Hypothesis Testing: Part 2
Dr. Qiuzhuang Sun
STAT3023
Simple vs. composite: UMP tests
Two-sided tests
Now consider testing H0 : θ = θ0 vs. H1 : θ ̸= θ0. It can be shown
(proof omitted) that if the family has monotone likelihood ratio
then the power functions of the UMP 1-sided tests are strictly
monotone.
Each of the 1-sided UMP test works well for their stated 1-sided
H1. However, for the 2-sided H1, there exists θ such that the tests
are biased in the sense that
Eθ [δ0(X)] < Eθ0 [δ0(X)]. 1
Simple vs. composite: UMPU tests
An unbiased test has power function no small than α for all θ
under H1.
For 1-parameter exponential families with PDF
fθ(x) = h(x)eθT(x)−A(θ), a UMP-Unbiased (UMPU) test exists.
Theorem: For a 1-parameter exponential family with sufficient
statistic T(X), a UMPU test at level α for testing H0 : θ = θ0 vs.
H1 : θ ̸= θ0 exists and is given by:
δ(X) =

1, T(X) > c2 or T(X) < c1
γi, T(X) = ci, i = 1, 2
0, c1 < T(X) < c2
where ci,γi, i = 1, 2, are chosen such that Eθ0 [δ(X)] = α and
Eθ0 [T(X)δ(X)] = Eθ0 [T(X)]Eθ0 [δ(X)] = αEθ0 [T(X)].
2
Simple vs. composite: UMPU tests
Example: Suppose we can observe a sample X ∼ exp(θ), with
PDF fθ(x) = 1θ e
−x/θ, x > 0. Determine the UMPU test for testing
H0 : θ = 1 vs. H1 : θ ̸= 1.
3
Simple vs. composite: UMPU tests
Example: Suppose X ∼ exp(θ), fθ(x) = 1θ e−x/θ, x > 0. Determine
the UMPU test for testing H0 : θ = 1 vs. H1 : θ ̸= 1.
4
Simple vs. Composite
ˆ We have mainly presented methods in this case applicable to
1-parameter exponential families.
ˆ The general properties of MLE also lead to generally
applicable testing methods.
ˆ Suppose we have X = (X1, . . . ,Xn) iid with common fθ(·) for
the parametric family {fθ(·) : θ ∈ Θ}, Θ ∈ R.
ˆ Recall the most power test for H0 : θ = θ0 vs. H1 : θ = θ1 is
given by the Neyman-Pearson likelihood ratio (NPLR) test
statistic
∏ni=1 fθ1(Xi)
∏ni=1 fθ0(Xi)
,
which requires knowing both θ0 and θ1.
ˆ If we are testing H0 : θ = θ0 vs. H1 : θ ∈ Θ\{θ0}, we could
try to first “estimate” a θ1 value and plug it into the NPLR
statistic. 5
Simple vs. Composite: GLRT
Definition: The Generalised likelihood ratio test (GLRT) for
testing H0 : θ = θ0 vs. H1 : θ ∈ Θ\{θ0} uses the statistic:
∏ni=1 fθˆ(Xi)
∏ni=1 fθ0(Xi)
,
where θˆ = argmaxθ∈Θ∏
n
i=1 fθ(Xi).
6
Simple vs. Composite: GLRT
The limiting distribution of the log-GLRT statistic under H0 is 12χ
2
1
(under some regularity conditions).
Sketch proof:
7
Simple vs. Composite: GLRT
(Proof continued)
8
Simple vs. Composite: GLRT
(Proof continued)
9
Composite vs. Composite
Suppose X ∼ fθ(x) for a 1-parameter family {fθ(·) : θ ∈ Θ},
Θ ∈ R. We are testing H0 : θ ∈ Θ0 vs. H1 : θ ∈ Θ\Θ0, Θ0 ⊆ Θ.
For certain composite H0, optimal tests exist.
Proposition: If the family has monotone likelihood ratio in a
statistic T(X), then the UMP test of H0 : θ ≤ θ0 vs. H1 : θ > θ0 is
of the same form as for H0 : θ = θ0 vs. H1 : θ > θ0:
δ(X) =

1, T(X) > c
γ, T(X) = c
0, T(X) < c,
where c,γ are chosen such that Eθ0 [δ(X)] = α.
10
Composite vs. Composite
Proposition: For a 1-parameter exponential family:
fθ(x) = h(x) exp(w(θ)T(x)−A(θ))
where w(θ) is strictly increasing in θ. We are testing
H0 : θ ≤ θ1 or θ ≥ θ2 (θ1 < θ2) vs. H1 : θ1 < θ < θ2. The UMPU
test exists and is of the form:
δ(X) =

1, c1 < T(X) < c2
γi, T(X) = γi, i = 1, 2
0, T(X) < c1 or T(X) > c2,
where ci,γi, i = 1, 2, are selected such that
Eθ1 [δ(X)] = Eθ2 [δ(X)] = α. (Such tests are of interest when trying
to show a new drug is “effectively equivalent” to some standard.) 11
Multivariate case
Suppose we have a family of distributions indexed by more than 1
parameter. The GLRT provides a useful general method of testing
H0 : θ ∈ Θ0 vs. H1 : θ ∈ Θ\Θ0. Compute the test statistic:
log
(
L(θˆ;X)
L(θˆ0;X)
)
where
θˆ = argmax
θ∈Θ
L(θ;X)
is the “unrestricted” MLE, and
θˆ0 = argmax
θ∈Θ0
L(θ;X).
12
Multivariate case
Many commonly used statistical tests are equivalent to the GLRT.
Example: 1-way ANOVA F-test. Consider Xij ∼ N(µi, σ2), for
i = 1, . . . , g being indexes for groups and j = 1, . . . ,ni, and all Xij
are independent. The total sample size is denoted by N = ∑gi=1 ni.
Consider testing H0 : µ1 = µ2 = · · · = µg vs. H1 : µi’s are not all
equal.
13
Multivariate case
(Example continued)
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Multivariate case
(Example continued)
15
Multivariate case
(Example continued)
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Multivariate case
Example: one-sided t-test. Let X1, . . . ,Xn be iid N(µ, σ2), and
Θ = {(µ, σ2) : µ ≥ 0, σ2 > 0}. Consider testing H0 : µ = 0 vs.
H1 : µ > 0.
17
Multivariate case
(Example continued)
18
Multivariate case
(Example continued)
19
Multivariate case
(Example continued)
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Multivariate case
(Example continued)
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Multivariate case
(Example continued)