Is this local martingale a true martingale?

Question

Using the Ito's formula I have shown that $X_t$ is a local martingale, because $dX_t=\dots dB_t$, where

$$X_t = (B_t+t)\exp\left(-B_t-\frac{t}{2}\right),$$ $B_t$ - is a standard Brownian motion

I would like to show it is a true martingale, so I am looking at these sets:

$$\mathcal{S}_1:=\{ X^{T_n}_t : n\geq 1\} \text{ or }\mathcal{S}_2:=\{X_T : \text{ T is a bounded stopping time} \}$$ And trying to show that either of them is UI. ($T_n$ are the stopping times reducing $X_t$)

I need some help with this step.

Edit

This is a homework exercise, which stipulates the usage of Ito's formula

Answer 1

let $f(x,t):= \exp{(-(x+t))}(x+t)$, then we are allowed to apply Itô's formula. We just need to calculate the following derivatives: $\frac{\partial f}{\partial x},\frac{\partial f}{\partial t},\frac{\partial^2 f}{\partial t^2}$ since the second component of $f$ is continuous and increasing, thus has finite variation and any continuous function of finite variation has zero quadratic variation (and you have to use Cauchy-Schwarz.)

Here's what I get:

$\frac{\partial f}{\partial x} = \exp{(-(x+t))}-\exp{(-(x+t))}(t+x)$

$\frac{\partial f}{\partial t} = \exp{(-(x+t))}-\frac{1}{2}\exp{(-(x+t))}(t+x)$

$\frac{\partial^2 f}{\partial t^2} = -2\exp{(-(x+t))}+\exp{(-(x+t))}(t+x)$

Therefore:

$f(B_s,t)=\int_0^t{[\exp{(-(B_s+s))}-\exp{(-(B_s+s))}(s+B_s)]dB_s}\\ + \int_0^t{[\exp{(-(B_s+s))}-\frac{1}{2}\exp{(-(B_s+s))}(s+B_s)]}ds +\int_0^t{[-\exp{(-(B_s+s))}+\frac{1}{2}\exp{(-(B_s+s))}(s+B_s)]d\langle B_s,B_s\rangle}= \int_0^t{[\exp{(-(B_s+s))}-\exp{(-(B_s+s))}(s+B_s)]dB_s}\\ + \int_0^t{[\exp{(-(B_s+s))}-\frac{1}{2}\exp{(-(B_s+s))}(s+B_s)]}ds +\int_0^t{[-\exp{(-(B_s+s))}+\frac{1}{2}\exp{(-(B_s+s))}(s+B_s)]ds}$

Comparing the $ds$ integral, this leads to:

$$f(B_s,t)=\int_0^t{[\exp{(-(B_s+s))}-\exp{(-(B_s+s))}(s+B_s)]dB_s}$$

Now I use the following Theorem, which can be found on this exercise sheet, Exercise 1a): http://www.math.ethz.ch/%7Egruppe3/HS11/MFF/MFF_2011_exercise08.pdf

And apply it to $M_t = B_t$ and $H_s=\exp{(-(B_s+s))}-\exp{(-(B_s+s))}(s+B_s)$

cheers

math

Answer 2

It seems to me that it is easy to show directly that $X_t$ is a martingale by verifying that $E[X_t \mid \mathcal{F}_s] = X_s$. (Here I assume that $B_t$ is a Brownian motion with respect to the filtration $\mathcal{F}_t$, and that you are trying to show $X_t$ is a martingale with respect to the same filtration.) One just writes $B_t = B_s + (B_t - B_s)$ and uses independence of increments. It helps to check that, for $N \sim N(0, \sigma^2)$ we have $$E[e^{-N}] = e^{\sigma^2/2}, \quad E[N e^{-N}] = -\sigma^2 e^{\sigma^2/2}.$$