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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.

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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.

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