Assessing Initial Variance Sensitivity in the Heston Call Formula
Summary
The document asks how the initial variance in the Heston stochastic-volatility model affects a European call price. It starts from the model’s characteristic-function pricing representation, where each probability term depends exponentially on initial variance through a coefficient. Differentiating under the integral produces an expression for the price sensitivity involving those coefficients and the characteristic function.
The question is whether that derivative has a definite sign. The response cautions that the real-valued integrand can be positive or negative over the integration range, so its sign cannot be read from the prefactor alone. It suggests evaluating the integral numerically for chosen Heston parameters and inspecting the resulting integrand. This is a preliminary diagnostic rather than a general result: it gives no parameter values, plots, computations, proof of monotonicity, or guidance on numerical integration accuracy. The sign may require further analysis for the specific model inputs and contract.
Key ideas
- The Heston call price representation depends on initial variance through characteristic-function coefficients.
- Differentiating the integral yields an expression for sensitivity to initial variance.
- A sign-changing real integrand prevents determining the derivative’s sign from the prefactor alone.
- The response suggests numerical inspection for selected parameter values, without establishing a general sign result.
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# Heston Model Sensitivity Qualitative Property
# Heston Model Sensitivity Qualitative Property
Consider the following Heston model: $$\begin{aligned} \mathrm{d}S_t&=rS_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}B_{1,t}\\ \mathrm{d}v_t&=-\kappa(v_t-\bar{v})\mathrm{d}t+\sigma_v\sqrt{v_t}\mathrm{d}B_{2,t} \end{aligned} $$ where $\mathrm{d}B_{1,t}\mathrm{d}B_{2,t}=\rho\mathrm{d}t$. Now, how can we estimate $v_0$'s effect on the the call option price?
My initial thought is taking derivative with respect to $v_0$. The option price is given by $$C=SP_1-Ke^{-rT}P_2$$ where $$P_j=\frac{1}{2}+\frac{1}{\pi}\int_0^{\infty}\mathrm{Re}\left(f_j\frac{e^{-i\phi \ln K}}{i\phi}\right)\mathrm{d}\phi$$ where $f_j=\exp(C_j+D_jv_0+i\phi x)$ is the characteristic function and $x=\ln S_T$. Now we can see that $$\frac{\partial P_j}{\partial v_0}=\frac{1}{\pi}\int_0^{\infty}\mathrm{Re}\left(f_jD_j\frac{e^{-i\phi \ln K}}{i\phi}\right)\mathrm{d}\phi$$ But then how do I know whether this derivative is positive or not?
An alternative way of thinking this problem is using risk-neutral pricing formula. Under $\mathbb{Q}$, we have $$\mathrm{d}x_t=\left(r-\frac{1}{2}v_t\right)\mathrm{d}t+\sqrt{v_t}\mathrm{d}B_{1,t}$$ where $x_t=\ln S_t$. Thus $$S_t=S_0\exp\left[\int_0^{t}\left(r-\frac{1}{2}v_s\right)\mathrm{d}s+\int_0^{t}\sqrt{v}_s\mathrm{d}B_{1,s}\right]$$ We can investigate $v_t$ then.
## Answer by KaiSqDist (score 1)
https://quant.stackexchange.com/a/78935
Here to pen my opinion - not exactly an answer but hopefully it helps with the thought process.
The area under the curve or an integral can be both positive and negative (depends if it is above or under the x-axis $\phi$). Therefore, we can ignore the $\frac{1}{\pi}$ coefficient and just focus on the integral.
There are quite a few uncertainties here, such as the value of the Heston parameters and there bounds of the area to be evaluated - from [0,$\infty$), so it might not be clear whether it is positive or negative.
My thoughts would be to first add some sample Heston parameters and try to plot out the curve on a graphing calculator such as Desmos to see if the area under the curve is positive or negative.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.