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Going through this article on Heston's model, where the variance evolves following the SDE \begin{equation} \label{sd1} d\sigma^2_t = \kappa \bigg( m - \color{red}{\sigma^2_t} \bigg)dt + \nu \sqrt {\sigma^2_t} dW_t \end{equation} with $\kappa, m, \nu$ being constants, and $W_t$ a Brownian Motion (corrected errata shown in red).

the author defines \begin{equation} \label{sd} M_t := \int_0^T \mathbb{E}[\sigma^2_s \vert \mathcal{F}_t ] ds \end{equation}

and then proceeds to claim (without further details) that \begin{equation} \label{sd2} dM_t = \nu \sqrt {\sigma^2_t} \bigg( \int_t^T \exp[-\kappa(s-t)] ds \bigg)dW_t \end{equation}

How can one use Itô's lemma to compute the differential? I thought about first defining $X_t := \mathbb{E}[\sigma^2_s \vert \mathcal{F}_t ]$ and computing $dX_t$, but I don't really know how to proceed.

Thanks for reading

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That is not the SDE for the Heston model - it violates the affine property in the drift term. In other words, the paper has a typo. The correct SDE is:

$$ d v_t = \kappa (m-v_t) dt + \nu \sqrt{v_t} dw_t $$ where $v_t := \sigma_t^2$ is the variance.

Let $\xi_t^T := \mathbb{E}_t [ v_T]$ denote the forward variance and see that

$$ \begin{align} \xi_{t}^{T} & = \mathbb{E}_t [ v_T] \\ & = \mathbb{E}_t \left[ v_t + \int_{t}^{T} \kappa (m-v_u) du + \int_{t}^{T} \nu \sqrt{v_u} dw_u \right] \\ & = v_t + \int_{t}^{T} \kappa (m- \xi_t^u ) d u \end{align} $$ In differential form (with respect to $T$) $$ d \xi_t^T = k (m-\xi_t^T) dT $$ Using the integrating factor method yields $\xi$ to be $$ \xi_t^T = m + e^{-\kappa (T-t)} ( \xi_t^t - m) $$ In differential format (with respect to $t$) $$ d \xi_t^T = e^{-\kappa (T-t)} \nu \sqrt{\xi_t^t} dw_t $$ Therefore the differential for $M$ (with respect to $t$) is $$ d M_t = \int_{0}^{T} d \xi_t^s ds = \nu \sqrt{v_t} \left[ \int_{0}^{T} e^{-\kappa (s-t)} ds \right] d w_t $$

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  • $\begingroup$ Thanks for the quick answer! The steps are really clear. Would you mind adding a few more detail in the solving of the differential equation using the integrating factor? Also the final result is missing the square root of v_t, as it comes from the previous equation where xi_t^t = v_t , right? $\endgroup$ – Guil Nov 25 '19 at 16:14
  • $\begingroup$ Yes you are correct, I was missing the square root in the final expression for v_t, I have corrected my typo now. I will leave the integrating factor derivation as an exercise for you to solve - it is not that difficult. If you get stuck, take a look at some "integrating factor examples" around on the exchange or on the web - there are plenty. $\endgroup$ – AXH Nov 25 '19 at 21:52

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