STAT200C: Hypothesis Testing
Zhaoxia Yu
Spring 2017
Some Definitions
• A hypothesis is a statement about a population parameter.
• The two complementary hypotheses in a hypothesis testing
are the null hypothesis (H0 : θ ∈ Ω0) and the alterantive
hypothesis (H1 : θ ∈ Ω1), where Ω0 and Ω1 are two disjoint
subsets of the full parameter space Ω.
• A hypothesis is called a simple hypothesis if the space is a
singleton set; else it is called a composite hypothesis.
1
Some Examples
• simple vs simple: H0; θ = θ0 vs H1 : θ = θ1;
• simple vs composite: H0 : θ = θ0 vs H1 : θ 6= θ0;
• composite vs composite: H0 : θ ≤ θ0 vs H1 : θ > θ0;
• composite vs composite H0 : θ ≥ θ0 vs H1 : θ < θ0.
2
Critical Function
• A test is determined by its critical function: φ(x) = the prob-
ability of rejecting H0 when x is the observed value of X.
• Example: Suppose X ∼ Binomial(5, θ), H0 : θ = 0.5, H1 : θ >
0.5. Consider two critical functions:
φ1(x) =
1 x = 50 x < 5
, φ2(x) =
1 x = 5
0.12 x = 40 x < 4
φ1(x) is a non-randomized test; φ2(x) is randomized test. A
randomized test can ensure a pre-fixed type I error rate; in
practice, it is not used often.
3
Errors in Hypothesis Testing
• Type I error: H0 is rejected when it is true.
• Type I error rate: α = Pr(reject H0|H0 is true)
• Type II error: H0 is failed to reject when it is not true.
• Type II error rate: 1− β = Pr(fail to reject H0|H1 is true)
4
Type I Error
• Type I error: H0 is rejected when it is true.
• Type I error rate: α = Pr(reject H0|H0 is true)
• α is called the significance level of a test.
5
Type I Error: An Example
• Consider the binomial example. Suppose θ = 0.5.
• For φ1(x), α = 1.0 ∗ Pr(X = 5|θ = 0.5) = 0.55 = 0.03125.
• For φ2(x), α = 1.0 ∗ Pr(X = 5|θ = 0.5) + 0.12 ∗ P (X = 4|θ =
0.5) = 0.03125 + 0.12 ∗ 5 ∗ 0.54 ∗ 0.5 = 0.05
6
Type II Error
• Type II error: H0 is failed to reject when it is not true.
• Type II error rate: 1− β = Pr(fail to reject H0|H1 is true)
• The power of a function β = 1−Pr(fail to reject H0|H1 is true)
7
Type II Error: An example
• Example. Consider the binomial example.
• Suppose θ = 0.6. β1 = 1.0 ∗ Pr(X = 5|θ = 0.6) = 0.65 =0.07776, β2 = 1.0 ∗ Pr(X = 5|θ = 0.6) + 0.12 ∗ Pr(X = 4|θ =0.6) = 0.108864
• Suppose θ = 0.7. β1 = 0.16807, β2 = 0.211288
• Suppose θ = 0.8. β1 = 0.32768, β2 = 0.376832
• Suppose θ = 0.9. β1 = 0.59049, β2 = 0.629856
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Power Function
• The power function of test φ: βφ(θ) = Eθ[φ(x)]. Let R be the
rejection region, then we can also define the power function
using: β(θ) = Pr(X ∈ R|θ) It is a measure of
– weakness (type I error) when θ ∈ Ω0
– strength (power) when θ ∈ Ω1
• A test is said to have a size α if supθ∈Θ0β(θ) = α.
• A test is said to have a level α if supθ∈Θ0β(θ) ≤ α.
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Example Let X = (X1, · · · , Xn) be a random sample fromN(θ, σ2) with σ2 known. H0 : θ = θ0 vs H1 : θ > θ0. Intuitively,we should reject H0 for large values of X − θ0. We can write therejection region as
φ(x) =
1 x−µ0
σ/√n> c
0 x−µ0σ/√n< c
The power function is
β(θ) = Pµ(X − θ0
σ/√n> c) = Pθ(
X − θσ/√n> c+
θ0 − µσ/√n
)
= Pθ(Z > c+θ0 − θσ/√n
)
Here β(θ) is an increasing function of θ. Homework: Write an Rfunction to choose the sample size n such that α = β(θ0) = 0.05and β(θ0 + σ) > 0.8 for θ > θ0 + σ.
10
Power Function: An Example Again, consider the
binomial example.
