8.4.∗The Sequential Probability Ratio Test 503- What is the probability of the procedure continuing indefinitely?
- What is the value of the power function of this test at each of the pointsθ=θ′
andθ=θ′′? - Ifθ′′is one of several values ofθspecified by an alternative composite hypoth-
esis, sayH 1 :θ>θ′, what is the power function at each pointθ≥θ′? - Since the sample sizeNis a random variable, what are some of the properties
of the distribution ofN? In particular, what is the expected valueE(N)of
N? - How does this test compare with tests that have a fixed sample sizen?
A course in sequential analysis would investigate these and many other problems.
However, in this book our objective is largely that of acquainting the reader with
this kind of test procedure. Accordingly, we assert that the answer to question 1
is zero. Moreover, it can be proved that ifθ=θ′or ifθ=θ′′,E(N) is smaller for
this sequential procedure than the sample size of a fixed-sample-size test that has
the same values of the power function at those points. We now consider question 2
in some detail.
In this section we shall denote the power of the test whenH 0 is true by the
symbolαand the power of the test whenH 1 is true by the symbol 1−β.Thus
αis the probability of committing a Type I error (the rejection ofH 0 whenH 0 is
true), andβis the probability of committing a Type II error (the acceptance ofH 0
whenH 0 is false). With the setsCnandBnas previously defined, and with random
variables of the continuous type, we then have
α=∑∞n=1∫CnL(θ′,n), 1 −β=∑∞n=1∫CnL(θ′′,n).Since the probability is 1 that the procedure terminates, we also have
1 −α=∑∞n=1∫BnL(θ′,n),β=∑∞n=1∫BnL(θ′′,n).If (x 1 ,x 2 ,...,xn)∈Cn,wehaveL(θ′,n)≤k 0 L(θ′′,n); hence, it is clear that
α=∑∞n=1∫CnL(θ′,n)≤∑∞n=1∫Cnk 0 L(θ′′,n)=k 0 (1−β).BecauseL(θ′,n)≥k 1 L(θ′′,n) at each point of the setBn,wehave
1 −α=∑∞n=1∫BnL(θ′,n)≥∑∞n=1∫Bnk 1 L(θ′′,n)=k 1 β.Accordingly, it follows that
α
1 −β≤k 0 ,k 1 ≤
1 −α
β, (8.4.5)