https://web.ma.utexas.edu/mediawiki/index.php?action=history&feed=atom&title=Optimal_stopping_problemOptimal stopping problem - Revision history2024-03-28T09:57:51ZRevision history for this page on the wikiMediaWiki 1.40.1https://web.ma.utexas.edu/mediawiki/index.php?title=Optimal_stopping_problem&diff=1012&oldid=previmported>Luis at 22:02, 12 March 20122012-03-12T22:02:25Z<p></p>
<table style="background-color: #fff; color: #202122;" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 17:02, 12 March 2012</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1">Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>The optimal stopping is a problem in the context of optimal control whose solution is obtained through the [[obstacle problem]].</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>The optimal stopping is a problem in the context of optimal <ins style="font-weight: bold; text-decoration: none;">[[stochastic </ins>control<ins style="font-weight: bold; text-decoration: none;">]] </ins>whose solution is obtained through the [[obstacle problem]].</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br/></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br/></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Description of the problem ==</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Description of the problem ==</div></td></tr>
</table>imported>Luishttps://web.ma.utexas.edu/mediawiki/index.php?title=Optimal_stopping_problem&diff=1011&oldid=previmported>Luis at 04:24, 29 January 20122012-01-29T04:24:44Z<p></p>
<table style="background-color: #fff; color: #202122;" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 23:24, 28 January 2012</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15">Line 15:</td>
<td colspan="2" class="diff-lineno">Line 15:</td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Derivation of the obstacle problem ==</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Derivation of the obstacle problem ==</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br/></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br/></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">In this setting, </del>the value of $\varphi(X_{\tau})$ depends on the initial point $x$ where the Levy process starts. We can call $u(x) = \mathbb E[\varphi(X_{\tau}) | X_0 = x]$. <del style="font-weight: bold; text-decoration: none;">Under this </del>choice, the function $u$ <del style="font-weight: bold; text-decoration: none;">satisfies </del>the <del style="font-weight: bold; text-decoration: none;">obstacle problem with </del>$<del style="font-weight: bold; text-decoration: none;">L</del>$ <del style="font-weight: bold; text-decoration: none;">being </del>the <del style="font-weight: bold; text-decoration: none;">generator </del>of the <del style="font-weight: bold; text-decoration: none;">Levy process </del>$<del style="font-weight: bold; text-decoration: none;">X_t</del>$.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Let us assume that </ins>the <ins style="font-weight: bold; text-decoration: none;">stochastic process $X_t$ is a [[Levy process]] with generator operator $L$. The expected </ins>value of $\varphi(X_{\tau})$ <ins style="font-weight: bold; text-decoration: none;">naturally </ins>depends on the initial point $x$ where the Levy process starts. We can call $u(x) = \mathbb E[\varphi(X_{\tau}) | X_0 = x]$.</div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The </ins>choice <ins style="font-weight: bold; text-decoration: none;">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.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">At those points $x$ where we would choose to continue</ins>, the function $u$ <ins style="font-weight: bold; text-decoration: none;">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.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">There are other points where we would choose to stop </ins>the <ins style="font-weight: bold; text-decoration: none;">process. At those points we are immediately given the value of the payoff function, thus </ins>$<ins style="font-weight: bold; text-decoration: none;">u(x) = \varphi(x)</ins>$<ins style="font-weight: bold; text-decoration: none;">. On the other hand, if we choose to stop it is because continuing would not improve </ins>the <ins style="font-weight: bold; text-decoration: none;">expected value </ins>of the <ins style="font-weight: bold; text-decoration: none;">payoff, therefore </ins>$<ins style="font-weight: bold; text-decoration: none;">Lu(x) \leq 0</ins>$ <ins style="font-weight: bold; text-decoration: none;">at those points (the function is a supersolution).</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Therefore we have derived the conditions of the [[obstacle problem]]</ins>.</div></td></tr>
</table>imported>Luishttps://web.ma.utexas.edu/mediawiki/index.php?title=Optimal_stopping_problem&diff=1010&oldid=previmported>Luis at 03:22, 29 January 20122012-01-29T03:22:22Z<p></p>
