2
$\begingroup$

I am curious about a calculation I saw in this question.

Specifically in this equation:

\begin{align*} &\ L(T_s, T_p, T_e) L(T_s, T_s, T_e) \\ =&\ L(t_0, T_p, T_e) L(t_0, T_s, T_e) e^{-\frac{\sigma_s^2}{2}(T_s-t_0) -\frac{\sigma_p^2}{2}(T_s-t_0) + \sigma_s\big(W_{T_s}^s -W_{t_0}^s\big) + \sigma_p\Big(\rho \big(W_{T_s}^s - W_{t_0}^s\big) + \sqrt{1-\rho^2}\big(W_{T_s}^p - W_{t_0}^p\big)\Big)}. \end{align*}

I'm trying to prove it using Ito's lemma and the dynamics:

\begin{align*} dL(t, T_s, T_e) &= \sigma_s L(t, T_s, T_e) d W_t^s,\\ dL(t, T_p, T_e) &= \sigma_p L(t, T_p, T_e)d\Big(\rho W_t^s + \sqrt{1-\rho^2}W_t^p\Big), \end{align*}

But when I apply Ito's lemma I end up computing

\begin{align*} dL(t, T_s, T_e)dL(t, T_p, T_e) &= \sigma_s \sigma_p L(t, T_s, T_e)L(t, T_p, T_e) \rho dt \end{align*}

And I don't know where the next term is coming from:

\begin{align*} -\frac{\sigma_s^2}{2}(T_s-t_0)-\frac{\sigma_p^2}{2}(T_s-t_0) \end{align*}

Can someone help me with the calculation? Thanks

$\endgroup$
  • 1
    $\begingroup$ Apply Ito's lemma to $\ln\left( L(t, T_p, T_e) L(t, T_s, T_e) \right)$. By the properties of the logarithm you'll get: $$ d\ln\left( L(t, T_p, T_e) L(t, T_s, T_e) \right) = d \ln L(t, T_p, T_e) + d\ln L(t, T_s, T_e) $$ For each member on the RHS compute the differential using Ito again, the terms you mention are related to the quadratic variations of the processes. Once done simply integrate overtime from $t_0$ to $T_s$ to get the solution. $\endgroup$ – Quantuple Nov 28 '17 at 8:23
  • $\begingroup$ @Quantuple I saw your answer earlier but I wasn't being able to comment. Thank you, using the logarithm I was able to achieve the result. Now I remember that when solving a geometric brownian motion one also has to use it. Do you know why? in the case I was working I was trying to achieve the solution using a multivariate Ito's lemma to the product which doesn't seemed wrong to me, but I came up with a different solution when compared to using the logarithm. $\endgroup$ – Aldo Shumway Nov 29 '17 at 6:07
  • 1
    $\begingroup$ Maybe this helps: quant.stackexchange.com/questions/35770/…. Applying bivariate Itô to the product is fine, but you won't get a SDE that can be integrated easily to obtain the solution. You need the log trick to help you in your endeavour. $\endgroup$ – Quantuple Nov 29 '17 at 9:08
  • $\begingroup$ @Quantuple it helped indeed. Thanks I appreciate it $\endgroup$ – Aldo Shumway Nov 30 '17 at 0:47

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.