6
$\begingroup$

Recently I started taking Stochastic calculus class, and I am struggling still with computation of stochastic integrals and I would appreciate any help with the following example:

We define: $$ Y_t = (B_{t∧T_\epsilon})^\alpha e^{- \lambda \int_0^{t∧T_\epsilon}\frac{ds}{B_s^2}} $$ We defined $T_\epsilon = \inf\{t \geq 1, B_t = \epsilon\}$,$B=(B_t)_t$ is a one-dimensional Brownian motion starting from 1, i.e. $B_0 = 1$, and $\lambda > 0$ and $\alpha \ne 0$ are real numbers.

I have to compute $$\mathbb{E}[\exp(- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}) ]$$ We are allowed to assume that Y is a martingale and that $\alpha(\alpha - 1) = 2 \lambda.$

I don't know how to approach this problem or how to start at all. I would appreciate any help with it.

So far I have shown that $T_\epsilon$ is unique stopping time but more than that I couldn't do.

My attempt at an answer:

Since $T_\epsilon$ is a stopping time and $Y_t$ is a martingale, then by Optional Stopping Theorem we have: $$\mathbb{E}[Y_T]=\mathbb{E}[Y_0]$$ for any stopping time $T$.

We can easly comptute $\mathbb{E}[Y_0],$ precisely:

$$\mathbb{E}[Y_0]= (B_0)^{\alpha}e^{-\lambda \int_{0}^{0}\frac{1}{B_s^2}}=(1)^\alpha e^0=1$$ Since as initially stated: $B_0 = 1$ Now for $T=T_\epsilon$ a stopping time we have: $$\mathbb{E}[Y_T]=\mathbb{E}[(B_{T_\epsilon})^\alpha e^{- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}}] = \mathbb{E}[\epsilon^\alpha e^{- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}}] = \epsilon^\alpha \mathbb{E}[e^{- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}}]$$ And this implies that $$\mathbb{E}[e^{- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}}] = \frac{1}{\epsilon^\alpha}$$.

My only question left is: Does $Y_t$ fulfils all the requirement for Optional Stopping Theorem? I would assume that $T_\epsilon$ is finite a.s., and $(B_{t∧T_\epsilon})$ is a bounded martingale, hence close, but what about the other part?

$\endgroup$

1 Answer 1

1
$\begingroup$

The trick to being able to use the Optional Stopping Theorem here is to realize that, in the equation $\alpha (\alpha - 1) = 2\lambda$, we can always choose $\alpha < 0$. This implies $Y_t = (B_{t \wedge T_{\epsilon}})^{\alpha}e^{-\lambda \int_0^{t \wedge T_{\epsilon}} \frac{ds}{B_s^2}} \le \epsilon^{\alpha}$ for all $t$, and hence $Y_t$ is a uniformly integrable martingale.

$T_{\epsilon}$ is, as you said, finite almost surely by the recurrence properties of Brownian Motion.

Therefore, your use of the Optional Stopping Theorem is justified, and we do have $\mathbb{E}[e^{-\lambda \int_0^{T_{\epsilon}} \frac{ds}{B_s^2}}] = \epsilon^{-\alpha}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .