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Let $X_t$be the solution to the SDE:

$dX_t=-X_tdt+dB_t$, $X_0=0$

Then $X_t$ is the Ornstein–Uhlenbeck process $X_t=e^{-t}\int_0^te^sdB_s$.

I want to calculate $\mathbb{E}[e^\tau X_\tau]$ when $\tau=\inf\{t\geq0:|X_t|=1-t^2\}$.

How can we treat the integral $\mathbb{E}[\int_0^\tau e^sdB_s]$?

My attempt for a solution:

We need to calculate $\mathbb{E}[e^{-\tau}X_\tau]=\mathbb{E}[\int_0^\tau e^sdB_s]$. If we can show that $M_t=\int_0^t e^sdB_s$ is a martingale, then we may be able to use the Optional Sampling Theorem.

Observe that: $\mathbb{E}\left<M,M\right>_t=\mathbb{E}\int_0^te^{2s}ds=\int_0^t\mathbb{E}[e^{2s}]ds=\int_0^te^{2s}ds=\frac{1}{2}e^{2t}-\frac{1}{2}<\infty$

Hence, $M$ is a martingale and so $\mathbb{E}[M_t]=\mathbb{E}[M_0]=0$

If we can show that $\tau$ is bounded, then from the BDG inequality, we can show that $\tau$ is optional for $M$. So the question is, is $\tau$ bounded?

If we set $g(t)=|X_t|-1+t^2$, then $g(0)=-1<0$ and $g(1)=|X_1|>0$. From the Intermediate value Theorem, there exists $t_0>0$ such that $g(t_0)=0$.

So $\tau$ must be bounded. Is that right?

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1 Answer 1

up vote 0 down vote accepted

Yes, that's correct (and really nice proof, by the way...). But why do you need BDG inequality? Since $\tau \leq 1$ almost surely and $M_{\tau} \in L^1$ you can apply the "usual" optional stopping theorem..

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I'm not quite sure why $M_\tau\in L^1$... If we have that $M_\tau$ is integrable, then the BDG inequality is indeed not needed. –  Nick Papadopoulos Dec 10 '12 at 22:23
    
$|X_{\tau}|=1- \tau^2 \leq 1$ and therefore $|M_{\tau}| = e^{\tau} \cdot |X_{\tau}| \leq e \cdot 1$ using that $\tau \leq 1$ a.s. –  saz Dec 11 '12 at 6:52

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