8.4.∗The Sequential Probability Ratio Test 501are known numbers. For this section, we denote the likelihood ofX 1 ,X 2 ,...,Xn
by
L(θ;n)=f(x 1 ;θ)f(x 2 ;θ)···f(xn;θ),a notation that reveals both the parameterθand the sample sizen. If we reject
H 0 :θ=θ′and acceptH 1 :θ=θ′′when and only when
L(θ′;n)
L(θ′′;n)≤k,wherek>0, then by Theorem 8.1.1 this is a best test ofH 0 againstH 1.
Let us now suppose that the sample sizenisnotfixed in advance. In fact,
let the sample size be a random variableNwith sample space{ 1 , 2 ,, 3 ,...}.An
interesting procedure for testing the simple hypothesisH 0 :θ=θ′against the simple
hypothesisH 1 :θ=θ′′is the following: Letk 0 andk 1 be two positive constants
withk 0 <k 1. Observe the independent outcomesX 1 ,X 2 ,X 3 ,...in a sequence, for
example,x 1 ,x 2 ,x 3 ,..., and compute
L(θ′;1)
L(θ′′;1),L(θ′;2)
L(θ′′;2),L(θ′;3)
L(θ′′;3),....The hypothesisH 0 :θ=θ′is rejected (andH 1 :θ=θ′′is accepted) if and only if
there exists a positive integernso thatxn=(x 1 ,x 2 ,...,xn) belongs to the set
Cn={
xn:k 0 <
L(θ′,j)
L(θ′′,j)<k 1 ,j=1,...,n− 1 ,and
L(θ′,n)
L(θ′′,n)≤k 0}. (8.4.1)
On the other hand, the hypothesisH 0 :θ=θ′ is accepted (andH 1 :θ=θ′′
is rejected) if and only if there exists a positive integernso that (x 1 ,x 2 ,...,xn)
belongs to the set
Bn={
xn:k 0 <
L(θ′,j)
L(θ′′,j)<k 1 ,j=1,...,n− 1 ,and
L(θ′,n)
L(θ′′,n)≥k 1}. (8.4.2)
That is, we continue to observe sample observations as long as
k 0 <
L(θ′,n)
L(θ′′,n)<k 1. (8.4.3)We stop these observations in one of two ways:- With rejection ofH 0 :θ=θ′as soon as
L(θ′,n)
L(θ′′,n)
≤k 0or- With acceptance ofH 0 :θ=θ′as soon as
L(θ′,n)
L(θ′′,n)≥k 1 ,