• β1(0.1) = 1e− 05, β2(0.1) = 6.4e− 05
• β1(0.3) = 0.00243, β2(0.3) = 0.005832
• β1(0.5) = 0.03125, β2(0.5) = 0.05
• β1(0.7) = 0.16807, β2(0.7) = 0.211288
• β1(0.9) = 0.59049, β2(0.9) = 0.629856
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An Example of Power Calculation
Let’s look at the power function for H0 : µ = µ0 vs H1 : µ > µ0
where µ is the mean of a N(µ, σ2) with known variance. The
likelihood ratio test rejects H0 when x = X−µ0σ/√n> Φ−1(1 − α).
The power function is
β(µ) = P (X − µ+ µ− µ0
σ/√n
> Φ−1(1− α))
= P (Z > Φ−1(1− α)−√nµ− µ0σ)
= 1−Φ(Φ−1(1− α)−√nµ− µ0
σ)
Suppose α, µ, and µ0 are given, for fixed n we can calculate
power and for fixed power we can calculate sample size n.
12
An Example of Power Calculation (contin-ued)
Let β = 0.8, α = 0.05, µ−µ0σ = 0.1, then
Φ(Φ−1(0.95)− 0.1√n) = 0.2
⇒ 0.1√n = Φ−1(0.95)−Φ−1(0.2)
⇒√n = 10 ∗ (1.6449 + 0.84)16
⇒ n ≈ 619
Question. What is the power of the test when n = 400?
β = P (X − µ0
σ/√n> Φ−1(0.95)) = P (
X − µσ/√n> Φ−1(0.95)− n
(µ− µ0)
σ/√n
)
= 1−Φ(1.6449− 2) ≈ 0.64
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Type I error rate vs Type II error rate / Power
• Consider the two tests for the binomial example
X ∼ Binomial(5,0.5)
φ1(x) =
1 x = 50 x < 5
, φ3(x) =
1 x > 00 x = 0
• Power functions:
β1(θ) = Pr(X = 5) = θ5, β3(θ) = 1−Pr(X = 0) = 1−(1−θ)5
• It can be shown that β1(θ) ≤ β3(θ).
14
Type I error rate vs Type II error rate / Power
• Test 3 achieves a smaller Type II error rate, but it has higher
Type I error rate
• How should one choose a test?
• One idea is to find the most powerful test among those with
a Type I error rate no greater than a pre-specified value, say
α.
15
Most Powerful Test: Simple vs Simple
• A test φ is said to be most power (MP) at level α for testingH0 : θ = θ0 vs H1 : θ = θ1 if
E0[φ(X)] ≤ α and E1[φ(X)] ≥ E1[ψ]
for any ψ so that E0[ψ(X)] ≤ α.
• In other words, we say φ is MP if it has the largest poweramong all tests with a level of α.
• How to find a MP test? Neyman-Pearson Lemma can beused to construct a most powerful level α test for simple vssimple hypotheses.
16
Neyman-Pearson Lemma
Neyman-Pearson Lemma (Most powerful level α test for H0 :
θ = θ0 vs H1 : θ = θ1)
• Existence. There exists a test φ and a constant k ≥ 0 such
that
E0[φ(x)] = α
and
φ(x) =
1 if f1(x) > kf0(x)0 if f1(x) < kf0(x)
where fj(x) = f(x|θj). [Note that nothing is being said about
φ on the set x : f1(x) = kf0(x).]
17
• Sufficiency If a test φ satisfies the conditions in Existence
for some k ≥ 0, then it is MP at level α for testing H0 : θ = θ0
vs H1 : θ = θ1.
• Necessity If φ is MP at level α for testing H0 : θ = θ0 vs
H1 : θ = θ1, then with probability 1 under both θ0 and θ1,
φ(x) =
1 if f1(x) > kf0(x)0 if f1(x) < kf0(x)
for some k.
• For α = 0 or 1, the proof is straightforward, provided that
k = 0 and ∞ are allowed. Now let 0 < α < 1.
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Proof: Existence
Let F (c) = P0[f1(X) ≤ cf0(X)] = P0[f1(X)/f0(X) ≤ c]. Then
F is the cdf of T (X) = f1(X)/f0(X) under θ0. [Note, under θ0,
f0(X) > 0 with probability 1].
Since F is cdf, F ↑, right-continuous: F (c)−F (c−0) = P0(T (X) =
c). Now let
α(c) = 1 − F (c) = P0(T (X) > c).
Then α(−∞) = 1, α(∞) = 0, α ↓, and right continuous: α(c−0)− α(c) = P0[T (X) = c]. Let c0 such that α(c0) ≤ α ≤ α(c0 − 0).
