Sequential likelihood-ratio test

Assume the Axiom of Choice, which states that every family of nonempty sets has a choice function. Let $(\mathrm{\Omega },\mathcal{F},({\mathcal{F}}_t{)}_{t\ge 0})$ be a discrete filtration on a common measurable space and let $P_0,P_1$ be probability measures on it. Assume that for every $t\ge 0$, the restrictions $P_0{\mathrm{\mid }}_{{\mathcal{F}}_t}$ and $P_1{\mathrm{\mid }}_{{\mathcal{F}}_t}$ are mutually absolutely continuous: each restriction assigns measure zero to exactly the same ${\mathcal{F}}_t$ events as the other. For each $t$, choose a nonnegative ${\mathcal{F}}_t$-measurable representative $L_t$ of the likelihood-ratio density $d(P_1{\mathrm{\mid }}_{{\mathcal{F}}_t})\mathrm{/}d(P_0{\mathrm{\mid }}_{{\mathcal{F}}_t})$ Likelihood ratio as a Radon–Nikodym derivative. The filtration and stopping-time conventions are those of Filtration and filtered probability space and Discrete stopping time.

Input thresholds $\alpha ,\beta \in (0,1)$. At times $t=1,2,\dots$, compute $L_t$ and apply these instructions:

1. If $L_t\ge 1\mathrm{/}\alpha$, stop and decide $H_1$.
2. Otherwise, if $L_t\le \beta$, stop and decide $H_0$.
3. Otherwise continue to the next observation.

The two boundaries are disjoint because $\beta <1<1\mathrm{/}\alpha$. The output time is $\tau =\inf \{t\ge 1:L_t\ge 1\mathrm{/}\alpha \text{ or }{L}_t\le \beta \},\inf \mathrm{\varnothing }=\mathrm{\infty }\mathrm{.}$ If $\tau =\mathrm{\infty }$, the procedure has no finite terminal decision. It does not itself assert that a crossing occurs.