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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!!

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  • $\begingroup$ You mentionned "undiscounted call price". Are you sure that you indeed started from this? $\endgroup$ – JejeBelfort May 23 '17 at 10:20
  • $\begingroup$ @JejeBelfort Yes, because in order to obtain equation 2.6 I need to substitute equation 2.5 (which is the undiscounted call price) in 2.4. Don't you agree with me? $\endgroup$ – AndrewTG May 23 '17 at 10:24
  • $\begingroup$ Unfortunately I dont have the book... I will try to get it in order to look at this. $\endgroup$ – JejeBelfort May 23 '17 at 10:35
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    $\begingroup$ You may need to provide more background information, for example, how is the Heston model defined? What are equation 2.5 and 2.6 etc., so that, for people without that book can understand your question. $\endgroup$ – Gordon May 23 '17 at 13:16
  • $\begingroup$ @Gordon you are right! $\endgroup$ – AndrewTG May 23 '17 at 13:48
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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 $P_1$.

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  • $\begingroup$ Maybe I'm missing something, but in my opinion this is not the point. The aim is to find a functional form for $P_0$ and $P_1$. My post was a part of the derivation of the Heston model by J. Gatheral, in that model the volatility is assumed to be stochastic. So it is clear that $P_0$ and $P_1$ are not the classical Black-Scholes probabilities. It is assumed that the price of a European Call is of the form of Eq 2.5, the unknown indeed are $P_0$ and $P_1$, you cannot make assumptions as you did by setting them equal to the undiscount price of digital option under different probability measures. $\endgroup$ – AndrewTG May 25 '17 at 14:15
  • $\begingroup$ @AndrewTG: I know you are working with the Heston stochastic variance model. But $P_0$ and $P_1$ are still defined as I have shown in the answer, which are model independent. To derive PDEs for $P_0$ and $P_1$ under the Heston stochastic variance model, you need some option values that are related to $P_0$ and $P_1$, and then substitute into (2.4). As I said, you cannot directly substitute (2.5) to (2.4) to have (2.6), and I do not think Jim is correct here. $\endgroup$ – Gordon May 25 '17 at 15:18
  • $\begingroup$ @AndrewTG: One more note, I do not define both $P_0$ and $P_1$ to be the value of a digital option under the different measure. In fact, $P_0$ is the undiscounted value of a digital option, while $P_1$ is the value of an option with a payoff as I defined in the answer, and it is not for a digital option. $\endgroup$ – Gordon May 25 '17 at 15:39
  • $\begingroup$ Gordon is correct. Eq. (2.6) is not a necessary condition of Eq. (2.4) and (2.5) as claimed by Jim Gatheral, namely "Substituting the proposed solution (2.5) into equation (2.4) implies that $P_0$ and $P_1$ must satisfy the equation (2.6)". It is only a sufficient decomposition able to give the correct final result including the initial condition. Jim Gatheral is not a mathematician. He is not very meticulous with derivation and mathematical phrases. $\endgroup$ – Hans Feb 8 '18 at 7:55
  • $\begingroup$ @Gordon: You are welcome. Also, by the linearity of the PDE, you can always subtract a function satisfying one PDE and the associated boundary value and derive the PDE and its boundary condition for the remainder. $\endgroup$ – Hans Feb 8 '18 at 17:30

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