Sequential likelihood-ratio tests — Examples

The one-observation likelihood ratio is $\lambda (1)=(3\mathrm{/}4)\mathrm{/}(1\mathrm{/}4)=3$ and $\lambda (0)=(1\mathrm{/}4)\mathrm{/}(3\mathrm{/}4)=1\mathrm{/}3$. Both laws give positive mass to both outcomes, so they are equivalent.

For $t\ge 1$, set $G_t={\prod }_{i=1}^t\lambda (X_i)$ and choose the algorithm's representatives to be $L_t=G_t$, with $L_0=1$. By [F3], $G_t$ is a version of the finite-time density. It is finite and strictly positive because each factor is either $3$ or $1\mathrm{/}3$, so for every $A\in {\mathcal{F}}_t$, $P_1(A)={\int }_A{G}_{t}d{P}_0$ implies $P_1(A)=0$ exactly when $P_0(A)=0$. At $t=0$, both restrictions to the trivial sigma-algebra assign zero to $\mathrm{\varnothing }$ and one to $\mathrm{\Omega }$, so they are equivalent and $L_0=1$ is their density. The algorithm's finite-time equivalence hypotheses therefore hold for every $t\ge 0$. Also, each nonempty atom of ${\mathcal{F}}_t$ is a fiber for a binary word and has positive $P_0$-probability by [F1]; consequently any other ${\mathcal{F}}_t$-measurable density representative agrees with $G_t$ everywhere, since it agrees $P_0$-almost surely.

The log increment is $Z(1)=\log 3$ and $Z(0)=-\log 3$. Hence ${\mathbb{E}}_{Q_0}\mathrm{\mid }Z\mathrm{\mid }={\mathbb{E}}_{Q_1}\mathrm{\mid }Z\mathrm{\mid }=\log 3<\mathrm{\infty }$, ${\mathbb{E}}_{Q_0}Z=-(1\mathrm{/}2)\log 3<0$, and ${\mathbb{E}}_{Q_1}Z=(1\mathrm{/}2)\log 3>0$. The hypotheses of [F5] hold by steps 1.1 and 2.1, so its general termination theorem gives almost-sure stopping under both laws.

Put $S_t={\sum }_{i=1}^t{X}_i$. By step 2.1 and the values in step 1.1, $L_t={\prod }_{i=1}^t\lambda (X_i)=3^{S_t}(1\mathrm{/}3{)}^{t-S_t}=3^{2S_t-t}$.

The boundaries are $4$ and $1\mathrm{/}4$. Starting a two-observation block at $L=1$, the first observation gives $L=3$ or $1\mathrm{/}3$, both strictly between the boundaries. A repeated outcome in the second observation gives $L=9$ for $11$ or $L=1\mathrm{/}9$ for $00$, crossing the upper or lower boundary; a mixed pair multiplies by $1$ and returns to $L=1$. Induction over successive disjoint pairs shows that the procedure stops at the first pair with equal outcomes.

Under either hypothesis, a pair is mixed with probability $2(1\mathrm{/}4)(3\mathrm{/}4)=3\mathrm{/}8$. The disjoint pairs are independent by [F1], so the probability that the first $k$ pairs are all mixed is $(3\mathrm{/}8{)}^k\to 0$. Thus the pair calculation also verifies almost-sure stopping.

Under $P_0$, each disjoint pair is $11$ with probability $1\mathrm{/}16$, $00$ with probability $9\mathrm{/}16$, and mixed with probability $3\mathrm{/}8$. The event of a type-I error is that the first pair that is not mixed is $11$, so $P_0(\text{type-I error})={\sum }_{k=1}^{\mathrm{\infty }}(3\mathrm{/}8{)}^{k-1}(1\mathrm{/}16)=(1\mathrm{/}16)\mathrm{/}(1-3\mathrm{/}8)=1\mathrm{/}10$. Under $P_1$, the probability of $00$ is $1\mathrm{/}16$ and of a mixed pair is again $3\mathrm{/}8$, so the event of a type-II error gives the same geometric sum and probability $1\mathrm{/}10$.

Since $1\mathrm{/}10\le 1\mathrm{/}4=\alpha =\beta$, these exact terminal errors satisfy both bounds in [F4]; the hypotheses for that theorem's procedure hold by step 2.1.