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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?

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    $\begingroup$ What are the parameters you're using for the Black-Scholes and Heston models, respectively. Actually your question would benefit from including in both model's definition in order to be able to "visualize" the difference more clearly. $\endgroup$ – SRKX Dec 16 '15 at 2:00
  • $\begingroup$ for example Heston( V0=0.2,rho=-0.4,kappa=3.5,theta=0.04,xi= 0.35) BS(sigma = sqrt(theta Heston)), same Risk Free Rate for both $\endgroup$ – Michal Dec 16 '15 at 11:59
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    $\begingroup$ It is incorrect to use BS(sigma=sqrt(theta Heston))$. In fact, they are not comparable. Generally, you need to calibrate the Heston parameters from the market, that is, from the Black-Scholes prices. Then they should be the same. On the other hand, if you have the Heston parameters, and the price from teh Heston model, then you can calibrate the volatility parameter for BS. In both cases, they are the same. $\endgroup$ – Gordon Dec 16 '15 at 13:38
  • $\begingroup$ Could you please explain how the volatility for BS should be derived given I have the Heston parameters and Heston price? $\endgroup$ – Michal Dec 16 '15 at 14:13
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    $\begingroup$ An exemplary set of parameters and the resulting option values can be found in several places, for example the book The Heston Model by Rouah $\endgroup$ – noob2 Dec 16 '15 at 22:06
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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).

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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 $v=\sigma_{\color{red}{BC}}^2\,.$


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

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

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    $\begingroup$ Why $\sigma_{BC}$ rather than $\sigma_{BS}$? $\endgroup$ – will Nov 28 '16 at 10:32
  • $\begingroup$ I don't Understand your question. $\endgroup$ – user16651 Nov 28 '16 at 10:36
  • $\begingroup$ you have $\sigma_H$ for Heston vol, i would have expected $sigma_{BS}$ for Scholes vol, but you use BC everywhere. $\endgroup$ – will Nov 28 '16 at 10:40
  • $\begingroup$ If $\lambda=\sigma_H=0$ and $v_0=\theta=\sigma^2_{BC}$ then we can say the BC model is a especial case of Heston model . $\endgroup$ – user16651 Nov 28 '16 at 10:45
  • $\begingroup$ Sure, but i think it's a little misleading - why BC? what does it stand for? $\endgroup$ – will Nov 28 '16 at 10:51

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