Let X1,X2,…,Xn be Random Sample with PDFf(x∣θ), where θ∈Ω={θ′,θ′′}, and L(θ∣n)=∏i=1nf(xi∣θ) be the Likelihood Function
Consider a simple test H0:θ=θ′,H1:θ=θ′′
We observe likelihood ratio sequentially i.e. L(θ′′∣1)L(θ′∣1),L(θ′′∣2)L(θ′∣2),…, and reject H0 if and only if
x=(x1,x2,…,xn)∈Cn, where Cn={(x∣∀j={1,2,…,n−1},k0<L(θ′′∣j)L(θ′∣j)<k1∧L(θ′′∣n)L(θ′∣n)≤k0}, and
do not reject H0 if and only if
x=(x1,x2,…,xn)∈Bn, where Bn={(x∣∀j={1,2,…,n−1},k0<L(θ′′∣j)L(θ′∣j)<k1∧L(θ′′∣n)L(θ′∣n)≥k1}
i.e. we continue to observe as long as k0<L(θ′′∣j)L(θ′∣j)<k1, and stop otherwise.
Choice of k0 and k1
Let α,β be type 1 and 2 error, respectively, then
α=i=1∑∞∫CnL(θ′∣n),β=i=1∑∞∫BnL(θ′′∣n),1−α=i=1∑∞∫BnL(θ′∣n),1−β=i=1∑∞∫CnL(θ′′∣n)
Hence,
α=i=1∑∞∫CnL(θ′∣n)≤i=1∑∞∫Cnk0L(θ′′∣n)=k0(1−β)1−α=i=1∑∞∫BnL(θ′∣n)≥i=1∑∞∫Bnk1L(θ′′∣n)=k1β
Therefore,
1−βα<k0<k1<β1−α
By the inequality, we choose k0:=1−βα,k1:=β1−α
