# How to derive a pricing PDE for an asset that follows a mean-reverting process?

I want to derive a Black-Scholes type partial differential equation to price options on an asset that follows a mean-reverting process (Schwartz model).

My attempt follows the methodology of deriving the Black-Scholes PDE but using a mean-reverting process to describe the asset instead of a geometric Brownian motion:

Let $$S$$ follow a mean-reverting stochastic process $$S = \kappa(\mu-\ln S)S dt + \sigma SdW$$ and let $$V=V(S,t)$$ denote the value of the option. From Itô's lemma we have $$dV=\left(\frac{\partial V}{\partial t}+\kappa(\mu-\ln S)S\frac{\partial V}{\partial S}+\frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2}\right)dt+ \sigma S \frac{\partial V}{\partial S} dW_t.$$

Let's perform a delta hedge, i.e. construct a portfolio $$\Pi=-V+\frac{\partial V}{\partial S} S$$. We see that $$d\Pi = \left(\frac{\partial V}{\partial t}+\frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2}\right)dt,$$ and since the portfolio $$\Pi$$ does not involve any risk, it must earn the risk-free interest rate, i.e. $$d\Pi = r\Pi dt= r\left(-V+\frac{\partial V}{\partial S} \right)dt.$$ Thus, we will have a PDE of the form $$\frac{\partial V}{\partial t} +rS\frac{\partial V}{\partial S} +\frac{1}{2}\sigma^2 S^2 \frac{\partial ^2V}{\partial S^2}-rV=0,$$ which is the regular Black-Scholes PDE.

Is this correct, or where do I go wrong here?

I believe that the PDE should be $$\frac{\partial V}{\partial t} +\kappa\left(\mu - \lambda-\ln S\right)S\frac{\partial V}{\partial S} +\frac{1}{2}\sigma^2 S^2 \frac{\partial ^2V}{\partial S^2}-rV=0,$$ where $$\lambda$$ is the market price of risk. This form of the PDE can be found in this post, for example.

• I think the step you are missing is the transformation of your mean-reverting process to an equivalent martingale measure, which is where the $\lambda$ is coming from. See also the reference you have quoted. Jun 30 at 11:27
• @Raskolnikov : Yes, but even though if I performed the delta hedge to the risk-neutralized process $$dS = \kappa(\mu-\lambda-\ln S)S dt+\sigma S dW^*,$$ where $W^*$ is the Brownian motion under the equivalent martingale measure, the drift term of this process still wouldn't show in the pricing PDE. I am missing the steps on how to derive the PDE in that form. Jun 30 at 12:05
• @user57127 I think your derivation is correct, for the stocks, the drift is canceled out. I am not very familiar with commodities, but isn't it that you hedge some commodities derivative with the other derivative rather than the underlying? If so, then the mean reversion can be seen as the process under risk neutral measure. Because of that the pricing pde for stocks is different than the ones for commodities
– emot
Jun 30 at 12:19

In a Black-Scholes-Merton-style hedge portfolio, we'd get:

$$dS_t=\kappa\left(\mu-\ln S_t\right)S_tdt+\sigma S_t dW_t$$

with a hedged portfolio

$$\Pi_t\equiv V_t-\Delta_tS_t$$

and

$$d\Pi_t=\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial S}dS+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}dS^2-\Delta_tdS_t$$

as usual, the portfolio is hedged iff $$\Delta_t=\frac{\partial V}{\partial S}$$ at all times. Then:

\begin{align} d\Pi_t&=\frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}dS^2=r\left(V_t-\Delta_tS_t\right)dt\\ \Rightarrow rV_t&=\frac{\partial V}{\partial t}+rS_t\frac{\partial V}{\partial S}+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}dS^2 \end{align}

...and $$\mu,\kappa$$ do not show up as the physical-world drift component does not have any say in this perfectly hedged (i.e. risk-neutral) world. HTH?

• Thank you for your answer, @Kermittfrog. So, should I understand this so that in the risk-neutral world, the drift is $\kappa(\mu-\lambda-\ln S)$, and which, in the risk-neutral world, equals $r$? And just replacing $r$ with $\kappa(\mu-\lambda-\ln S)$, the two PDEs coincide? Jun 30 at 12:14

