3
$\begingroup$

Consider two Brownian motions $$X_{1t}=\mu t+\sigma_1B_{1t}$$ and $$X_{2t}=\mu t+\sigma_2B_{2t}.$$ Here $B_{1t}$ and $B_{2t}$ are uncorrelated. Let $\tau_1$ and $\tau_2$ be the stopping times: \begin{eqnarray} \tau_i=\inf\{t>0:X_{it}=-x_i\} \end{eqnarray}

I'm looking for the following expectation $\mathbb{E}_{x_1,x_2}[X_{2\tau_1}\mathbb{1}_{\{\tau_1<\tau_2\}}]$. I have started with the following \begin{eqnarray} \mathbb{E}_{x_1,x_2}[X_{2\tau_1}\mathbb{1}_{\{\tau_1<\tau_2\}}]&=&\mathbb{E}_{x_1,x_2}[(\mu \tau_1+\sigma_2B_{2\tau_1})\mathbb{1}_{\{\tau_1<\tau_2\}}]=\mu\mathbb{E}_{x_1,x_2}[\tau_1\mathbb{1}_{\{\tau_1<\tau_2\}}]+\sigma_2\mathbb{E}_{x_1,x_2}[B_{2\tau_1}\mathbb{1}_{\{\tau_1<\tau_2\}}], \end{eqnarray} where, I guess, based on hitting time distributions the first term could be derived using this \begin{eqnarray}\mathbb{E}_{x_1,x_2}\left[ \tau_1 \mathbb{1}_{\tau_1<\tau_2}\right]&=& \int^{\infty}_{0}t \frac{x_1}{\sigma_1\sqrt{2 \pi t^3}}\mathrm{e}^{-\frac{(x_1+\mu t)^2}{2 \sigma_1^2 t}} dt \int^{\infty}_{t} \frac{x_2}{\sigma_2\sqrt{2 \pi s^3}}\mathrm{e}^{-\frac{(x_2+\mu t)^2}{2 \sigma_2^2 s}} ds. \\&=&\int^{\infty}_{0} \frac{ x_1}{\sigma_1\sqrt{2\pi t}}e^{- \frac{(x_1+\mu t)^2}{2\sigma^2_1t}}\left(\Phi\left(\frac{x_2+\mu t}{\sigma_2 \sqrt{t}}\right)-\mathrm{e}^{-\frac{2\mu x_2 }{\sigma_2^2}} \Phi\left(\frac{-x_2+\mu t}{\sigma_2 \sqrt{t}}\right)\right)dt, \notag \end{eqnarray} with $\Phi$ is the standard normal CDF.

Now I'm not entirely sure what to do with the second term, namely, $\mathbb{E}_{x_1,x_2}[B_{2\tau_1}\mathbb{1}_{\{\tau_1<\tau_2\}}]$. Could anyone help?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.