9
$\begingroup$

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 $0$.

$\endgroup$
9
  • 1
    $\begingroup$ The futures price at maturity is the same as the underlying spot price. What payout you are calculate the mean? The mean of the option payout is the undiscounted option price, it will certainly depend on the maturity. $\endgroup$
    – Gordon
    Jun 6, 2018 at 16:45
  • $\begingroup$ If your option price does not change with time to expiry, then you are doing something wrong in your implementation. As @Gordon mentions the mere fact of discounting should have an impact on the option prices of different maturities. $\endgroup$ Jun 6, 2018 at 17:16
  • $\begingroup$ @Daneel: We can skip the discounting to makes things as easy as possible. $\endgroup$ Jun 7, 2018 at 12:46
  • $\begingroup$ @Gordon: Thanks a lot for your details explanation. Maybe I am doing somewthing wrong, but there is still the same problem: In my eyes the mean reversion makes sure that the standard deviation of the simulated path is always the same (at least at the long end). Is this true or wrong? $\endgroup$ Jun 7, 2018 at 12:53
  • $\begingroup$ @JoergVanAken: That is true for $S_T$, but not for $(S_T-K)^+$. In addition, your SDE has some typos. Can you make the change? $\endgroup$
    – Gordon
    Jun 7, 2018 at 13:11

1 Answer 1

7
$\begingroup$

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*}

$\endgroup$
4
  • 1
    $\begingroup$ There are a couple of little errors. The formula for $\sigma_f $ should have $T-t $ instead of $T $ as we haven't assumed $t=0$. More seriously in the equation for $d_{1,2} $ both of the $T $ are spurious. The $\sigma_f $ calculated is the standard deviation of the distribution of $S_T $, not an implied vol. The implied vol would be divided by $\sqrt {T-t} $. $\endgroup$
    – q.t.f.
    Jun 7, 2018 at 21:45
  • 1
    $\begingroup$ @q.t.f.: I am very happy that you have pointed this out. I will make the changes shortly. $\endgroup$
    – Gordon
    Jun 8, 2018 at 13:57
  • $\begingroup$ Interestingly, it would appear that for high speeds of mean reversion, the value of the option is independent of the underlying price. Hence, the option Delta is zero for cases of high speeds of mean reversion. That is, if $k(T-t)$ is "large," then $f(t,T)$ is effectively independent of $S_t$, therefore $d_1$ and $d_2$ are also effecrively independent of $S_t$, and the value of the option is independent of $S_t$. Can this possible be the case? $\endgroup$
    – Mark Viola
    Oct 8, 2023 at 18:43
  • $\begingroup$ Well, think about it. With a really high mean reversion and you starting a bit away from your long-term average, your process is just immediately adjusting towards it, wherever it starts. If that effect is very strong, it doesn't really matter where you started, if the mean reversion is barely present, it does still matter where you were. $\endgroup$
    – not_sure95
    Oct 13, 2023 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.