Most Powerful Tests
Let f (x|θ), x = (x1 , x2 , . . . , xn ), θ ∈ Θ, be the joint pdf/pmf of X = (X1 , . . . , Xn )
˜
˜
˜
(the parameter θ can be vector-valued, i.e., θ ≡ θ). We want to test the hypothesis
˜
H0 : θ = θ0
vs
H1 : θ = θ1
where θ0 , θ1 ∈ Θ, θ0 6= θ1 .
Definition: A test function ϕ(x) is called a most powerful (MP) test of size α if
˜
1. Eθ0 ϕ(X ) = α.
˜
2. Eθ1 ϕ(X ) ≥ Eθ1 ϕ̄(X ) holds for any other test rule ϕ̄(x) with Eθ0 ϕ̄(X ) ≤ α.
˜
˜
˜
˜
It turns out that a MP test does exist, as described below.
Theorem: (Neyman-Pearson Lemma) Let f (x|θ), θ ∈ Θ, be the joint pdf/pmf
˜
of X1 , . . . , Xn . Then for testing H0 : θ = θ0 vs H1 : θ = θ1 , a MP test of size α
exists for all α ∈ [0, 1] and is given by


1
if f (x|θ1 ) > kf (x|θ0 )



˜
˜

γ
if f (x|θ1 ) = kf (x|θ0 )
ϕ(x) =
(1)
˜
˜

˜



if f (x|θ1 ) < kf (x|θ0 )
0
˜
˜
where γ ∈ [0, 1] and 0 ≤ k ≤ ∞ are constants satisfying
Eθ0 ϕ(X ) = α.
˜
(2)
Remarks:
• Let L(θ) ≡ f (x|θ), the likelihood function at θ. Then the MP test in (1)
˜
rejects H0 for all x such that the likelihood L(θ1 ) of θ1 is “large” compared
˜
to the likelihood of L(θ0 ) of θ0 ; the “largeness” factor k is determined by (2).
• In general, the choice of (γ, k) satisfying (2) is not unique.
• If the distribution of f (x|θ1 )/f (x|θ0 ) under θ0 is continuous, then we may set
˜
˜
γ = 0.
1
Example: Let X1 , . . . , Xn be a random sample from Gamma(α = 3, θ), θ > 0. Find
a MP test of size α for H0 : θ = θ0 vs H1 : θ = θ1 (where 0 < θ0 < θ1 ).
Example: Let X1 , . . . , Xn be iid Binomial(2, p), 0 < p < 1. Find a MP test of size
α for H0 : p = p0 vs H1 : p = p1 (where 0 < p1 < p0 < 1) when
(i) p0 = 0.6, n = 10, α = 0.0003
(ii) p0 = 0.6, n = 10, α = 0.001
(iii) p0 , α, n are arbitrary (but fixed)
2