## The Annals of Mathematical Statistics

- Ann. Math. Statist.
- Volume 41, Number 4 (1970), 1206-1216.

### On a Scheduling Problem in Sequential Analysis

#### Abstract

This paper reconsiders the usual sequential decision problem of choosing between two simple hypotheses $H_0$ and $H_1$, in terms of iirv when there is a time delay, assumed to have a known exponential distribution, in obtaining observations. A basic assumption underlying much of the current analysis, is that results of taking observations are considered as immediately available. In other words, it is assumed that there is no time delay between the decision to take an observation and obtaining the result of the observation. This, of course, can be a tremendous limitation to the applicability of the theory. In actuality, such time lags can be substantial and taking an observation often involves experimentation. One important example, where this time delay often considerably inhibits the use of sequential analysis is medical experimentation. Here, a long time may elapse between the application and the result of a treatment. The theory of sequential analysis is considered, explicitly taking account of time lags. At any point in time the decision maker must decide whether to stop and take action now or to continue and schedule more experiments. If he continues he must also decide how many more experiments to schedule. The problem basically then is to find an optimal procedure incorporating a stopping, scheduling and terminal action rule. There is an interesting interplay among these three; and optimal stopping rules, currently used for some problems, may not be optimal when scheduling factors are considered. The usual losses related to decision errors are specified and linear cost assumptions, with regard to amount of experimentation and time until a final decision, are made. Time until a terminal decision is an important variable. If it is decided to continue observation then scheduling many experiments will result in a small expected waiting time until the next result. However, this advantage must be balanced with the cost of scheduling these experiments. Finally, all the previous must be weighed with the loss of taking immediate action with only the currently available information. Bayes procedures are derived. The information state, at any time, because of the exponential assumption, will be described by $(n, \pi)$ where $\pi$ is the current posterior probability of $H_0$ and $n$ the numbers of results outstanding from tests already scheduled. As indicated in Section 2, when a known exponential time delay distribution is assumed, possible decision changes should be made only when test results are obtained. In Section 3, the usually used Sequential Probability Ratio Test (SPRT) stopping rule is studied. Here there are two values $0 < \pi_1 \leqq \pi_2 < 1$ and SPRT specifies that $(n, \pi)$ is a continuation state as long as $\pi_1 < \pi < \pi_2$. It is shown that there is a bounded function $z(\pi)$ such that the optimal scheduling quantity, for a continuation state $(n, \pi)$, is $y(n, \pi) = \max \lbrack 0, z(\pi) - n\rbrack$. That is, if $n > z(\pi)$ we schedule no experiments. On the other hand, if $n \leqq z(\pi)$ then $z(\pi) -n$ more experiments are scheduled. The functional equation approach of Dynamic Programming is used and provides a computational method for approximating $z(\pi)$. The general case, where the problem is to find an optimal stopping and scheduling rule, is studied in Section 4. Various results about the optimal stopping region, in the $(n, \pi)$ plane, are derived and it is shown that the optimal procedure stops with probability one. The optimal stopping region is a kind of generalized SPRT described by functions $0 < \pi_1(n) \leqq \pi_2(n) < 1$ such that $(n, \pi)$ is a continuation state as long as $\pi_1(n) < \pi < \pi_2(n)$. Also, it is shown that there exist two bounded functions $z_1(\pi) \leqq z_2(\pi) \leqq \mathbf{M} < \infty$ such that if $(n, \pi)$ is a continuation point the optimal scheduling quantity is $y(n, \pi) = z_1(\pi) - n$ if $n \leqq z_1(\pi)$, and $y(n, \pi) = 0$ if $n \geqq z_2(\pi)$. When a continuous approximation to the problem is used, allowing $n$ to take on a continuous range of values, a stronger result is proven. Here, the two functions $z_1(\pi)$ and $z_2(\pi)$ may be taken as equal. The results for optimal scheduling rules have similarities to some problems studied in Inventory theory, see [5].

#### Article information

**Source**

Ann. Math. Statist., Volume 41, Number 4 (1970), 1206-1216.

**Dates**

First available in Project Euclid: 27 April 2007

**Permanent link to this document**

https://projecteuclid.org/euclid.aoms/1177696895

**Digital Object Identifier**

doi:10.1214/aoms/1177696895

**Mathematical Reviews number (MathSciNet)**

MR278453

**Zentralblatt MATH identifier**

0203.51803

**JSTOR**

links.jstor.org

#### Citation

Ehrenfeld, S. On a Scheduling Problem in Sequential Analysis. Ann. Math. Statist. 41 (1970), no. 4, 1206--1216. doi:10.1214/aoms/1177696895. https://projecteuclid.org/euclid.aoms/1177696895