Optimal stopping problem

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The optimal stopping is a problem in the context of optimal stochastic control whose solution is obtained through the obstacle problem.

Description of the problem

The setting is the following. There is a stochastic process $X_t$ and we have the choice of stopping it at any time $\tau$. When we stop, we are given a quantity $\varphi(X_\tau)$. The problem is to choose the optimal stopping time that would maximize the value of the expected value of the final payoff $\varphi(X_\tau)$.

The choice of the stopping time $\tau$ has to be made in terms of the information that we have up to time $\tau$ only. In probabilistic technical terms, $\tau$ has to be measurable with respect to the filtration associated with the stochastic process $X_t$.

The problem may have some extra constraints. We may be forced to stop before an expiration time T or as soon as $X_t$ exits a domain $D$.

Connection with American options

In finance, an option gives an agent the possibility to buy or sell a given asset or basket of assets in the future. The payoff of this option is a random variable that will depend on the value of these assets at the moment the option is exercised. In the American market, the options can be exercised any time until their expiration time $T$. If we model the price of the assets by a stochastic process $X_t$, the optimal choice of the moment to exercise the option in order to maximize the expected payoff corresponds to the optimal stopping problem.

This is a highly simplified model for the pricing of American options. In financial mathematics there are other factors that enter into consideration (mostly related to risk).

Derivation of the obstacle problem

Let us assume that the stochastic process $X_t$ is a Levy process with generator operator $L$. The expected value of $\varphi(X_{\tau})$ naturally depends on the initial point $x$ where the Levy process starts. We can call $u(x) = \mathbb E[\varphi(X_{\tau}) | X_0 = x]$.

The choice of when to stop depends on the current position of $X_t$ only. This is a simple consequence of the Markovian property of Levy processes, or in layman's terms, from the fact that the future of a Levy process does not depend on the past but only on the current position. Thus, there are some points $x$ where we would choose to stop, and others where we would choose to continue.

At those points $x$ where we would choose to continue, the function $u$ will satisfy the PDE from the generator operator $Lu(x) = 0$. If we choose to continue it is because this choice is better than stopping. So necessarily $u(x) \geq \varphi(x)$ at these points.

There are other points where we would choose to stop the process. At those points we are immediately given the value of the payoff function, thus $u(x) = \varphi(x)$. On the other hand, if we choose to stop it is because continuing would not improve the expected value of the payoff, therefore $Lu(x) \leq 0$ at those points (the function is a supersolution).

Therefore we have derived the conditions of the obstacle problem.

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