How should option prices differ when using the Heston versus the Black-Scholes model?

Question

I am running Monte Carlo simulations for a European Call using Heston Model and I am trying to compare them with prices calculated using Black-Scholes formula. I am not quite sure if the prices I get from the Heston model are correct.

• Is there any regularity here?
• What should be the relation between the prices? (should a price of a European call from Heston be greater or smaller than from Black-Scholes?)

Can anyone explain what the relation should be?

Answer 1

If $\sigma_{H}\ne 0$ and $v_0\ne \theta\ne \sigma^2_{BC}$ then prices are different in BC and Heston model.

Especial case

In the Black-Scholes model the dynamics of $S_t$ under risk neutral measure follow the stochastic process $$dS_t=(r-q)S_tdt+\sigma_{\color{red}{BC}}S_tdW^{\mathbb{Q}}(t)\tag 1$$ on the other hand $$dS_t=(r-q)S_t+\sqrt{v_t}S_tdW^{\mathbb{Q}}_1(t)\\ \quad dv_t=\kappa(\theta-v_t)dt+\sigma_{\color{red}{H}}\sqrt{v_t}dW^{\mathbb{Q}}_2(t)\tag 2$$ where $d[W^{\mathbb{Q}}_1(t)\,,\,W^{\mathbb{Q}}_2(t)]=\rho dt$. In CRR model we have $$\text{Var}[v_t\big{|}v_0]=\frac{v_0\sigma_{\color{red}{H}}^2e^{-\kappa t}}{\kappa}\left(1-e^{-\kappa t}\right)+\frac{\theta\sigma_{\color{red}{H}}^2}{2\kappa}\left(1-e^{-\kappa t}\right)^2\tag 3$$

Now if we set $\sigma_\color{red}{H}=0$ ,then $$\text{Var}[v_t|v_0] = 0\tag 4$$ This will produce volatility that is time-varying, but $\color{red}{\text{deterministic}}$. Indeed
$$dv_t=\kappa(\theta-v_t)dt\tag 5$$ or $$v'_t+\kappa v_t=\kappa\theta\tag 6$$ Equation $(6)$ is a linear ordinary differential equation. We can show easily $$v_t=\theta+c\,e^{-\kappa t}\quad ,\quad c\in\mathbb{R}\tag 7$$ Set $v_0=\theta=\sigma_{\color{red}{BC}}^2$. Therefore $c=0$ and $v=\sigma_{\color{red}{BC}}^2\,.$

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$\color{red}{\text{Warning}\,!}$

In the Heston Model we have \begin{align} C(t\,,{{S}_{t}},{{v}_{t}},K,T)={{S}_{t}}{{P}_{1}}-K\,{{e}^{-r\tau }}{{P}_{2}}\tag 8 \end{align} where,for $j=1,2$

\begin{align} & \mathbb{P}_j({{x}_{t}}\,,\,{{v}_{t}}\,;\,\,{{x}_{T}},\ln K)=\frac{1}{2}+\frac{1}{\pi }\int\limits_{0}^{\infty }{\operatorname{Re}\left( \frac{{{e}^{-i\phi \ln K}}{{f}_{j}}(\phi ;t,x,v)}{i\phi } \right)}\,d\phi \tag 9 \\ & {{f}_{j}}(\phi \,;{{v}_{t}},{{x}_{t}})=\exp [{{C}_{j}}(\tau ,\phi )+{{D}_{j}}(\tau ,\phi ){{v}_{t}}+i\phi {{x}_{t}}]\tag {10} \\ \end{align}

