Deriving the solution for European call option in the Heston Model

Question

I'm deriving the solution for European call option in the Heston Model. I follow the original paper by Heston and Fabrice Douglas Rouah's derivations in his book The Heston Model and Its Extensions in Matlab and C#. However, I'm having troubles understanding a few steps - I have 3 questions.

The hedging portfolio in the Heston Model consists of an option, $V = V(S,v,t)$, $\Delta$ stocks and $\phi$ units of the option to hedge volatility, $U(S,v,t)$, and has the value: \begin{align*} \Pi = V + \Delta S + \phi U, \end{align*} where the change in the value of the portfolio in the time interval, $dt$ is given by: \begin{align} \label{HestonPort} d\Pi = dV + d\Delta S + d\phi U. \end{align}

Next, I want to obtain the process followed by $dV$. Rouah writes that, one must apply Itô's lemma to $V$, and that one must differentiate $V$ wrt $t,S$ and $v$, and create a second-order Taylor expansion. This results in: \begin{align*} dV = \frac{\partial V}{\partial t}dt + \frac{\partial V}{\partial S}dS + \frac{\partial V}{\partial v}dv + \frac{1}{2}vS^2\frac{\partial^2V}{\partial S^2}dt + \frac{1}{2}v\sigma^2\frac{\partial^2V}{\partial v^2}dt + \sigma \rho v S \frac{\partial^2 V}{\partial S \partial v}dt. \end{align*}

1. I don't understant this step. Why do I need to create a second-order Taylor expansion? And why do I differentiate $V$ wrt $t,S$ and $v$? I understand Itô's lemma as in the derivation of the Black Scholes model - is this some kind of extensions of Itô's? Or how should I know that I need the second-order Taylor expansion?

Later in the derivations, Heston writes that for at a European call option he "guesses a solution of the form": $$ C(S,v,t) = SP_1 - Ke^{-rT}P_2. $$ (page 330, equation 10). This is analogy with the Black-Scholes formula. The first term is the present value of the spot asset upon optimal exercise, and the second term is the present value of the strike-price payment. Both of these terms must satisfy the PDE given by:

\begin{align} \label{HestonPDE} \begin{split} & \frac{\partial U}{\partial t} + \frac{1}{2}vS^2\frac{\partial^2U}{\partial S^2} + \sigma \rho v S \frac{\partial^2 U}{\partial S \partial v} + \frac{1}{2}v\sigma^2\frac{\partial^2U}{\partial v^2} \\ - &rU + rS \frac{\partial U}{\partial S} + \left[ \kappa(\theta - v) - \lambda(S,v,t) \right] \frac{\partial U}{\partial v} = 0. \end{split} \end{align}

Substituting the proposed solution into the original PDE shows that P1 and P2 must satisfy:

\begin{align} \label{PPDE} \frac{\partial P_j}{\partial t} + \rho \sigma v \frac{\partial^2 P_j}{\partial v \partial x} + \frac{1}{2} v \frac{\partial^2 P_j}{\partial x^2} + \frac{1}{2} \sigma^2 v \frac{\partial^2 P_j}{\partial v^2} + (r+u_j v) \frac{\partial P_j}{\partial x} + (a-b_j v) \frac{\partial P_j}{\partial v} = 0, \end{align}

2. I can see that Heston "guess" is similar to the Black Scholes equation - but how can he "guess" this solutions? Can this guess be derived from the PDE?
3. Why is it that the two terms from the guessed solution also must satisfy the PDE? And why is it nessecary to derive a PDE for P1 and P2?

Thanks in advance!

Answer 1

Itô's Lemma

The standard version of Itô's Lemma applies to a single Itô process $\text{d}X_t=\mu(t,X_t)\mathrm{d}t+\sigma(t,X_t)\mathrm dW_t$. Then, $$\mathrm{d}f(t,X_t) = \left(f_t+\mu(t,X_t)f_x + \frac{1}{2}\sigma(t,X_t)^2f_{xx}\right)\mathrm{d}t+\sigma(t,X_t)f_x\mathrm dW_t.$$ Let $\text{d}Y_t=m(t,Y_t)\mathrm{d}t+s(t,Y_t)\mathrm dW_t^{(2)}$ be a second Itô process with $\mathrm dW_t\mathrm dW_t^{(2)}=\rho\mathrm dt$. Then, \begin{align*} \mathrm{d}f(t,X_t,Y_t) = \bigg(& f_t+\mu(t,X_t)f_x+m(t,Y_t)f_y + \frac{1}{2}\sigma(t,X_t)^2f_{xx}+\rho\sigma(t,X_t)s(t,Y_t)f_{xy} \\ &+ \frac{1}{2}s(t,Y_t)^2f_{yy}\bigg)\mathrm{d}t+\sigma(t,X_t)f_x\mathrm dW_t+s(t,Y_t)f_y\mathrm dW_t^{(2)}. \end{align*} Alternatively, we can write $$\mathrm{d}f= \left(f_t+ \frac{1}{2}\sigma(t,X_t)^2f_{xx}+\rho\sigma(t,X_t)s(t,Y_t)f_{xy}+ \frac{1}{2}s(t,Y_t)^2f_{yy}\right)\mathrm{d}t+f_x\mathrm dX_t+f_y\mathrm dY_t.$$ Note:

• The proof for this version is also based on a Taylor polynomial and thus it resembles the corresponding second order, two-dimensional expansion.
• Itô's Lemma can be further generalised to functions of more variables, $f(t,X^{(1)}_t,...,X^{(n)}_t)$, complex valued functions and functions that are not smooth, see this answer. It can also be generalised to jump processes and more general integrators.

