Optimal Stopping

An Optimal Stopping Problem is an Markov Decision Process where there are two actions: $a=0$ meaning to stop, and $a=1$ meaning to continue. Here there are two types of costs

$c(x,a)=\begin{cases} \kappa(x),& \text{for }a=0\quad\textit{ (the stopping cost)}\\ c(x),& \text{for }a=1\quad\textit{ (the continuation cost)}, \end{cases}$

This defines a stopping problem.

Assuming that time is finite, the Bellman equation is

$C_s(x) = \min \left\{ k(x) , c(x) + \mathbb{E}_x [C_{s-1}(\hat{X})] \right\}$

for $s\in \mathbb N$ and $C_0(x)=k(x)$ .

Def [OLSA rule] In the one step lookahead (OSLA) rule we stop when ever $x\in S$ where

$S=\{ x: k(x)\leq c(x) + \mathbb{E}_x[k(\hat{X})]\}.$

We call $S$ the stopping set. In words, you stop whenever it is better stop now rather than continue one step further and then stop.

Def [Closed Stopping Set] We say the set $S\subset {\mathcal X}$ is closed, it once inside that said you cannot leave, i.e.

$P_{xy}=0,\qquad \forall x\in S, y\notin S.$

Prop 1. If, for the finite time stopping problem, the set $S$ given by the one step lookahead rule is closed then the one step lookahead rule is an optimal policy.

Proof. Given the set $S$ is closed, we argue that if $C_{s-1}(x)=k(x)$ for $x\in S$ then $C_s(x)=k(x)$ :If $x\in S$ then since $S$ is closed $\hat{X} \in S$ . In otherwords $C_{s-1}(\hat{X}) =k(\hat{X})$ . Therefore, in this case, Bellman’s equation becomes

$C_s(x) = \min \{ k(x), c(x) + \mathbb E_x [ C_{s-1}(\Hat{X})] \} = \min \{ k(x) , c(x) + \mathbb E_x \big[ k(\hat{X} ) \big] \} = k(x).$

The last inequality above follows by the definition of $x\in S$ .

We now proceed by induction. The OSLA rule is optimal for $s=1$ steps, since OSLA is exactly the optimal policy for one step.

Suppose that the result is holds for upto $s-1$ steps. Now consider the Optimal Stopping Problem with $s$ steps. If $x\in S$ then $C_s(x)=k(x)$ . So it is better to stop. If $x\notin S$ , then clearly it’s better to continue. $\square$

Ex. [The Secretary Problem]. There are $N$ candidates for a secretary job. You interview candidates sequentially. After each interview, you must either accept or reject the candidate. We assume each candidate has the rank: $1,2,...,N$ And arrive for interview uniformly at random. Find the policy that maximises the probability that you hire the best candidate.

Ans. All that matters at each time is if the current candidate is the best so far. At time $t$ let

$x_t = \begin{cases} 1, & \text{if candidate is best so far,}\\ 0, & \text{otherwise,} \end{cases} \quad a_t = \begin{cases} 1, & \text{if candidate accepted,}\\ 0, & \text{if reject.} \end{cases}$

Since is uniform random where the best candidate is $\mathbb P ( t\text{ is the best } | x_t=1 ) = \frac{{1}/{N}}{{1}/{t}} = \frac{t}{N}.$

Under the chosen policy, we let

$R_t = \mathbb P (\text{ select the best candidate } | \text{ rejected first } t )$

be our reward function. Now

$\begin{aligned} x_t = 1,\text{ w.p. } \frac{1}{t} \implies \begin{cases} a_t = 1 & \text{ gives expected reward } \frac{t}{N} \\ a_t = 0 & \text{ gives expected reward } R_{t+1} \end{cases} \\ x_t = 0, \text{ w.p. } 1-\frac{1}{t} \implies \begin{cases} a_t = 1 & \text{ gives expected reward } 0 \\ a_t = 0 & \text{ gives expected reward } R_{t+1}. \end{cases}\end{aligned}$

Thus the Bellman equation for the above problem is

$\begin{aligned} R_t & = \underbrace{ \frac{1}{t} \max \left\{ \frac{t}{N} , R_{t+1} \right\} }_{ \max\{ 1/N , R_{t+1}/t \} } + \left( 1-\frac{1}{t} \right) \underbrace{ \max \left\{ 0, R_{t+1} \right\} }_{ R_{t+1} } \\ & = \max \left\{ \frac{1-t}{t} R_{t+1} + \frac{1}{N} R_{t+1} \right\} \\ & = \begin{cases} R_{t+1}, & \text{ if } R_{t+1} \geq \frac{t}{N}, \\ \frac{t-1}{t} R_{t+1} + \frac{1}{N}, & \text{ if } R_{t+1} < \frac{t}{N}. \end{cases}\end{aligned}$ Notice that $R_t \geq R_{t+1}$ . Let $t^*$ be the smallest $t$ such that $R_{t^*} \geq (t^*-1)/N$ . Starting from $R_{N+1}=0$ note that so long as $latex R_{t+1}<\frac{t}{N}$ holds in second case in the above expression, we have that

$\frac{R_t}{t-1} = \frac{R_{t-1}}{t} + \frac{1}{N} \frac{1}{t-1}$

Thus

$\frac{R_{t^*}}{t^*-1} = \sum_{t=t^*}^N \frac{R_t}{t-1} - \frac{R_{t+1}}{t} = \sum_{t=t^*}^N \frac{1}{N} \frac{1}{t-1}.$

Thus our condition for the optimal $t^*$ is to take the smallest $t^*$ such that

$\sum_{t=t*}^N \frac{1}{t-1} \geq 1.$

In other words, the optimal policy is to interview the first $t^*$ candidates and then accept the next best candidate.

