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In [[mathematics]], the theory of '''optimal stopping''' is concerned with the problem of choosing a time to take a particular action, in order to [[Optimization (mathematics)|maximise]] an expected reward or minimise an expected cost. Optimal stopping problems can be found in areas of [[statistics]], [[economics]], and [[mathematical finance]] (related to the pricing of [[American options]]). A key example of an optimal stopping problem is the [[secretary problem]]. Optimal stopping problems can often be written in the form of a [[Bellman equation]], and are therefore often solved using [[dynamic programming]].
 
==Definition==
 
===Discrete time case===
 
Stopping rule problems are associated with two objects:
<ol>
<li>A sequence of random variables <math>X_1, X_2, \ldots</math>, whose joint distribution is something assumed to be known
<li>A sequence of 'reward' functions <math>(y_i)_{i\ge 1}</math> which depend on the observed values of the random variables in 1.:
<br><math>y_i=y_i (x_1, \ldots ,x_i)</math>
</ol>
 
Given those objects, the problem is as follows:
* You are observing the sequence of random variables, and at each step <math>i</math>, you can choose to either stop observing or continue
* If you stop observing at step <math>i</math>, you will receive reward <math>y_i</math>
* You want to choose a [[stopping rule]] to maximise your expected reward (or equivalently, minimise your expected loss)
 
===Continuous time case===
 
Consider a gain processes <math>G=(G_t)_{t\ge 0}</math> defined on a [[filtration (mathematics)|filtered]] [[probability space]] <math>(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge 0},\mathbb{P})</math> and assume that <math>G</math> is [[adapted process|adapted]] to the filtration. The optimal stopping problem is to find the [[stopping time]] <math>\tau^*</math> which maximizes the expected gain
:<math> V_t^T = \mathbb{E} G_{\tau^*} = \sup_{t\le \tau \le T} \mathbb{E} G_\tau </math>
where <math>V_t^T</math> is called the [[value function]]. Here <math>T</math> can take value <math>\infty</math>.
 
A more specific formulation is as follows. We consider an adapted strong [[Markov process]] <math>X = (X_t)_{t\ge 0}</math> defined on a filtered probability space <math>(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge 0},\mathbb{P}_x)</math> where <math>\mathbb{P}_x</math> denotes the [[probability measure]] where the [[stochastic process]] starts at <math>x</math>. Given continuous functions <math>M,L</math>, and <math>K</math>, the optimal stopping problem is
:<math> V(x) = \sup_{0\le \tau \le T} \mathbb{E}_x \left( M(X_\tau) + \int_0^\tau L(X_t) dt + \sup_{0\le t\le\tau} K(X_t) \right). </math>
This is sometimes called the MLS (which stand for Mayer, Lagrange, and supremum, respectively) formulation.<ref name="opt2006">{{cite doi|10.1007/978-3-7643-7390-0}}</ref>
 
==Solution methods==
There are generally two approaches of solving optimal stopping problems.<ref name=opt2006/> When the underlying process (or the gain process) is described by its unconditional [[finite-dimensional distribution|finite dimensional distributions]], the appropriate solution technique is the martingale approach, so called because it uses [[Martingale (probability theory)|martingale]] theory, the most important concept being the [[Snell envelope]]. In the discrete time case, if the planning horizon <math>T</math> is finite, the problem can also be easily solved by [[dynamic programming]].
 
When the underlying process is determined by a family of (conditional) transition functions leading to a Markovian family of transition probabilities, very powerful analytical tools provided by the theory of [[Markov process]]es can often be utilized and this approach is referred to as the Markovian method. The solution is usually obtained by solving the associated [[Free boundary problem|free-boundary problems]] ([[Stefan problem]]s).
 
