# ECSE 506: Stochastic Control and Decision Theory

Example: Optimal choice of the best alternative

A decision maker (DM) wants to select the best alternative from a set of $$T$$ alternatives. The DM evaluates the alternatives sequentially. After evaluating alternative $$t$$, the DM knows whether alternative $$t$$ was the best alternative so far or not. Based on this information, the DM must decide whether to choose alternative $$t$$ and stop the search, or to permanently reject alternative $$t$$ and evaluate remaining alternatives. The DM may reject the last alternative and not make a choice at all. All alternatives are equally likely to be the best. Find the optimal choice strategy that maximize the probability of picking the best alternative.

This optimization problem is known by different names including secretary problem (in which the alternatives correspond to finding the best candidate as a secretary), marriage problem (in which the alternatives correspond of find the best spouse), Googol (in which the alternatives consist of finding the biggest number), parking problem (in which the alternatives correspond to finding the nearest parking spot) and so on

We can view this an optimal stopping problem with binary state $$X_t$$. $$X_t = 1$$ means that alternative $$t$$ is the best alternative so far. Thus, $$\{X_t\}_{t \ge 1}$$ is an independent process with $$\PR(X_t = 1) = 1/t$$. The continuation reward is zero. The DM receives a stopping reward only if the current alternative is best, i.e., the current alternative is best so far ($$X_t = 1$$) and better than all future alternative ($$X_\tau = 0, \tau > t$$). Thus, the expected stopping reward conditioned on $$X_t$$ is $s_t(x) = \IND\{ x = 1 \} \cdot \PR( X_{t+1:T} = 0 | X_t = x ) = x \cdot \frac tT.$ Thus, the optimal strategy is given by the solution of the followind dynamic program.

Dynamic program \begin{align*} V_{T+1}(x) &= 0 \\ V_t(x) &= \max\bigg\{ x \cdot \frac tT, \EXP[ V_{t+1}(X_{t+1}) ] \bigg\} \end{align*}

# 1 Structure of optimal policy

Lemma

Define $L_t = V_t(0) = \frac t{t+1} V_{t+1}(0) + \frac 1{t+1}V_{t+1}(1).$ Then, $V_t(1) = \max\bigg\{ \frac tT, L_t \bigg\}$ and, therefore, $L_t - L_{t+1} = \bigg[ \frac 1T - \frac {L_{t+1}}{t+1} \bigg]^+ \quad \text{with } L_T = 0.$

Note that it is never optimal to select an alternative if it is not the best so far (i.e., $$X_t = 0$$). Thus, we can completely characterize an optimal strategy by solving for $$\{L_t\}_{t=1}^T$$ in a backward manner.

#### Proof

The result follows immediately from the definition of $$L_t$$.

Theorem
1. There exists a critical time $$t_0$$, $$t_0 < T$$, such that it is optimal to reject all alternatives until $$t_0 - 1$$ and then select the first alternative that is superior to all previous ones, if it occurs.

2. The critical time is the smallest integer $$t$$ such that $\sum_{k=t}^{T-1} \frac 1k < 1.$

3. The value function are given by $L_t = \begin{cases} \displaystyle \frac tT \sum_{k=t}^{T-1} \frac 1k, & \text{for } t \ge t_0, \\ L_{t_0}, & \text{for } t < t_0. \end{cases}$

4. For large $$T$$, $$t_0 \approx T/e$$ and the probability of selecting the best candidate is approximately $$1/e$$.

#### Proof

Note that $$L_t - L_{t+1} \ge 0$$. Thus, $$L_t$$ is decreasing with time.

$$V_t(1) = \max\{t/T, L_t\}$$ where the first term is increasing with time and the second term is decreasing with time. Thus, the critical time $$t_0$$ is the first time when $$t/T \ge L_t$$. Since $$L_T = 0$$ and $$T/T = 1$$, such a $$t_0 < T$$.

For any $$t < t_0$$ (i.e., $$t/T < L_t$$), $L_{t-1} = L_t + \bigg[ \frac 1T - \frac{L_t}{t} \bigg]^+ = L_t.$

For any $$t \ge t_0$$ (i.e., $$t/T \ge L_t$$), we have $$(t+1)/T \ge L_{t+1}$$. Therefore, $L_{t} = L_{t+1} + \bigg[ \frac 1T - \frac{L_{t+1}}{t+1} \bigg]^+ = L_{t+1} + \frac 1T - \frac{L_{t+1}}{t+1} = \frac tT \bigg[ \frac 1t + \frac{T}{t+1} L_{t+1} \bigg].$ The above $$L_t$$ can be shown to be equal to the form given above in point 3 by induction.

For large $$T$$, $\sum_{k=t}^{T-1} \frac 1k \approx \int_{t}^T \frac 1k dk = \log \frac Tt.$ Thus, $$t_0 \approx T/e$$. Moreover $V_1(0) = V_1(1) = L_1 = L_{t_0} \approx \frac{t_0}{T} = \frac 1e.$

# References

For a history of the secretary problem, see Ferguson (1989)

Ferguson, T.S. 1989. Who solved the secretary problem? Statistical science, 282–289.

This entry was last updated on 31 Mar 2020 and posted in MDP and tagged optimal stopping, structural results, threshold strategy.