1
$\begingroup$

how can I solve $\rho$ from the following:

$\int_0^T dV_t = \int_0^T \kappa (\theta - V_t) dt + \int_0^T \sigma \rho \sqrt{V_t} dW_t + \int_0^T \sigma \sqrt{1-\rho^2} \sqrt{V_t} dZ_t$,

where $W_t$ and $Z_t$ are uncorrelated Brownian motions and I know $V_t$ and all other paramaters i.e. $\sigma$, $\kappa$, $\theta$.

Thanks.

$\endgroup$
1
$\begingroup$

Note that $\mathrm d\langle V,W\rangle_t=\sigma\rho\sqrt{V_t}\mathrm dt$ hence, for every $t\gt0$, $$\rho=\frac1{\sigma t}\int_0^t\frac{\mathrm d\langle V,W\rangle_s}{\sqrt{V_s}}.$$ Likewise, for every $t\gt0$, $$\sqrt{1-\rho^2}=\frac1{\sigma t}\int_0^t\frac{\mathrm d\langle V,Z\rangle_s}{\sqrt{V_s}}.$$ Edit: Numerically, the integral involving the co-variation of $V$ and $W$ in the first formula translates into $$\int_0^t\frac{\mathrm d\langle V,W\rangle_s}{\sqrt{V_s}}\approx\sum_{k=1}^{n}\frac1{\sqrt{V_{t_k}}}\left(V_{t_k}-V_{t_{k-1}}\right)\left(W_{t_k}-W_{t_{k-1}}\right),$$ when the mesh $\|P\|$ of the partition $P=(t_k)_{0\leqslant k\leqslant n}$ of the interval $[0,t]$ goes to zero.

$\endgroup$
  • $\begingroup$ Maybe my question was unclear. I need a formula to get a numerical result of $\rho$. Or can I evaluate $d<V,W>_s$ numerically? $\endgroup$ – SchnitzelRaver Aug 31 '14 at 6:47
  • $\begingroup$ Sure, using the standard numerical scheme for co-variation processes, see Edit. $\endgroup$ – Did Aug 31 '14 at 7:17
  • $\begingroup$ Yes, but I do not observe the Brownian itself, hence I don't know its increment $ W_t_k - W_t_{k-1} $. $\endgroup$ – SchnitzelRaver Aug 31 '14 at 11:06
  • $\begingroup$ With each comment, we are facing more starkly the fact that you did not ask the question that is interesting you. Having no interest for a career in divination, and unless you decide to state clearly what your problem is, I think it is best for me to leave things as they are. $\endgroup$ – Did Aug 31 '14 at 11:11
  • $\begingroup$ I have found another way to numerically solve $\rho$. Thanks anyways. $\endgroup$ – SchnitzelRaver Aug 31 '14 at 12:27

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.