==A jump diffusion result==
Let <math>Y_t</math> be a [[Lévy process|Lévy]] diffusion in <math>\mathbb{R}^k</math> given by the [[Stochastic differential equation|SDE]]
:<math> dY_t = b(Y_t) dt + \sigma (Y_t) dB_t + \int_{\mathbb{R}^k} \gamma (Y_{t-},z)\bar{N}(dt,dz),\quad Y_0 = y </math>
where <math> B </math> is an  <math> m</math>-dimensional [[Brownian motion]], <math> \bar{N} </math> is an <math> l </math>-dimensional compensated [[Poisson random measure]], <math> b:\mathbb{R}^k \to \mathbb{R}^k </math>, <math> \sigma:\mathbb{R}^k \to \mathbb{R}^{k\times m} </math>, and <math> \gamma:\mathbb{R}^k \times \mathbb{R}^k \to \mathbb{R}^{k\times l} </math> are given functions such that a unique solution <math> (Y_t) </math> exists. Let <math> \mathcal{S}\subset \mathbb{R}^k </math> be an open set (the solvency region) and  
:<math> \tau_\mathcal{S} = \inf\{ t>0: Y_t \notin \mathcal{S} \} </math>
be the bankruptcy time. The optimal stopping problem is:
:<math>V(y) = \sup_{\tau \le \tau_\mathcal{S}} J^\tau (y) = \sup_{\tau \le \tau_\mathcal{S}} \mathbb{E}_y \left[ M(Y_\tau) + \int_0^\tau L(Y_t) dt \right]. </math>
It turns out that under some regularity conditions,<ref name="oksendal2007">{{cite doi|10.1007/978-3-540-69826-5}}</ref> the following verification theorem holds:
 
If a function <math>\phi:\bar{\mathcal{S}}\to \mathbb{R}</math> satisfies
* <math> \phi \in C(\bar{\mathcal{S}}) \cap C^1(\mathcal{S}) \cap C^2(\mathcal{S}\setminus \partial D) </math> where the continuation region is <math> D = \{y\in\mathcal{S}: \phi(y) > M(y) \} </math>,
* <math> \phi \ge M </math> on <math> \mathcal{S} </math>, and
* <math> \mathcal{A}\phi + L \le 0 </math> on <math> \mathcal{S} \setminus \partial D </math>, where <math> \mathcal{A} </math> is the [[Infinitesimal generator (stochastic processes)|infinitesimal generator]] of <math> (Y_t) </math>
then <math> \phi(y) \ge V(y) </math> for all <math> y\in \bar{\mathcal{S}} </math>. Moreover, if
* <math> \mathcal{A}\phi + L = 0 </math> on <math> D </math>
Then <math> \phi(y) = V(y) </math> for all <math> y\in \bar{\mathcal{S}} </math> and <math> \tau^* = \inf\{ t>0: Y_t\notin D\} </math> is an optimal stopping time.
 
These conditions can also be written is a more compact form (the [[Variational inequality|integro-variational inequality]]):
* <math> \max\left\{ \mathcal{A}\phi + L, M-\phi \right\} = 0 </math> on <math> \mathcal{S} \setminus \partial D. </math>
 
==Examples==
 
=== Coin tossing ===
(Example where <math>\mathbb{E}(y_i)</math> converges)
 
You have a fair coin and are repeatedly tossing it. Each time, before it is tossed, you can choose to stop tossing it and get paid (in dollars, say) the average number of heads observed.
 
You wish to maximise the amount you get paid by choosing a stopping rule.
If ''X''<sub>''i''</sub> (for ''i'' ≥ 1) forms a sequence of independent, identically distributed random variables with [[Bernoulli distribution]]
: <math>\text{Bern}\left(\frac{1}{2}\right),</math>
and if
: <math>y_i = \frac 1 i \sum_{k=1}^{i} X_k</math>
then the sequences <math>(X_i)_{i\geq 1}</math>, and <math>(y_i)_{i\geq 1}</math> are the objects associated with this problem.
 
