Deriving the Conditional Variance Integral in the Heston Model
Summary
The document explains how to derive the time differential of an integrated conditional expectation of variance in a Heston setting. It identifies an error in the variance process stated in the source material: the mean-reverting drift should be affine in variance, with the variance itself rather than its square appearing in the drift term. Correcting this matters because the stated form would violate the model’s affine structure.
The answer defines forward variance as the conditional expectation of future variance, takes conditional expectations in the stochastic differential equation, and solves the resulting mean-reverting equation using an integrating factor. Its time differential is then integrated across maturities to obtain the differential of the conditional variance integral, with a stochastic term weighted by an exponential decay factor. The derivation assumes the corrected Heston variance process and relies on exchanging differentiation and integration; the exchange does not discuss broader model conditions or boundary behavior.
Key ideas
- The variance drift in the presented source equation contains a typo; it should depend linearly on variance.
- Forward variance is the conditional expectation of variance at a future maturity.
- The conditional expectation equation follows by taking expectations of the variance process over the future interval.
- An integrating factor solves the mean-reverting forward variance equation.
- Integrating its time differential across maturities yields the differential of the conditional variance integral.
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# Computing Itô differential of conditional expectation process (Heston SDE)
# Computing Itô differential of conditional expectation process (Heston SDE)
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
## Answer by AXH (score 3, accepted)
https://quant.stackexchange.com/a/49898
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 $$ d M_t = \int_{0}^{T} d \xi_t^s ds = \nu \sqrt{v_t} \left[ \int_{0}^{T} e^{-\kappa (s-t)} ds \right] d w_t $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.