and

\begin{align} & {{C}_{j}}(\tau ,\phi )=(r-q)i\phi \,\tau +\frac{a}{{{\sigma_{\color{red}{H}} }^{2}}}{{\left( ({{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi +{{d}_{j}})\,\tau -2\ln \left(\frac{1-{{g}_{j}}{{e}^{{{d}_{j}}\tau }}}{1-{{g}_{j}}}\right) \right)}} \tag{11}\\ & {{D}_{j}}(\tau ,\phi )=\frac{{{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi +{{d}_{j}}}{{{\sigma_{\color{red}{H}} }^{2}}}\left( \frac{1-{{e}^{{{d}_{j}}\tau }}}{1-{{g}_{j}}{{e}^{{{d}_{j}}\tau }}} \right) \tag{12}\\ \end{align} such that \begin{align} & {{g}_{j}}=\frac{{{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi +{{d}_{j}}}{{{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi +{{d}_{j}}} \\ & {{d}_{j}}=\sqrt{{{({{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi )}^{2}}-{{\sigma_{\color{red}{H}} }^{2}}(2i{{u}_{j}}\phi -{{\phi }^{2}})} \\ & {{u}_{1}}=\frac{1}{2}\,,\,{{u}_{2}}=-\frac{1}{2}\,,\,a=\kappa \theta \,,\,{{b}_{1}}=\kappa +\lambda -\rho \sigma_{\color{red}{H}} \,,\,{{b}_{2}}=\kappa +\lambda \,,\ {{i}^{2}}=-1 \\ \end{align}

We can not simply substitute $\sigma_{\color{red}{H}} = 0$ into the pricing functions, because that will lead to division by zero in the expressions for $C_j(\tau,\phi)$ and $D_j(\tau,\phi)$.

With $\sigma_{\color{red}{H}}=0$, the Riccati equation reduces to the ordinary first-order differential equation in the Heston's article (1993) $$\frac{\partial {{D}_{j}}}{\partial \tau }={{p}_{j}}-{{b}_{j}}{{D}_{j}}\tag {13}$$

where $p_j=u_j i\phi-\frac 12 \phi^2$.The solution of this ODE is $$D_j(\tau ,\phi )=\frac{(i{{u}_{j}}\phi -\frac{1}{2}{{\phi }^{2}})(1-{{e}^{-{{b}_{j}}\tau }})}{{{b}_{j}}}\tag {14}$$ on other hand , Heston showed $$\frac{\partial {{C}_{j}}}{\partial \tau }=ri\phi +a{{D}_{j}}\tag{15}$$ substitute $(14)$ in $(15)$ and integrate to obtain $${{C}_{j}}(\tau ,\phi )\ =ri\phi \tau +\frac{a(i{{u}_{j}}\phi -\frac{1}{2}{{\phi }^{2}})}{{{b}_{j}}}\left( \tau -\frac{1-{{e}^{-{{b}_{j}}\tau }}}{{{b}_{j}}} \right)\tag{16}$$ In the case $j=2$ and $\lambda=0$, we have $$\begin{align} & {{D}_{2}}(\tau ,\phi )=-\frac{(i\phi +{{\phi }^{2}})(1-{{e}^{-\kappa \tau }})}{2\kappa } \\ & {{C}_{2}}(\tau ,\phi )\ =ri\phi \tau -\frac{\theta (i\phi +{{\phi }^{2}})}{2}\left( \tau -\frac{1-{{e}^{-\kappa \tau }}}{\kappa } \right) \\ \end{align} \tag {17}$$ We know $${{f}_{2}}(\phi ;{{\ln S}_{t}},{{v}_{t}})=\exp\left[i\phi\,{\ln S_t}+{{C}_{2}}(\tau \,,\,\phi )+{{D}_{2}}(\tau \,,\,\phi ){{v}_{t}}\right]$$

let $v_0=\theta=\sigma_{\color{red}{BC}}^2$, thus $$\color{red}{{{f}_{2}}=\exp\left( i\phi \left[\ln {{S}_{t}}+(r-\frac{1}{2}{{\sigma }_{BC}}^{2})\tau \right]-\frac{1}{2}{{\phi }^{2}}{{\sigma }_{BC}}^{2}\tau \right)=\mathbb{E}\left[\exp\left(i\,\phi\,\ln S_t\right)\right]\tag {18}}$$

Answer 2

The answer depends on the parameter settings, but you should always check the risk neutral densities. One of the several complaints against BS is the underlying GBM has lower tails so it underprices OTM (out-of-the-money) options as the market is more heavy tailed.

Heston log-return density is leptokurtic (higher tails and kurtosis). See http://fedc.wiwi.hu-berlin.de/xplore/tutorials/stfhtmlnode46.html

It depends on the setting of your parameters but usually Heston should give higher prices for OTM options for compatible parameter optimizations (i.e. both models' parameters are estimated using the same data).