Ex [The Secretary Problem, continued] Argue that as $N\rightarrow \infty$ , the optimal policy is to interview $e^{-1}$ of the candidates and then to accept the next best candidate.

Ans. From [[OS:Secretary]], the optimal condition is

$\sum_{t=t*}^N \frac{1}{t-1} \geq 1.$

We know that as $N\rightarrow \infty$

$\sum_{t=t^*}^N \frac{1}{t-1} \sim \int_{t^*}^N \frac{1}{t} dt = \log N - \log t^*.$

Thus $\log N - \log t^* \geq 1$ for $t^*/N =e^{-1}$ .

Optimal stopping in infinite time

We now give conditions for the one step look ahead rule to be optimal for infinite time stopping problems.

Prop. If the following two conditions hold

$K = \max_{x} k(x) < \infty$
$C= \min_{x} c(x) > 0$

then the One-Step-Lookahead-Rule is optimal.

Proof. Suppose that the optimal policy $\pi$ stops at time $\tau$ then

$\begin{aligned} (s+1) C \mathbb P ( \tau > s) \leq \mathbb E \left[ \left\{ \sum_{t=0}^{\tau-1} c(x_{t}) + k(x_{\tau}) \right\} \mathbb I [\tau > s] \right] \leq k(x_0) \leq K.\end{aligned}$

Therefore if we follow optimal policy $\pi$ but for the $s$ time horizon problem and stop at $s$ if $\tau \geq s$ then

$L(x) \leq L_s(x) \leq L(x) + K \mathbb P (\tau > s) \leq L(x) + \frac{K^2}{C(s+1)} \xrightarrow[s\rightarrow\infty]{} L(x)$

Thus $L_s(x) \rightarrow L(x)$ .

As before (for the finite time problem), it is no optimal to stop if $x\notin S$ and for the finite time problem $L_s(x) = k(x)$ for all $x\in S$ . Therefore, since $L_s(x) \rightarrow L(x)$ , we have that $L(x)=k(x)$ for all $x\in S$ and there for it is optimal to stop for $x\in S$ . $\square$

The one step lookahead rule is not always the correct solution to an optimal stopping problem.

Def. [Concave Majorant] For a function $r:\{0,...,N\}\rightarrow \mathbb R_+$ a concave majorant is a function $G$ such that

$G(x)\geq \frac{1}{2} G(x-1) + \frac{1}{2} G(x+1)$
$G(x) \geq r(x)$ .

Prop 3 [Stopping a Random Walk] Let $X_t$ be a symmetric random walk on $\{0,...,N\}$ where the process is automatically stopped at $0$ and $N$ . For each $x\in \{ 0,...,N\}$ , there is a positive reward of $r(x)$ for stopping. We are asked to maximize $\mathbb E [r(X_T)]$ where $T$ is our chosen stopping time. The optimal value function $V(x)$ is the minimal concave majorant, and that it is optimal to stop whenever $V(x)=r(x)$ .

Proof. The Bellman equation is

$R(x) = \max\left\{ r(x), \frac{1}{2} R(x-1) + \frac{1}{2} R(x+1) \right\}$

with $R(x)=r(0)$ and $R(N)=r(N)$ . Thus the optimal value function is a concave majorant.

We will show that the optimal policy is the minimal concave majorant of $r(x)$ . We do so by, essentially applying induction on value iteration. First $R_0(x)=0 \leq G(x)$ for any concave majorant of $r(x)$ . Now suppose that $R_{s-1}$ , the function reached after $s-1$ value iterations, satisfies $R_{s-1}(x)\leq G(x)$ for all $x$ , then

$\begin{aligned} R_s(x) & = \max\left\{ r(x) , \frac{1}{2} R_{s-1}(x-1) + \frac{1}{2} R_{s-1}(x+1) \right\} \\ & \leq \max\left\{ r(x) , \frac{1}{2} G(x-1) + \frac{1}{2} G(x+1) \right\} \\ & \leq \max\left\{ r(x), G(x) \right\} = G(x).\end{aligned}$

Since value iteration converges $R_s(x) \nearrow V(x)$ , where $V(x)$ satisfies $V(x) \leq G(x)$ , as required.

Finally observe that from the Bellman equation the optimal stopping rule is to stop whenever $V(x)=r(x)$ for the minimal concave majorant. $\square$

Optimal stopping in infinite time

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