Sequential Tests for the Mean of a Normal Distribution IV (Discrete Case)

The problem of sequentially testing whether the mean of a normal distribution is positive has been approximated by the continuous analogue where one must decide whether the mean drift of a Wiener-Levy process is positive or negative [3]. The asymptotic behavior of the solution of the latter problem has been studied as $t ightarrow infty$ and as $t ightarrow 0$ [1], [2], [4], [5]. The original (discrete) problem, can be regarded as a variation of the continuous problem where one is permitted to stop observation only at the discrete time points $t_0, t_0 + delta, t_0 + 2delta, cdots$. Especially since the numerical computation of the solution of the continuous version can be carried out by solving the discrete version for small $delta$, it is important to study the relationship between the solutions of the discrete and continuous problems. These solutions are represented by symmetric continuation regions whose upper boundaries are $ ilde x_delta(t)$ and $ ilde x(t)$ respectively. The main result of this paper is that egin{equation*} ag{(1.1)} ilde x_delta(t) = ilde x(t) + hat zsqrtdelta + o(sqrtdelta).end{equation*} This result involves relating the original problem to an associated problem and studying the limiting behavior of the solution of the associated problem. This solution corresponds to the solution of a Wiener-Hopf equation. Results of Spitzer [6], [7] can be used to characterize the solution of the Wiener-Hopf equation and yield $hat z$ as an integral, which, as Gordon Latta pointed out to the author, is equal to $zeta(frac{1}{2})/(2pi)^{frac{1}{2}} = -.5824$. The associated problem referred to above is the following. A Wiener-Levy process $Z_t$ starting at a point $(z, t), t < 0$ is observed at a cost of one per unit time. If the observation is stopped before $t = 0$, there is no payoff. If $t = 0$ is reached, the payoff is $Z^2_0$ if $Z_0 < 0$ and 0 if $Z_0 geqq 0$. Stopping is permitted at times $t = -1, -2, cdots$.