Optimal stopping

Stopping rule problems are associated with two objects:

1. A sequence of random variables ${\displaystyle X_{1},X_{2},\ldots }$, whose joint distribution is something assumed to be known
2. A sequence of 'reward' functions ${\displaystyle (y_{i})_{i\geq 1}}$ which depend on the observed values of the random variables in 1:

   ${\displaystyle y_{i}=y_{i}(x_{1},\ldots ,x_{i})}$

Given those objects, the problem is as follows:

• You are observing the sequence of random variables, and at each step ${\displaystyle i}$, you can choose to either stop observing or continue
• If you stop observing at step ${\displaystyle i}$, you will receive reward ${\displaystyle y_{i}}$
• You want to choose a stopping rule to maximize your expected reward (or equivalently, minimize your expected loss)

Consider a gain process ${\displaystyle G=(G_{t})_{t\geq 0}}$ defined on a filtered probability space ${\displaystyle (\Omega ,{\mathcal {F}},({\mathcal {F}}_{t})_{t\geq 0},\mathbb {P} )}$ and assume that ${\displaystyle G}$ is adapted to the filtration. The optimal stopping problem is to find the stopping time ${\displaystyle \tau ^{*}}$ which maximizes the expected gain

${\displaystyle V_{t}^{T}=\mathbb {E} G_{\tau ^{*}}=\sup _{t\leq \tau \leq T}\mathbb {E} G_{\tau }}$

where ${\displaystyle V_{t}^{T}}$ is called the value function. Here ${\displaystyle T}$ can take value ${\displaystyle \infty }$.

A more specific formulation is as follows. We consider an adapted strong Markov process ${\displaystyle X=(X_{t})_{t\geq 0}}$ defined on a filtered probability space ${\displaystyle (\Omega ,{\mathcal {F}},({\mathcal {F}}_{t})_{t\geq 0},\mathbb {P} _{x})}$ where ${\displaystyle \mathbb {P} _{x}}$ denotes the probability measure where the stochastic process starts at ${\displaystyle x}$. Given continuous functions ${\displaystyle M,L}$, and ${\displaystyle K}$, the optimal stopping problem is

${\displaystyle V(x)=\sup _{0\leq \tau \leq T}\mathbb {E} _{x}\left(M(X_{\tau })+\int _{0}^{\tau }L(X_{t})dt+\sup _{0\leq t\leq \tau }K(X_{t})\right).}$

This is sometimes called the MLS (which stand for Mayer, Lagrange, and supremum, respectively) formulation.[4]

(Example where ${\displaystyle \mathbb {E} (y_{i})}$ 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_i (for i ≥ 1) forms a sequence of independent, identically distributed random variables with Bernoulli distribution

${\displaystyle {\text{Bern}}\left({\frac {1}{2}}\right),}$

and if

${\displaystyle y_{i}={\frac {1}{i}}\sum _{k=1}^{i}X_{k}}$

then the sequences ${\displaystyle (X_{i})_{i\geq 1}}$, and ${\displaystyle (y_{i})_{i\geq 1}}$ are the objects associated with this problem.

(Example where ${\displaystyle \mathbb {E} (y_{i})}$ does not necessarily converge)

You have a house and wish to sell it. Each day you are offered ${\displaystyle X_{n}}$ for your house, and pay ${\displaystyle k}$ to continue advertising it. If you sell your house on day ${\displaystyle n}$, you will earn ${\displaystyle y_{n}}$, where ${\displaystyle y_{n}=(X_{n}-nk)}$.

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

In this example, the sequence (${\displaystyle X_{i}}$) is the sequence of offers for your house, and the sequence of reward functions is how much you will earn.

(Example where ${\displaystyle (X_{i})}$ 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 ${\displaystyle R_{1},\ldots ,R_{n}}$ (n is some large number) are the ranks of the objects, and ${\displaystyle y_{i}}$ is the chance you pick the best object if you stop intentionally rejecting objects at step i, then ${\displaystyle (R_{i})}$ and ${\displaystyle (y_{i})}$ 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).

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.[6]

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 ${\displaystyle r}$ be the risk-free interest rate and ${\displaystyle \delta }$ and ${\displaystyle \sigma }$ be the dividend rate and volatility of the stock. The stock price ${\displaystyle S}$ follows geometric Brownian motion

${\displaystyle S_{t}=S_{0}\exp \left\{\left(r-\delta -{\frac {\sigma ^{2}}{2}}\right)t+\sigma B_{t}\right\}}$

under the risk-neutral measure.

When the option is perpetual, the optimal stopping problem is

${\displaystyle V(x)=\sup _{\tau }\mathbb {E} _{x}\left[e^{-r\tau }g(S_{\tau })\right]}$

where the payoff function is ${\displaystyle g(x)=(x-K)^{+}}$ for a call option and ${\displaystyle g(x)=(K-x)^{+}}$ for a put option. The variational inequality is

${\displaystyle \max \left\{{\frac {1}{2}}\sigma ^{2}x^{2}V''(x)+(r-\delta )xV'(x)-rV(x),g(x)-V(x)\right\}=0}$

for all ${\displaystyle x\in (0,\infty )\setminus \{b\}}$ where ${\displaystyle b}$ is the exercise boundary. The solution is known to be[7]

• (Perpetual call) ${\displaystyle V(x)={\begin{cases}(b-K)(x/b)^{\gamma }&x\in (0,b)\\x-K&x\in [b,\infty )\end{cases}}}$ where ${\displaystyle \gamma =({\sqrt {\nu ^{2}+2r}}-\nu )/\sigma }$ and ${\displaystyle \nu =(r-\delta )/\sigma -\sigma /2,\quad b=\gamma K/(\gamma -1).}$
• (Perpetual put) ${\displaystyle V(x)={\begin{cases}K-x&x\in (0,c]\\(K-c)(x/c)^{\tilde {\gamma }}&x\in (c,\infty )\end{cases}}}$ where ${\displaystyle {\tilde {\gamma }}=-({\sqrt {\nu ^{2}+2r}}+\nu )/\sigma }$ and ${\displaystyle \nu =(r-\delta )/\sigma -\sigma /2,\quad c={\tilde {\gamma }}K/({\tilde {\gamma }}-1).}$

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.