Why the Heston Call Price Decomposition Produces Two PDEs

# The Heston Solution For European Option - Jim Gatheral


# The Heston Solution For European Option - Jim Gatheral












I have this equation (Eq. (2.4) "The Volatility Surface - A Practitioner's Guide" by Jim Gatheral (Ed. 2006)): $$-\frac{\partial C(v, x, \tau)}{\partial \tau}+\frac{1}{2}v \frac{\partial^2 C(v,x,\tau)}{\partial x^2}-\frac{1}{2}v\frac{\partial C(v,x,\tau)}{\partial x}+\frac{1}{2}v\eta^2\frac{\partial^2 C(v,x,\tau)}{\partial v^2}+\rho\eta v\frac{\partial^2 C(v,x,\tau)}{\partial v \partial x}-\lambda(v-\bar{v})\frac{\partial C(v,x,\tau)}{\partial v}=0$$ Where $x:=ln{\frac{F_{t,T}}{k}}$ ($F_{t,T}$ is the forward price) and $\tau=T-t$. Assuming that:$$C(x,v,\tau)=K\{e^xP_1(x,v,\tau)-P_0(x,v,\tau)\}$$ Where the above equation correspont to Eq 2.5 of "The Volatility Surface - A Practitioner's Guide" by Jim Gatheral (Ed. 2006). By substituing the last equation in the previous one, J. Gatheral obtains: $$-\frac{\partial P_j(v, x, \tau)}{\partial \tau}+\frac{1}{2}v \frac{\partial^2 P_j(v,x,\tau)}{\partial x^2}-(\frac{1}{2}-j)v\frac{\partial P_j(v,x,\tau)}{\partial x}+\frac{1}{2}v\eta^2\frac{\partial^2 P_j(v,x,\tau)}{\partial v^2}+\rho\eta v\frac{\partial^2 P_j(v,x,\tau)}{\partial v \partial x}+(a-b_jv)\frac{\partial P_j(v,x,\tau)}{\partial v}=0$$ For $j=0,1$, where $a=\lambda \bar{v}, b_j= \lambda - j\rho \eta$. This is Eq 2.6 of the referred book. Now, my problem is the following. When I substitute 2.5 in 2.4, I obtain the following: $$k\{-\frac{ \partial P_0(v,x,\tau)}{\partial \tau}+\frac{1}{2}v\frac{\partial ^2 P_0(v,x,\tau)}{\partial x^2}-\frac{1}{2}v\frac{\partial P_0(v,x,\tau)}{\partial x}+\frac{1}{2}v\eta^2\frac{\partial^2 P_0(v,x,\tau)}{\partial v^2}+\rho\eta v\frac{\partial^2 P_0(v,x,\tau)}{\partial v \partial x}-\lambda(v-\bar{v})\frac{\partial P_0(v,x,\tau)}{\partial v}= ke^x \{ -\frac{\partial P_1(x,v,\tau)}{\partial \tau}-\frac{\partial x}{\partial \tau}P_1(x,v,\tau)+\frac{1}{2}v\frac{P_1(v,x,\tau)}{\partial x^2}+\frac{1}{2}v\frac{\partial P_1(v,x,\tau)}{\partial x}+\frac{1}{2}v\eta^2\frac{\partial^2 P_1(v,x,\tau)}{\partial v^2}+\rho\eta v\frac{\partial^2 P_1(v,x,\tau)}{\partial v \partial x}+(a-b_jv)\frac{\partial P_1(v,x,\tau)}{\partial v}\}$$. First question:

As one can see I have obtained an equation in $P_0({x,v,\tau})$ and $P_1(x,v, \tau)$. J. Gatheral obtains two equations. In order to obtain the same result as him, I have set $k=1$ and $F_{t,T}=0$ to obtain a PDE in $P_0(x,v,\tau)$ and then I have set $k=0$ and $F_{t,T}=1$ to obtain a PDE in $P_1(x,v,\tau)$. Is it correct? Am I allowed to do that? If yes, why?

Second question:

When I take the derivative of the undiscounted call price with respect to $\tau$ from equation 2.5, I obtain the following: $$\frac{\partial C(x,v,\tau )}{\partial \tau} = K\{e^x\frac{\partial P_1(x,v,\tau )}{\partial \tau} + e^x\frac{\partial x}{\partial \tau} P_1 (x,v,\tau)-\frac{\partial P_0(x,v,\tau )}{\partial \tau}\}=K\{e^x\frac{\partial P_1(x,v,\tau )}{\partial \tau} + e^xr P_1 (x,v,\tau)-\frac{\partial P_0(x,v,\tau )}{\partial \tau}\}$$ Which in my opinion is correct, given that $x$ depend on $\tau$ thanks to $F_{t,T}$. However, I am not able to obtain equation 2.6 because the term $e^xr P_1 (x,v,\tau)$ is not there (I don't see another term which allows me to make a simplification). What am I missing here?

Thanks guys!!

## Answer by Gordon (score 2)

https://quant.stackexchange.com/a/34388

I do not think that Equation $(2.5)$ can be directly substituted into $(2.4)$ to obtain equations of the form given by $(2.6)$ for both $P_0$ and $P_1$. In fact, since Equation $(2.4)$ is satisfied for any European option price, two different option prices are substituted into Equation $(2.4)$ to obtain the respective equations.

Specifically, note that \begin{align*} P_0 = \mathbb{E}(1_{\{S_T >K\}}\mid \mathcal{F}_t), \end{align*} and \begin{align*} P_1 = \mathbb{E}^S(1_{\{S_T >K\}}\mid \mathcal{F}_t), \end{align*} where $\mathbb{E}$ is the expectation operator under the risk-neutral probability measure, while $\mathbb{E}^S$ is the expectation operator under the probability measure with the stock price process $S$ as the numeraire.

As $P_0$ is the undiscounted price of a digital option, under a deterministic interest rate setting, we can substitute $P_0$ into $(2.4)$, to obtain the equation for $P_0$. As for $P_1$, we consider the option with Payoff, at maturity $T$, of the form \begin{align*} S_T 1_{\{S_T >K\}}. \end{align*} The undiscounted value is given by \begin{align*} \frac{B_t}{P(t, T)}\mathbb{E}\left(\frac{S_T 1_{\{S_T >K\}}}{B_T}\mid \mathcal{F}_t\right) &= \frac{S_t}{P(t, T)}\mathbb{E}^S(1_{\{S_T >K\}}\mid \mathcal{F}_t)\\ &=F_{t, T}\mathbb{E}^S(1_{\{S_T >K\}}\mid \mathcal{F}_t)\\ &=Ke^xP_1,\tag{1} \end{align*} where $B_t=e^{\int_0^t r_s ds}$ is the money-market account value at time $t$, $P(t, T)$ is the price of a zero-coupon bond with maturity $T$ and unit face value, and $F_{t, T}=\frac{S_t}{P(t, T)}$ is the forward price. Now, we substitute $(1)$ into Equation $(2.4)$ to obtain the equation for $P_1$.