=== House selling ===
(Example where <math>\mathbb{E}(y_i)</math> does not necessarily converge)
 
You have a house and wish to sell it. Each day you are offered <math>X_n</math> for your house, and pay <math>k</math> to continue advertising it. If you sell your house on day <math>n</math>, you will earn <math>y_n</math>, where <math>y_n = (X_n - nk)</math>.
 
You wish to maximise the amount you earn by choosing a stopping rule.
 
In this example, the sequence (<math>X_i</math>) is the sequence of offers for your house, and the sequence of reward functions is how much you will earn.
 
=== Secretary problem ===
{{Main|Secretary problem}}
(Example where <math>(X_i)</math> is a finite sequence)
 
You are observing a sequence of objects which can be ranked from best to worst. You wish to choose a stopping rule which maximises your chance of picking the best object.
 
Here, if <math>R_1, \ldots, R_n</math> (''n'' is some large number, perhaps) are the ranks of the objects, and <math>y_i</math> is the chance you pick the best object if you stop intentionally rejecting objects at step i, then <math>(R_i)</math> and <math>(y_i)</math> are the sequences associated with this problem. This problem was solved in the early 1960s by several people. An elegant solution to the secretary problem and several modifications of this problem is provided by the more recent [[odds algorithm]]
of optimal stopping (Bruss algorithm).
 
=== Search theory ===
{{Main|Search theory}}
Economists have studied a number of optimal stopping problems similar to the 'secretary problem', and typically call this type of analysis 'search theory'. Search theory has especially focused on a worker's search for a high-wage job, or a consumer's search for a low-priced good.
 
=== Option trading ===
In the trading of [[Option (finance)|options]] on [[financial market]]s, the holder of an [[Option style|American option]] is allowed to exercise the right to buy (or sell) the underlying asset at a predetermined price at any time before or at the expiry date. Therefore the valuation of American options is essentially an optimal stopping problem. Consider a classical [[Black-Scholes]] set-up and let <math> r </math> be the [[risk-free interest rate]] and <math> \delta </math> and <math> \sigma </math> be the dividend rate and volatility of the stock. The stock price <math> S </math> follows geometric Brownian motion
:<math> S_t = S_0 \exp\left\{ \left(r - \delta - \frac{\sigma^2}{2}\right) t + \sigma B_t \right\} </math>
under the risk-neutral measure. When the option is perpetual, the optimal stopping problem is
:<math> V(x) = \sup_{\tau} \mathbb{E}_x \left[ e^{-r\tau} g(S_\tau) \right] </math>
where the payoff function is <math> g(x) = (x-K)^+ </math> for a call option and <math> g(x) = (K-x)^+ </math> for a put option. The variational inequality is
:<math> \max\left\{ \frac{1}{2} \sigma^2 x^2 V''(x) + (r-\delta) x V'(x) - rV(x), g(x) - V(x) \right\} = 0</math>
for all <math>x \in (0,\infty)\setminus \{b\}</math>
where <math> b </math> is the exercise boundary. The solution is known to be<ref name="karatzas1998">{{cite doi|10.1007/b98840}}</ref>
* (Perpetual call) <math> V(x) = \begin{cases} (b-K)(x/b)^\gamma & x\in(0,b) \\ x-K & x\in[b,\infty)  \end{cases} </math> where <math> \gamma = (\sqrt{\nu^2 + 2r} - \nu) / \sigma</math> and <math> \nu = (r-\delta)/\sigma - \sigma / 2, \quad b = \gamma K / (\gamma - 1). </math>
* (Perpetual put) <math> V(x) = \begin{cases} K - x & x\in(0,c] \\(K-c)(x/c)^\tilde{\gamma} & x\in(c,\infty)  \end{cases} </math> where <math> \tilde{\gamma} = -(\sqrt{\nu^2 + 2r} + \nu) / \sigma </math> and <math> \nu = (r-\delta)/\sigma - \sigma / 2, \quad c = \tilde{\gamma} K / (\tilde{\gamma} - 1). </math>
If the expiry date is finite, the problem is associated with a 2-dimensional free-boundary problem with no known closed-form solution. Various numerical methods can be used.
 