<table style="background-color: #fff; color: #202122;" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 22:22, 28 January 2012</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1">Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Given </del>a [[<del style="font-weight: bold; text-decoration: none;">Levy process</del>]] $X_t$ we <del style="font-weight: bold; text-decoration: none;">consider </del>the <del style="font-weight: bold; text-decoration: none;">following problem. We want to find the optimal </del>stopping time $\tau$ to maximize the expected value of $\varphi(X_<del style="font-weight: bold; text-decoration: none;">{</del>\tau<del style="font-weight: bold; text-decoration: none;">}</del>)$ <del style="font-weight: bold; text-decoration: none;">for </del>some given <del style="font-weight: bold; text-decoration: none;">(</del>payoff<del style="font-weight: bold; text-decoration: none;">) function </del>$<del style="font-weight: bold; text-decoration: none;">\varphi</del>$.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The optimal stopping is </ins>a <ins style="font-weight: bold; text-decoration: none;">problem in the context of optimal control whose solution is obtained through the </ins>[[<ins style="font-weight: bold; text-decoration: none;">obstacle problem</ins>]]<ins style="font-weight: bold; text-decoration: none;">.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Description of the problem ==</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The setting is the following. There is a stochastic process </ins>$X_t$ <ins style="font-weight: bold; text-decoration: none;">and </ins>we <ins style="font-weight: bold; text-decoration: none;">have </ins>the <ins style="font-weight: bold; text-decoration: none;">choice of </ins>stopping <ins style="font-weight: bold; text-decoration: none;">it at any </ins>time $\tau$<ins style="font-weight: bold; text-decoration: none;">. When we stop, we are given a quantity $\varphi(X_\tau)$. The problem is </ins>to <ins style="font-weight: bold; text-decoration: none;">choose the optimal stopping time that would </ins>maximize <ins style="font-weight: bold; text-decoration: none;">the value of </ins>the expected value of <ins style="font-weight: bold; text-decoration: none;">the final payoff </ins>$\varphi(X_\tau)$<ins style="font-weight: bold; text-decoration: none;">.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">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 $X_t$.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The problem may have </ins>some <ins style="font-weight: bold; text-decoration: none;">extra constraints. We may be forced to stop before an expiration time T or as soon as $X_t$ exits a domain $D$.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Connection with American options ==</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">In finance, an option gives an agent the possibility to buy or sell a </ins>given <ins style="font-weight: bold; text-decoration: none;">asset or basket of assets in the future. The </ins>payoff <ins style="font-weight: bold; text-decoration: none;">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 </ins>$<ins style="font-weight: bold; text-decoration: none;">X_t</ins>$<ins style="font-weight: bold; text-decoration: none;">, the optimal choice of the moment to exercise the option in order to maximize the expected payoff corresponds to the optimal stopping problem.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">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)</ins>.</div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Derivation of the obstacle problem ==</ins></div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br/></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br/></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 $X_t$.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 $X_t$.</div></td></tr>
</table>imported>Luishttps://web.ma.utexas.edu/mediawiki/index.php?title=Optimal_stopping_problem&diff=1009&oldid=previmported>Luis at 18:52, 28 January 20122012-01-28T18:52:25Z<p></p>
<table style="background-color: #fff; color: #202122;" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 13:52, 28 January 2012</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1">Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">This is an excerpt from the main article on the [[optimal stopping problem]].</del></div></td><td colspan="2" class="diff-side-added"></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 $\varphi$.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 $\varphi$.</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br/></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br/></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 $X_t$.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 $X_t$.</div></td></tr>
</table>imported>Luishttps://web.ma.utexas.edu/mediawiki/index.php?title=Optimal_stopping_problem&diff=1008&oldid=previmported>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..."2012-01-28T18:52:10Z<p>Created page with "This is an excerpt from the main article on the <a href="/mediawiki/index.php/Optimal_stopping_problem" title="Optimal stopping problem">optimal stopping problem</a>. Given a <a href="/mediawiki/index.php/Levy_process" class="mw-redirect" title="Levy process">Levy process</a> $X_t$ we consider the following problem. We want to find the optimal stopp..."</p>
<p><b>New page</b></p><div>This is an excerpt from the main article on the [[optimal stopping problem]].<br />
<br />
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 $\varphi$.<br />
<br />
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 $X_t$.</div>imported>Luis