Example: Heston's stochastic volatility model. Let \begin{align*} \text{d}S_t&=\mu S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm dW_t \\ \text{d}v_t&=\kappa(\bar{v}-v_t)\mathrm{d}t+\xi\sqrt{v_t}\mathrm dW_t^{(2)}, \end{align*} where $\mathrm dW_t\mathrm dW_t^{(2)}=\rho\mathrm dt$. Then, $$\mathrm{d}f(t,S_t,v_t) = \left(f_t+\mu S_t f_S+\kappa (\bar{v}-v_t)f_v + \frac{1}{2}v_tS_t^2f_{SS}+\rho\xi v_t S_tf_{Sv} + \frac{1}{2}\xi^2v_tf_{vv}\right)\mathrm{d}t+\sqrt{v_t} S_t f_S\mathrm dW_t+\xi \sqrt{v_t}f_v\mathrm dW_t^{(2)}.$$

From here, we can proceed as in your notes, similar to the Black-Scholes derivation. Instead of a plain delta hedge, we need a simultaneous delta and vega hedge to eliminate the risk from the stock and the variance risk.

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Guessing the Solution

Firstly, you often make ``good guesses'' to solve PDEs. After some (many?) years, one gains experience with PDEs and sometimes can indeed guess the functional form of the solution. In the case of the Heston model: the Black-Scholes option call formula carries a lot of economic intuition (price of asset-or-nothing call and cash-or-nothing call), see this answer. Options on zero-coupon bond option also have a similar functional form. Thus, it's a reasonable guess to assume that the Black-Scholes functional form carries through to stochastic volatility model.

In fact, the numéraire change technique from Geman et al. (1995) tells us that option prices cannot only be written as sum of digital options but also as sum of exercise probabilities, \begin{align*} C(S;K,T) = Se^{-qT}\mathbb{S}[\{S_T\geq K\}] - Ke^{-rT}\mathbb{Q}[\{S_T\geq K\}], \end{align*} where $\mathbb{Q}$ is the standard risk-neutral measure and $\mathbb{S}$ is the stock measure. So, Heston's guess is sensible.

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Heston's PDE

After we guess $C=SP_1-Ke^{-rT}P_2$, we have, for example, $$\frac{\partial}{\partial S} C= P_1+S\frac{\partial}{\partial S}P_1 -Ke^{-rT}\frac{\partial}{\partial S}P_2$$ and $$\frac{\partial}{\partial t} C= S\frac{\partial}{\partial t}P_1 -Ke^{-rT}\frac{\partial}{\partial t}P_2.$$ If you plug all of this into the actual PDE for $C$ (alongside the other necessary partial derivatives), then you get two PDEs for $P_1$ and $P_2$.

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Alternative Derivation

I provide a quick alternative derivation for Heston's formula using numéraire changes. Recall \begin{align*} C(S;K,T) = Se^{-qT}\mathbb{S}[\{S_T\geq K\}] - Ke^{-rT}\mathbb{Q}[\{S_T\geq K\}]. \end{align*}

Gil-Pelaez's (1951) inversion formula states that for any probability measure $\mathcal{P}$, \begin{align*} \int_0^\infty \Re\left(\frac{e^{-i\ln(K)u}\varphi_{\ln(S_T)}^\mathcal{P}(u)}{iu}\right)\mathrm{d}u = \pi\left(\mathcal{P}\big[\{S_T\geq K\}\big] - \frac{1}{2}\right), \end{align*} where $\varphi_{X}^\mathcal{P}(u)=\mathbb{E}^\mathcal{P}[e^{iu X}]$ is the characteristic function of an integrable random variable $X$ under $\mathcal{P}$. If $X$ has a probability density function, then $\varphi$ is the Fourier transform of this density.

A numéraire change gives $$\varphi_{\ln(S_T)}^\mathbb{S}(u)=\mathbb{E}^\mathbb{S}[e^{iu \ln(S_T)}] = \mathbb{E}^\mathbb{Q}\left[\frac{\text{d}\mathbb{S}}{\mathrm d\mathbb{Q}}e^{iu \ln(S_T)}\right]=\mathbb{E}^\mathbb{Q}\left[\frac{S_T}{\mathbb{E}^\mathbb{Q}[S_T]}e^{iu \ln(S_T)}\right]=\frac{\varphi_{\ln(S_T)}^\mathbb{Q}(u-i)}{\varphi_{\ln(S_T)}^\mathbb{Q}(-i)}.$$

You can combine of all this and arrive at Heston's formula, all expressed in terms of a single characteristic function, $\varphi_{\ln(S_T)}^\mathbb{Q}$, \begin{align*} \mathbb{Q}\big[\{S_T\geq K\}\big] &= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty \Re\left(\frac{e^{-i\ln(K)u}\varphi(u)}{iu}\right)\mathrm{d}u, \\ \mathbb{S}\big[\{S_T\geq K\}\big] &= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty \Re\left(\frac{e^{-i\ln(K)u}\varphi(u-i)}{iu\varphi(-i)}\right)\mathrm{d}u, \end{align*} where $\varphi$ is the standard Heston characteristic function of $\ln(S_T)$ under $\mathbb{Q}$, which you find in many textbooks.

Note

• These formulae actually apply to all models with known characteristic function (most stochastic volatility models and exponential Lévy processes).
• If you know more about Fourier methods, you'll recognise these formulae as equivalent to Bakshi and Madan's (2000) formula and Bates' (2006) formula. They are also a special case of Lewis' (2001) formula which, in turn, nests Carr and Madan's (1999) approach.