==See also==
*[[Stochastic control]]
*[[Markov decision process]]
 
==References==
{{Reflist}}
* T. P. Hill. "[http://www.americanscientist.org/issues/feature/2009/2/knowing-when-to-stop/1 Knowing When to Stop]". ''American Scientist'', Vol. 97, 126-133 (2009). (For French translation, see [http://www.pourlascience.fr/ewb_pages/f/fiche-article-savoir-quand-s-arreter-22670.php cover story] in the July issue of ''Pour la Science'' (2009))
* [http://www.math.ucla.edu/~tom/Stopping/Contents.html Optimal Stopping and Applications], retrieved on 21 June 2007
*Thomas S. Ferguson. "Who solved the secretary problem?" ''Statistical Science'', Vol. 4.,282-296, (1989)
* [[F. Thomas Bruss]]. "Sum the odds to one and stop." ''Annals of Probability'', Vol. 28, 1384–1391,(2000)
* F. Thomas Bruss. "The art of a right decision: Why decision makers want to know the odds-algorithm." ''Newsletter of the European Mathematical Society'', Issue 62, 14-20, (2006)
*R. Rogerson, R. Shimer, and R. Wright (2005), 'Search-theoretic models of the labor market: a survey'. ''Journal of Economic Literature'' 43, pp.&nbsp;959–88.
{{Use dmy dates|date=September 2010}}
 
==External links==
* [http://www.spotlightmind.com/optimal-search Neil Bearden's Optimal Search Page]
 
{{DEFAULTSORT:Optimal Stopping}}
[[Category:Mathematical optimization]]
[[Category:Mathematical finance]]
[[Category:Decision theory]]
[[Category:Sequential methods]]
[[Category:Dynamic programming]]

Revision as of 08:04, 1 May 2013

In mathematics, the theory of optimal stopping is concerned with the problem of choosing a time to take a particular action, in order to maximise an expected reward or minimise an expected cost. Optimal stopping problems can be found in areas of statistics, economics, and mathematical finance (related to the pricing of American options). A key example of an optimal stopping problem is the secretary problem. Optimal stopping problems can often be written in the form of a Bellman equation, and are therefore often solved using dynamic programming.

Definition

Discrete time case

Stopping rule problems are associated with two objects:

  1. A sequence of random variables , whose joint distribution is something assumed to be known
  2. A sequence of 'reward' functions which depend on the observed values of the random variables in 1.:

Given those objects, the problem is as follows:

Continuous time case

Consider a gain processes defined on a filtered probability space and assume that is adapted to the filtration. The optimal stopping problem is to find the stopping time which maximizes the expected gain

where is called the value function. Here can take value .

A more specific formulation is as follows. We consider an adapted strong Markov process defined on a filtered probability space where denotes the probability measure where the stochastic process starts at . Given continuous functions , and , the optimal stopping problem is

This is sometimes called the MLS (which stand for Mayer, Lagrange, and supremum, respectively) formulation.[1]

Solution methods

There are generally two approaches of solving optimal stopping problems.[1] When the underlying process (or the gain process) is described by its unconditional finite dimensional distributions, the appropriate solution technique is the martingale approach, so called because it uses martingale theory, the most important concept being the Snell envelope. In the discrete time case, if the planning horizon is finite, the problem can also be easily solved by dynamic programming.

When the underlying process is determined by a family of (conditional) transition functions leading to a Markovian family of transition probabilities, very powerful analytical tools provided by the theory of Markov processes can often be utilized and this approach is referred to as the Markovian method. The solution is usually obtained by solving the associated free-boundary problems (Stefan problems).

A jump diffusion result

Let be a Lévy diffusion in given by the SDE

where is an -dimensional Brownian motion, is an -dimensional compensated Poisson random measure, , , and are given functions such that a unique solution exists. Let be an open set (the solvency region) and

be the bankruptcy time. The optimal stopping problem is:

It turns out that under some regularity conditions,[2] the following verification theorem holds:

If a function satisfies

then for all . Moreover, if

Then for all and is an optimal stopping time.

