H_{t}$ be a bounded continuous and $\textbf{F}^{B}_{t}$ an adapted process. $B$ Brownian motion.

Show that $M_{t}= \exp\left(-\int^{t}_{0}H_{s}dB_{s} -\frac{1}{2}\int^{t}_{0}H^{2}_{s}ds\right)$ is a martingale. Find $M_{t}$ quadratic variation and $\left\langle M,B\right\rangle_{t}$

Some help would be appreciated.

• i compute $\textbf{E}(M_{t}|\textbf{F}_{s})$ but i cant get $M_{s}$
– John
Jun 22, 2014 at 16:34
• Computing the conditional expectation $\mathbb{E}(M_t \mid \mathcal{F}_s)$ is rather difficult. Try instead to apply Itô's formula in order to show that $M_t -M_0 = \int_0^t N_s \, dB_s$ for some process $(N_s)$. This implies automatically that $(M_t)$ is a (local) martingale.
– saz
Jun 22, 2014 at 16:38
• ok, I understand
– John
Jun 22, 2014 at 17:25
• Isn't this a standard application exercise aimed at helping you check that you master the notions of the corresponding chapter? If it is, to get freely a full solution here (instead of chewing this up a little bit by yourself) is to run the danger of NOT acquiring the content of this chapter.
– Did
Aug 22, 2014 at 12:10

Note that

$$X_t := - \int_0^t H_s \, dB_s - \frac{1}{2} \int_0^t H_s^2 \, ds$$

defines an Itô process. Therefore we may apply Itô's formula to $f(x) := e^x$:

\begin{align*} f(X_t)-f(X_0) &= \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s \\ \Leftrightarrow M_t - 1 &= - \int_0^t M_s H_s \, dB_s - \frac{1}{2} \int_0^t M_s H_s^2 \, ds + \frac{1}{2} \int_0^t M_s H_s^2 \, ds \\ &= -\int_0^t M_s H_s \, dB_s \tag{1} \end{align*}

Here we have used that $M_t = f(X_t) = f'(X_t)=f''(X_t)$ and that the quadratic variation process $\langle X \rangle_t$ is given by $$\langle X \rangle_t = \int_0^t H_s^2 \, ds.$$

Since $H$ is bounded, the right-hand side of $(1)$ is a martingale. Consequently, $(M_t)_{t \geq 0}$ is a martingale. On the other hand, if we set $g(x) := e^{2x}$, then $g(X_t)= M_t^2$ and a similar calculation yields

$$M_t^2 - 1 = -2 \int_0^t M_s^2 H_s \, dB_s + \int_0^t M_s^2 H_s^2 \, ds.$$

Therefore, it follows from the definition of quadratic variation that

$$\langle M \rangle_t = \int_0^t M_s^2 H_s^2 \, ds.$$

Moreover, $(1)$ shows

$$\langle M,B \rangle_t = -\int_0^t M_s H_s \, ds.$$