Optimal stopping problem: Difference between revisions
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imported>Luis (Created page with "This is an excerpt from the main article on the optimal stopping problem. Given a Levy process $X_t$ we consider the following problem. We want to find the optimal stopp...") |
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Given a [[Levy process]] $X_t$ we consider the following problem. We want to find the optimal stopping time $\tau$ to maximize the expected value of $\varphi(X_{\tau})$ for some given (payoff) function $\varphi$. | Given a [[Levy process]] $X_t$ we consider the following problem. We want to find the optimal stopping time $\tau$ to maximize the expected value of $\varphi(X_{\tau})$ for some given (payoff) function $\varphi$. | ||
In this setting, the value of $\varphi(X_{\tau})$ depends on the initial point $x$ where the Levy process starts. We can call $u(x) = \mathbb E[\varphi(X_{\tau}) | X_0 = x]$. Under this choice, the function $u$ satisfies the obstacle problem with $L$ being the generator of the Levy process $X_t$. | In this setting, the value of $\varphi(X_{\tau})$ depends on the initial point $x$ where the Levy process starts. We can call $u(x) = \mathbb E[\varphi(X_{\tau}) | X_0 = x]$. Under this choice, the function $u$ satisfies the obstacle problem with $L$ being the generator of the Levy process $X_t$. | ||
Revision as of 12:52, 28 January 2012
Given a Levy process $X_t$ we consider the following problem. We want to find the optimal stopping time $\tau$ to maximize the expected value of $\varphi(X_{\tau})$ for some given (payoff) function $\varphi$.
In this setting, the value of $\varphi(X_{\tau})$ depends on the initial point $x$ where the Levy process starts. We can call $u(x) = \mathbb E[\varphi(X_{\tau}) | X_0 = x]$. Under this choice, the function $u$ satisfies the obstacle problem with $L$ being the generator of the Levy process $X_t$.