Let $$\{r_t, \, t\ge 0\}$$ be the interest rate process. For maturity $$T$$ and $$0\le t \le T$$, note that \begin{align*} V(S_t, t) = e^{\int_0^t r_s ds}\,\mathbb{E}\left(e^{-\int_0^T r_s ds}V(S_T, T) \mid \mathscr{F}_t\right), \end{align*} where $$\mathbb{E}$$ is the expectation under the risk-neutral probability measure. Then $$M_t = e^{-\int_0^t r_s ds}V(S_t, t)$$ is a martingale, for $$0\le t \le T$$. Moreover, note that \begin{align*} dM_t &= e^{-\int_0^t r_s ds}dV - r_t e^{-\int_0^t r_s ds}V dt\\ &=e^{-\int_0^t r_s ds}\left[\bigg(-r_t V_t + \frac{\partial V}{\partial t}+\kappa(\mu-\ln S)S\frac{\partial V}{\partial S}+\frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} \bigg)dt + \sigma S \frac{\partial V}{\partial S} dW_t \right] \end{align*} Then, \begin{align*} \frac{\partial V}{\partial t}+\kappa(\mu-\ln S)S\frac{\partial V}{\partial S}+\frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2}-r_t V_t=0. \end{align*}

Regarding the application to futures option valuation, see, for example, this question. For application to a bond price, see here.

The issue in your derivation is that $$\Pi=-V+\frac{\partial V}{\partial S} S$$ is not a self-financing portfolio. See discussions in this question.

• When I follow derivation from your link i.e. constructing the hedging portfolio with $\Delta_t^1$ and $\Delta_t^2$ then I end up with: \begin{align*} \frac{\partial C}{\partial t} + r S_t \frac{\partial C}{\partial S} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2 -rC = 0. \end{align*} rather than the equation with $k(u-lnS)$. In this post you assume the drift $k(u-lnS)$ is already in risk neutral measure?
– emot
Jul 1 at 6:26
• If you are doing risk neutral pricing in a complete market then the drift of any tradable underlying must be the risk free rate. It doesn't matter if the underlying is mean reverting or not in it real world drift. You can mess with the volatility function though if you believe there are bounds to the possible range of prices (see quadratic local vol FX models for example) Jul 1 at 7:34
• Thank you for your answer! I hadn't come across this approach in deriving the pricing PDEs before. This seems to work also for deriving the (regular) Black-Scholes PDE if we use the risk-neutral process $dS=rSdt+\sigma S dW^*$ for the stock. I also faced the same problem as @emot when trying to construct the hedging portfolio $\Pi_t = \Delta_t^1 S_t + \Delta_t^2 V_t$, as in the post you linked. Is it possible to derive the PDE with the coefficient $\kappa(\mu-\lambda-\ln S)S$ by the delta-hedge approach using any kind of a hedging portfolio? Jul 1 at 8:20
• @sound wave Not really. As far as I know the choice of "a" is the model assumption. In the interest rates world there are many models that have different "a" function (vasicek, cir, hull-white, etc). Which one of them is correct? Probably none of them. You should check stability of the parameters day to day. You find parameters by calibration to options and if they are stable i.e. $\kappa$ and $\mu$ don't change much day to day, the model is fine.
– emot
Jul 8 at 12:12
• @sound wave why do we want stable parameters? Because if we hedge one contract with the other, the only change in value should be explained by the greeks, not by the change of the parameters, because change in parameters is unhedgable.
– emot
Jul 8 at 12:13

(Just a complementary note)

A variable change $$X = \ln S$$ ($$S =\exp X$$) gives:

$$dS/S = dX + \sigma^2/2 dt,$$

making $$X$$ an Ornstein-Uhlenbeck process

$$dX = \kappa[(\mu - \sigma^2/(2\kappa)) - X] dt + \sigma dW,$$

which allows direct calculations of the time-$$0$$ price of the commodity futures with last trading date $$T$$, $$F(S_0,T)$$:

$$F(S_0,T)=E_0[S_T]=\exp \left( E_0[X_T] +V_0[X_T]/2 \right)$$

$$= \exp \left( e^{-\kappa T} \ln S_0 + (1- e^{-\kappa T})(\mu - \sigma^2/(2\kappa)) + \sigma^2/(2\kappa) (1- e^{-2\kappa T})\right)$$

One can then check that $$F(S_0,T)$$ solves:

\begin{align*} -\frac{\partial F}{\partial T}+\kappa(\mu-\ln S)S\frac{\partial F}{\partial S}+\frac{1}{2} \sigma^2 S^2 \frac{\partial^2 F}{\partial S^2}=0, \; \; F(S,0)=S \end{align*}

This model was first applied in the commodity space, and enhanced to two and three dimensions, by Eduardo Schwartz here.

Also, the market price of risk $$\lambda$$ indeed simply adjusts $$\mu$$ to $$\mu-\lambda$$ both in the underlying's SDE and derivative's PDE.