Why the Heston Variance Integral Cannot Be Simplified by Differential Algebra
Summary
The question attempts to find the distribution of the stochastic term in a Heston-style asset process by substituting an integral expression for the variance process, taking a square root, and applying Brownian differential identities. The response identifies a fundamental problem in that manipulation: increments at different times cannot be treated as though their products obey the same-time covariation rule. In particular, the correlation relation between the two Brownian drivers does not justify replacing a product involving an earlier-time increment and a later-time increment with a correlation times the time step.
The answer clarifies the notation by giving the two Brownian motions distinct labels, making the time indices and differentials easier to distinguish. It does not derive the correct distribution or expectation of the stochastic integral, so it should be read as a warning about the attempted calculation rather than a complete solution. The broader lesson is to use Itô calculus and its adapted-integrand rules carefully; ordinary algebraic operations on infinitesimal differentials can produce invalid conclusions about stochastic integrals.
Key ideas
- The proposed reduction treats Brownian increments from different times as if they shared a same-time covariation.
- Correlated Brownian motions satisfy a covariation rule at the same time, not for arbitrary cross-time increments.
- Using distinct labels for the asset and variance Brownian drivers helps expose the time-index error.
- The response diagnoses the derivation but does not provide the stochastic integral’s correct distribution.
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# How calculate expectation and variation of stochastic integral Based on Heston model?
# How calculate expectation and variation of stochastic integral Based on Heston model?
I was calculated Heston volatility model. But I think it is wrong.
$dS_t = \mu dt + \sqrt V_t dW_t^s$ $dV_t = k(\theta - V_t)dt + \sigma \sqrt V_t dW_t^v$. $dW^s_t dW^v_t = \rho dt$
take integral to calculate stock price process,
$\int_0^t dS_s = \int_0^t\mu dt + \int_0^t \sqrt V_s dW_s^s$ then, $S_t-S_0 = \mu t + \int_0^t \sqrt V_s dW_s^s$
### My derivation
We have to calculate the second term $\int_0^t \sqrt V_s dW_s^s$. since, volatility process know as CIR process can express blow.
$V_t=e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right)+\sigma e^{-\kappa t} \int_0^t e^{\kappa s} \sqrt{V_s} d W_s^v$
thus take a root and multiple $dW_t^s$ then, $\sqrt V_t dW_t^s = \sqrt {(e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right)+\sigma e^{-\kappa t} \int_0^t e^{\kappa s} \sqrt{V_s} d W_s^v)} dW_t^s$.
$= \sqrt {(e^{-\kappa t} V_0 (dW^s_t)^2+\theta\left(1-e^{-\kappa t}\right)(dW^s_t)^2+\sigma e^{-\kappa t} \int_0^t e^{\kappa s} \sqrt{V_s} d W_s^v(dW^s_t)^2)}$
since, $(dW_t)^2 = dt , dW_t^s dW_t^v = \rho dt \text{ and } dt dW_t^s\rho = 0$
$\sqrt V_t dW_t^s = \sqrt {(e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right)} dW_t^s$.
therefore,
$\int_0^t \sqrt V_s dW_s^s = (\sqrt {(e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right))} \int_0^t dW_t^s \sim \mathcal{N}(0,e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right))$
Please let me know where the wrong process is. thank you.
## Answer by Feiyeung Chen (score 1)
https://quant.stackexchange.com/a/74611
Using $s$ for labelling $W$ and as an integrand caused ambiguity. Here I have relabelled $W^s \rightarrow W^1; \qquad W^v \rightarrow W^2$ $$ dW_s^2 (dW_t^1)^2 = \underbrace{dW_s^2 dW_t^1}_{\neq \rho \ d t \ !} dW_t^1 $$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.