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I'm confused about the impact that a mean reverting stock price process has on the value of an option on it.

Several sources say that there is indeed an impact on the price of an option:

Option pricing and mean reversion

Lo and Wang (1995)

Yet, another source seems to say that mean reversion has no impact on the price of an option:

"The drift term of the process has no impact on the price of a call option, since we know that under the correct pricing measure we need the discounted stock price to have zero drift. This is achieved by changing the drift of the original process, rendering any initial drift term irrelevant"

  • Mark Joshi, Quant Job Interview Questions and Answers.

So I guess my ultimate question is, if the stock price follows the following process:

$$ dS_t=\alpha(\mu-S_t)dt+\sigma S_tdZ$$

Is the price of an option on the stock just equal to the BSM price where $\sigma_{BSM} = \sigma$?

It would make sense to me that there is no effect, because the replicating portfolio argument still works, and we end up with the same PDE and boundary conditions, which would give the same price.

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    $\begingroup$ Mark Joshi was taking about SDE's such as $dS_t = \mu S_t dt + \sigma S_t dZ$ not the equation you have. $\endgroup$
    – Alex C
    Commented Jun 6, 2019 at 19:07
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    $\begingroup$ I'm copied the formula he gave in the book exactly. For reference, I'm looking at problem 3.43 in the book (version 1.01) if you can find it online. $\endgroup$
    – pauliewalnuts
    Commented Jun 6, 2019 at 19:35
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    $\begingroup$ Although it is counterintuitive , I think Joshi is correct. Whatever the instantaneous drift, there is some local change of measure you can do to make the risk neutral drift equal the risk free rate. However I don’t see what’s wrong with the references that say otherwise. Great question. $\endgroup$
    – dm63
    Commented Jun 6, 2019 at 23:12
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    $\begingroup$ Don’t the references unequivocally contradict Joshi? Clearly the answer cannot both be yes and no... $\endgroup$
    – pauliewalnuts
    Commented Jun 6, 2019 at 23:18
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    $\begingroup$ A similar question is answered here. For a process of the form $dS_t/S_t = \mu(S_t,t) \, dt + \sigma \, dZ_t$ change of measure leads to a price in the form of Black-Scholes. What changes is how the volatility parameter is used in the option formula. With mean reversion volatility no longer scales like the square root of time, so using $\sigma \sqrt{T}$ for expiration time $T$ in the formula will be incorrect. $\endgroup$
    – RRL
    Commented Jun 7, 2019 at 3:07

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Joshi is correct. The no arbitrage argument implies that the stock price instantaneous return under the risk neutral measure is equal to the short rate, and the girsanov theorem implies that the instantaneous volatility $\sigma$ is the same under the historical measure and under the risk neutral measure, so under the risk neutral measure the stock price is a GBM with drift $r$ and volatility $\sigma$ and the option pricing formula is BS with volatility $\sigma$ even when the stock process is mean reverting under the historical measure.

However if the stock process is mean reverting under the historical measure, then an historical estimation of $\sigma$ must take into account the mean reversion and cannot rely on simply computing the standard deviation of $\delta t$ returns. For instance with a 100% mean reversion a 20% historical annual standard deviation would translate into approximately 30% instantaneous volatility. This is in essence the topic of the Lo and Wang paper, computing options prices with mean reversion under the historical measure assuming the historical standard deviation is known.

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  • $\begingroup$ So @Antoine Conze, in your example the correct volatility to use in the BS formula is 30%, which contradicts the “adjusted volatility “ formula given in some references , which would give a lower volatility ? $\endgroup$
    – dm63
    Commented Jun 7, 2019 at 10:32
  • $\begingroup$ @dm63 if the process is mean reverting under the risk neutral measure (which cannot be the case for a no dividend stock price) and the mean reversion function is such that the terminal price is log normal than yes its "BS formula" equivalent volatility is lower than its instantaneous volatility. But here I believe we were considering a mean reverting stock price under the historical measure, which becomes GBM under the risk neutral measure. So its historical 1 year standard deviation is lower than its instantaneous volatility. $\endgroup$
    – Antoine Conze
    Commented Jun 7, 2019 at 19:35

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