These conditions can also be written is a more compact form (the integro-variational inequality):

Examples

Coin tossing

(Example where converges)

You have a fair coin and are repeatedly tossing it. Each time, before it is tossed, you can choose to stop tossing it and get paid (in dollars, say) the average number of heads observed.

You wish to maximise the amount you get paid by choosing a stopping rule. If Xi (for i ≥ 1) forms a sequence of independent, identically distributed random variables with Bernoulli distribution

and if

then the sequences , and are the objects associated with this problem.

House selling

(Example where does not necessarily converge)

You have a house and wish to sell it. Each day you are offered for your house, and pay to continue advertising it. If you sell your house on day , you will earn , where .

You wish to maximise the amount you earn by choosing a stopping rule.

In this example, the sequence () is the sequence of offers for your house, and the sequence of reward functions is how much you will earn.

Secretary problem

Mining Engineer (Excluding Oil ) Truman from Alma, loves to spend time knotting, largest property developers in singapore developers in singapore and stamp collecting. Recently had a family visit to Urnes Stave Church. (Example where is a finite sequence)

You are observing a sequence of objects which can be ranked from best to worst. You wish to choose a stopping rule which maximises your chance of picking the best object.

Here, if (n is some large number, perhaps) are the ranks of the objects, and is the chance you pick the best object if you stop intentionally rejecting objects at step i, then and are the sequences associated with this problem. This problem was solved in the early 1960s by several people. An elegant solution to the secretary problem and several modifications of this problem is provided by the more recent odds algorithm of optimal stopping (Bruss algorithm).

Search theory

Mining Engineer (Excluding Oil ) Truman from Alma, loves to spend time knotting, largest property developers in singapore developers in singapore and stamp collecting. Recently had a family visit to Urnes Stave Church. Economists have studied a number of optimal stopping problems similar to the 'secretary problem', and typically call this type of analysis 'search theory'. Search theory has especially focused on a worker's search for a high-wage job, or a consumer's search for a low-priced good.

Option trading

In the trading of options on financial markets, the holder of an American option is allowed to exercise the right to buy (or sell) the underlying asset at a predetermined price at any time before or at the expiry date. Therefore the valuation of American options is essentially an optimal stopping problem. Consider a classical Black-Scholes set-up and let be the risk-free interest rate and and be the dividend rate and volatility of the stock. The stock price follows geometric Brownian motion

under the risk-neutral measure. When the option is perpetual, the optimal stopping problem is

where the payoff function is for a call option and for a put option. The variational inequality is

for all where is the exercise boundary. The solution is known to be[3]

If the expiry date is finite, the problem is associated with a 2-dimensional free-boundary problem with no known closed-form solution. Various numerical methods can be used.

See also

References

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

  • T. P. Hill. "Knowing When to Stop". American Scientist, Vol. 97, 126-133 (2009). (For French translation, see cover story in the July issue of Pour la Science (2009))
  • Optimal Stopping and Applications, retrieved on 21 June 2007
  • Thomas S. Ferguson. "Who solved the secretary problem?" Statistical Science, Vol. 4.,282-296, (1989)
  • F. Thomas Bruss. "Sum the odds to one and stop." Annals of Probability, Vol. 28, 1384–1391,(2000)
  • F. Thomas Bruss. "The art of a right decision: Why decision makers want to know the odds-algorithm." Newsletter of the European Mathematical Society, Issue 62, 14-20, (2006)
  • R. Rogerson, R. Shimer, and R. Wright (2005), 'Search-theoretic models of the labor market: a survey'. Journal of Economic Literature 43, pp. 959–88.

30 year-old Entertainer or Range Artist Wesley from Drumheller, really loves vehicle, property developers properties for sale in singapore singapore and horse racing. Finds inspiration by traveling to Works of Antoni Gaudí.

External links