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.
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.