Pricing Options on a Mean-Reverting Log-Price Process

# Option pricing and mean reversion


# Option pricing and mean reversion












In different books one can find a formula for option pricing when we assume that $\ln(S)$ follows a mean reversion process

$$ dS_t/S_t=\kappa(\theta-\ln(S_t))dt+\sigma dZ$$

If we calculate an adjusted volatility

$$\hat{\sigma}=\sigma\sqrt{\frac{1-e^{-2\kappa T}}{2\kappa}}$$

we can use the standard Black-Scholes formula (see for example "The complete guide to option pricing formulas" from Espen Gaarder Haug, page 410).

This has the effect that the price for the option increases with the time to expiry.

But for me there are now two problems:

(1) This is not intuitive because the distribution of the prices in the futures is almost the same if the price is really mean reverting (assume $T \in \{1,2,3\}$)

(2) If I do a Monte Carlo based on a discretization of the stochastic process and calculate the mean of the payout I get the expected result: The option price does not change with time to expiry

How does this fit together?

P.S.: For simplicity we assume that the risk free interest rate $r$ is $0$.

## Answer by Gordon (score 8)

https://quant.stackexchange.com/a/40188

From the SDE \begin{align*} \frac{dS_t}{S_t}= k(\theta-\ln S_t) dt + \sigma dW_t, \end{align*} where $\{W_t,\, t\ge 0\}$ is a standard Brownian motion, we obtain that \begin{align*} d(e^{kt}\ln S_t) = ke^{kt} \Big(\theta -\frac{1}{2k}\sigma^2\Big) dt + \sigma e^{kt} dW_t. \end{align*} Then, \begin{align*} \ln S_T = e^{-k(T-t)} \ln S_t + \Big(\theta -\frac{1}{2k}\sigma^2\Big)\Big(1-e^{-k(T-t)} \Big)+\sigma \int_t^T e^{-k(T-s)} dW_s. \end{align*} Moreover, for $0 \le t \le T$, the futures price at time $t$ is given by \begin{align*} f(t, T) &= E(S_T\,|\, \mathcal{F}_t)\\ &=\exp\bigg(e^{-k(T-t)} \ln S_t + \Big(\theta -\frac{1}{2k}\sigma^2\Big)\Big(1-e^{-k(T-t)} \Big)+ \frac{\sigma^2}{4k} \Big(1-e^{-2k(T-t)} \Big) \bigg). \end{align*} Note that, $f(T, T) = S_T$, and \begin{align*} df(t, T) &= \sigma e^{-k(T-t)}f(t, T)dW_t,\\ f(0, T) &= \exp\bigg(e^{-kT} \ln S_0 + \Big(\theta -\frac{1}{2k}\sigma^2\Big)\Big(1-e^{-kT} \Big)+ \frac{\sigma^2}{4k} \Big(1-e^{-2kT} \Big) \bigg). \end{align*} For $0\le t \le T$, let \begin{align*} \sigma_{t,T}^f &= \sqrt{\frac{1}{T-t}\int_t^T \sigma^2 e^{-2k(T-s)} ds}\\ &=\sigma\sqrt{\frac{1-e^{-2k(T-t)}}{2k(T-t)}}. \end{align*} Then, the price, at time $0\le t \le T$, of a European exercise style call option with payoff $$(S_T-K)^+,$$ at maturity $T$, is given by \begin{align*} e^{-r(T-t)}\big[f(t, T)\Phi(d_1) - K\Phi(d_2) \big], \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable, and \begin{align*} d_{1, 2} = \frac{\ln\frac{f(t, T)}{K} \pm \frac{(\sigma_{t,T}^f)^2}{2} (T-t)}{\sigma_{t,T}^f \sqrt{T-t}}. \end{align*}