1
$\begingroup$

I came across with the following problem:

For the Ornstein-Uhlenbeck process $(X_t, 0\leq t\leq T)$ with initial condition $X_0 = x$, find the stopping time $\tau$ that maximizes $\mathbb{E}[e^{-r\tau}(S - X_\tau)^+]$, for $r\geq 0$ and $S > 0$.

I am aware that, if the process was a geometric Brownian motion instead of a Ornstein-Uhlenbeck, that problem gets translated into the optimal exercising of an American put option, where $T$, $S$, and $r$ happen to be the maturity date, strike price, and discount rate (which is taken to be equal to the interest rate).

I wonder if the problem is still relevant from a financial perspective. I know that an Ornstein-Uhlenbeck process could be used for modeling interest rates, and in that case, it is known as the Vasicek model or the Hull-White model, but the way it is used in the problem suggests that it models the stock price rather than the interest rate. One could argue that the problem may represent the optimal exercising of an American-style interest rate option, but still, the interest rate is already considered in the exponential discount.

I guess that the question could be boiled down to: Is an Ornstein-Uhlenbeck process used to model stock prices? Can you think of a reasonable application of the problem described above?

Just in case, I leave here the dynamics of the Ornstein-Uhlenbec process: $$ \mathrm{d}X_t = a(b - X_t)\mathrm{d}t + c\mathrm{d}W_t, \quad X_0 = x,\quad a > 0, b \in \mathbb{R}, c > 0,\quad 0\leq t\leq T. $$

$\endgroup$
2
  • $\begingroup$ As you say, OU processes can be used to model mean reverting variables like interest rates, variances, inflation etc. Moreover, OU is popular for real options models: you can model the output price of a firm's good by an OU process (competition will ensure that the price doesn't go 0 or infinity). The firm's production and (dis)investment options are then (perpetual) options written on that OU process. $\endgroup$
    – Kevin
    Feb 24 at 18:13
  • $\begingroup$ Thanks for the comment. I kind of get your point, but still don't know why the price of a firm's good may be modeled by means of OU process. Or is it just an arbitrary decision? $\endgroup$
    – Aguazz
    Feb 25 at 18:12

1 Answer 1

0
$\begingroup$

If one insists on first principles then OU processes should be limited to variables that do not describe prices of trades assets. This limits OU to -as you say- interest rates for example.

A process that exhibits mean reversion under the risk-neutral measure cannot really be used to model the price process of an asset like a stock or an FX rate in an arbitrage-free model. That's because $$ S_te^{(\delta-r)t}\quad\text{ resp . }\quad X_te^{(r_f-r_d)\,t} $$ must be martingales which allows only SDEs $$ \frac{dS_t}{S_t}=(r-\delta)\,dt+\sigma\,dW_t\quad\text{ resp . }\quad\frac{dX_t}{X_t}=(r_f-r_d)\,dt+\sigma\,dW_t $$ I use here constant parameters only for simplicity. The point is that quite generally a mean reverting drift is not possible in these equations. At least not under the risk-neutral measure if one insists on a school book arbitrage free model.

Having said that: It is quite possible that there are asset classes that are badly described by such SDEs. Commodity prices could be such which exhibit all kinds of weird features such as seasonality, perhaps even mean reversion.

In the end the model will be a best practice compromise between first principles and realistic behaviour.

$\endgroup$
1
  • $\begingroup$ It is true that if one sticks to arbitrage-free assumptions the OU process may find no place in modeling a stock price. But I think that one could come across arbitrage situations. I was thinking about pair trading. Does it make sense to you that one could combine two American options, on different positions, and written on assets that exhibit mean-reverting properties, and then end up with an American option with the spread as the underlying asset? The spread could arguably be modeled with an OU process, as it is typically done in pair trading. $\endgroup$
    – Aguazz
    Feb 25 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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