1
$\begingroup$

We are given the following SDE:

$dX_t=X_tdt+\sqrt{2}X_tdB_t, \quad X_0=1,$

and

$F(x,t)=e^{-t}x,\quad t\geq0,\; x\in\mathbb{R}.$

We are asked to apply Ito's formula to $F(t,X_t)$ for $t\geq0$ and determine a continuous local martingale $(M_t)_{t\geq0}$ (starting at $0$) and a continuous bounded variation process $(A_t)_{t\geq0}$ such that $F(t,X_t)=M_t+A_t$ for $t\geq0$.

If I am correct, $M_t=\int_0^tF_x(s,X_s)dX_s=\int_0^te^{-s}ds+\sqrt{2}\int_0^te^{-s}dB_s$, $t\geq0$

Now, we need to show that $M_t$ is a martingale and compute $\langle M,M\rangle_t$ and $\mathbb{E}[e^{-\tau}X_\tau]$ when $\tau=\inf\{t\geq0:X_t=2-t\}$ but I don't know how! Any help would be appreciated!

  • 0
    Thank you very much! The proof is a little demanding but it was really helpful!2012-12-01

1 Answers 1

0

It's not difficult to see that

$X_t := \exp \left(\sqrt{2} B_t \right)$

solves the given SDE. (You can either use Itô's formula to check it or use some standard methods for linear SDE's to obtain this solution.) Moreover, by Ito's formula:

$f(t,X_t)-\underbrace{f(0,x_0)}_{x_0} = \sqrt{2} \int_0^t e^{-s} \cdot X_s \, dB_s + \int_0^t e^{-s} \cdot X_s + (-e^{-s} \cdot X_s) \, ds \\ = \sqrt{2} \int_0^t e^{-s} \cdot e^{\sqrt{2} B_s} \,dB_s \\ \Rightarrow f(t,X_t) = \underbrace{\sqrt{2} \int_0^t e^{\sqrt{2} B_s-s} \,dB_s}_{M_t} + \underbrace{x_0}_{A_t}$

where $x_0=1$. Let

$g(s,w) := \sqrt{2} \cdot e^{\sqrt{2} B(s,w)-s}$

Then $g \in L^2(\lambda_T \otimes \mathbb{P})$, i.e.

$\int_0^T \int_\Omega g(s,w)^2 \, d\mathbb{P} \, ds <\infty$

There is a general result which says that this condition implies that $M_t$ is a martingale (and not only a local one). Moreover,

$\langle M,M \rangle_t = \int_0^t |g(s,w)|^2 \, ds$

(see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 14.13).

Concerning the integral $\mathbb{E}(e^{-\tau} \cdot X_\tau)$: Remark that

$\tau = \inf\{t \geq 0; X_t=2-t\} = \inf\{t \geq 0; \sqrt{2} B_t = \ln(2-t)\}$

Now let

$\sigma := 2\tau = \inf\{t \geq 0; \underbrace{\sqrt{2} B_{\frac{t}{2}}}_{=:W_t} = \ln (2-t/2)\}$

where $(W_t)_{t \geq 0}$ is again a Brownian Motion (scaling property). Thus

$\mathbb{E}(e^{-\tau} \cdot X_\tau) = \mathbb{E}(e^{-\tau+\sqrt{2} B_\tau}) = \mathbb{E}(e^{-\frac{\sigma}{2}+W_\sigma}) \stackrel{\ast}{=} 1$

In $(\ast)$ we applied the exponential Wald identity (see remark).

Remark Exponential Wald identity: Let $(W_t)_{t \geq 0}$ a Brownian motion and $\sigma$ a $\mathcal{F}_t^W$-stopping time such that $\mathbb{E}e^{\sigma/2}<\infty$, then $\mathbb{E}(e^{W_\sigma-\frac{\sigma}{2}})=1$. (see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 5.14)

  • 0
    Right, since $2\sqrt{2}B_s\tilde\ N(0,(2\sqrt{2})^2s)\ \Longrightarrow\ e^{2\sqrt{2}B_s}\tilde\ lnN(0,(2\sqrt{2})^2s)$2012-12-08