# 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.

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.