In mathematics, the theory of optimal stopping[1] [2] or early stopping[3] 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.
Stopping rule problems are associated with two objects:
X1,X2,\ldots
(yi)i\ge
yi=yi(x1,\ldots,xi)
Given those objects, the problem is as follows:
i
i
yi
Consider a gain process
G=(Gt)t\ge
(\Omega,l{F},(l{F}t)t\ge,P)
G
\tau*
| T | |
| V | |
| t |
=E
| G | |
| \tau* |
=\supt\leEG\tau
| T | |
| V | |
| t |
T
infty
X=(Xt)t\ge
(\Omega,l{F},(l{F}t)t\ge,Px)
Px
x
M,L
K
V(x)=\sup0\leEx\left(M(X\tau)+
| \tau | |
| \int | |
| 0 |
L(Xt)dt+\sup0\leK(Xt)\right).
There are generally two approaches to solving optimal stopping problems. 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
T
When the underlying process is determined by a family of (conditional) transition functions leading to a Markov family of transition probabilities, powerful analytical tools provided by the theory of Markov processes can often be utilized and this approach is referred to as the Markov method. The solution is usually obtained by solving the associated free-boundary problems (Stefan problems).
Let
Yt
Rk
dYt=b(Yt)dt+\sigma(Yt)dBt+
| \int | |
| Rk |
\gamma(Yt-,z)\bar{N}(dt,dz), Y0=y
B
m
\bar{N}
l
b:Rk\toRk
\sigma:Rk\toRk x
\gamma:Rk x Rk\toRk x
(Yt)
l{S}\subsetRk
\taul{S}=inf\{t>0:Yt\notinl{S}\}
V(y)=
| \sup | |
| \tau\le\taul{S |
If a function
\phi:\bar{l{S}}\toR
\phi\inC(\bar{l{S}})\capC1(l{S})\capC2(l{S}\setminus\partialD)
D=\{y\inl{S}:\phi(y)>M(y)\}
\phi\geM
l{S}
l{A}\phi+L\le0
l{S}\setminus\partialD
l{A}
(Yt)
\phi(y)\geV(y)
y\in\bar{l{S}}
l{A}\phi+L=0
D
\phi(y)=V(y)
y\in\bar{l{S}}
\tau*=inf\{t>0:Yt\notinD\}
These conditions can also be written is a more compact form (the integro-variational inequality):
max\left\{l{A}\phi+L,M-\phi\right\}=0
l{S}\setminus\partialD.
(Example where
E(yi)
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
| Bern\left( | 1 |
| 2 |
\right),
yi=
| 1 | |
| i |
| i | |
| \sum | |
| k=1 |
Xk
(Xi)i\geq
(yi)i\geq
(Example where
E(yi)
You have a house and wish to sell it. Each day you are offered
Xn
k
n
yn
yn=(Xn-nk)
You wish to maximise the amount you earn by choosing a stopping rule.
In this example, the sequence (
Xi
See main article: Secretary problem. (Example where
(Xi)
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
R1,\ldots,Rn
yi
(Ri)
(yi)
See main article: 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.
A special example of an application of search theory is the task of optimal selection of parking space by a driver going to the opera (theater, shopping, etc.). Approaching the destination, the driver goes down the street along which there are parking spaces – usually, only some places in the parking lot are free. The goal is clearly visible, so the distance from the target is easily assessed. The driver's task is to choose a free parking space as close to the destination as possible without turning around so that the distance from this place to the destination is the shortest.[7]
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
r
\delta
\sigma
S
St=S0\exp\left\{\left(r-\delta-
| \sigma2 | |
| 2 |
\right)t+\sigmaBt\right\}
When the option is perpetual, the optimal stopping problem is
V(x)=\sup\tauEx\left[e-r\taug(S\tau)\right]
g(x)=(x-K)+
g(x)=(K-x)+
max\left\{
| 1 | |
| 2 |
\sigma2x2V''(x)+(r-\delta)xV'(x)-rV(x),g(x)-V(x)\right\}=0
x\in(0,infty)\setminus\{b\}
b
V(x)=\begin{cases}(b-K)(x/b)\gamma&x\in(0,b)\ x-K&x\in[b,infty)\end{cases}
\gamma=(\sqrt{\nu2+2r}-\nu)/\sigma
\nu=(r-\delta)/\sigma-\sigma/2, b=\gammaK/(\gamma-1).
V(x)=\begin{cases}K-x&x\in(0,c]\\(K-c)(x/c)\tilde{\gamma}&x\in(c,infty)\end{cases}
\tilde{\gamma}=-(\sqrt{\nu2+2r}+\nu)/\sigma
\nu=(r-\delta)/\sigma-\sigma/2, c=\tilde{\gamma}K/(\tilde{\gamma}-1).
On the other hand, when 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, however, be used. See Black–Scholes model#American options for various valuation methods here, as well as Fugit for a discrete, tree based, calculation of the optimal time to exercise.
(For French translation, see cover story in the July issue of Pour la Science (2009).)