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Proof: Existence (continued)
Define
φ(x) =
1 if T (x) > k [eqt f1(x) > kf0(x)]
α−α(c0)α(c0−0)−α(c0) if T (x) = k [eqt f1(x) = kf0(x)]
0 if T (x) < k [eqt f1(x) < kf0(x)]
[If F is continuous at c0 , defined 0/0 = 1]
The existence holds with k = c0, because
E0[φ(X)] = P0[T (X) > c0]
+α− α(c0)
α(c0 − 0)− α(c0)P0[T (X) = c0]
= α(c0) + α− α(c0) = α
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Proof: Sufficiency
Let φ be a test satisfying the conditions in Existence and let φ∗
be an arbitrary level α test i.e., E0[φ∗(X)] ≤ α. We shall show
that
βφ(θ1) = E1[φ(X)] ≥ E1[φ∗(X)] = β∗φ(θ1)
Let S+ = x : φ(x)− φ∗(x) > 0 and S− = x : φ(x)− φ∗(x) < 0.Then for x ∈ S+,
φ(x) > φ∗(x) ≥ 0⇒ f1(x) ≥ kf0(x) [ eqt T (x) ≥ k]
and for x ∈ S−,
φ(x) < φ∗(x) ≤ 1⇒ f1(x) ≤ kf0(x) [ eqt T (x) ≤ k]
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Proof: Sufficiency (continued)
Thus
[φ(x)− φ∗(x)][f1(x)− kf0(x)]
≥ 0 ∨x ∈ S+ ∪ S−= 0 ∨x 6∈ S+ ∪ S−
Hence
E1[φ(X)]− E1[φ∗(X)] =∫
[φ(x)− φ∗(x)]f1(x)dx
=∫
[φ(x)− φ∗(x)][f1(x)− kf0(x)]dx
+ k∫
[φ(x)− φ∗(x)]f0(x)dx
≥ 0 + kE0[φ(x)− E0[φ∗(x)]]≥ 0
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Proof: Necessity
Suppose that φ∗ is MP at level α. Let φ be a test satisfies theconditions in Existence. Let S+ and S− be as above. The seton which φ∗ violates the conditions is
S = (S+ ∪ S−) ∩ x : f1(x) 6= kf0(x)Clearly S = x : φ(x) 6= φ∗(x), f1(x) 6= kf0(x).
We shall show that
P0(S) = P1(S) = 0
Suppose the distribution is continuous. In this case, it is enoughto show that ∫
Sdx = 0
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Proof: Necessity (continued)
Note that [φ(x) − φ∗(x)][f1(x) − kf0(x)] > 0 for ∨x ∈ S. Hence∫S dx > 0 implies that∫[φ(x)−φ∗(x)][f1(x)−kf0(x)]dx =
∫S
[φ(x)−φ∗(x)][f1(x)−kf0(x)] > 0,
i.e., ∫[φ(x)− φ∗(x)]f1(x)dx > k
∫[φ(x)− φ∗(x)]f0(x)
> k(α− E0[φ∗(X)]) > 0
The last inequality is true because E0[φ∗(X)] ≤ α. Thus, E1[φ(X)] >
E1[φ∗(X)], contradicting the MP level α property of φ∗.
Note: the proof of discrete distributions is similar - just replace∫with
∑.
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A Corollary
Corollary. Let β = E1[φ(X)] where φ is a MP test at level α ∈(0,1) for H0 : θ = θ0 vs H1 : θ = θ1. Then β > α.
Proof. For let φ0(x) ≡ α, the test that always reject H0 withprobability α, irregardless the value of x. It is obvious that
E0[φ0(X)] = E1[φ0(X)] = α
i.e., both the Type I error rate and power of φ0(x) is α.
Since φ0(x) is level α test but it does not satisfy the Neyman-Pearson conditions, it is not a MP test (by the necessity), whereasφ(x) is. Therefore,
β = E1[φ(X)] > E1[φ0(X)] = α
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Remark 1: On x : f1(x) = c0f0(x),
φ(x) =α− α(c0)
α(c0 − 0)− α(c0)=
α− 1 + F (c0)
1− F (c0 − 0)− 1 + F (c0)=
α− 1 + F (c0)
F (c0)− F (c0)− 0
where F is the cdf of T (X) = f1(X)/f0(X). Suppose F underθ0 is discontinuous at c0, then
(1) F (c0−0) < 1−α = F (c0)→ φ(x) = 0 on x : f1(x) = c0f0(x)
(2) F (c0−0) = 1−α < F (c0)→ φ(x) = 1 on x : f1(x) = c0f0(x)
(3) F (c0 − 0) < 1 − α < F (c0) → φ(x) = γ(x) ∈ [0,1] on x :f1(x) = c0f0(x) for any γ(x) satisfying
∫x:f1(x)=c0f0(x) γ(x)f0(x)dx =
α − α(c0) is MP at level α for H0 : θ = θ0 vs H1 : θ = θ1. Thuswe have a unique MP test at level α in case (1) or (2), but notin case (3).
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Remark 2: A MP level α test for H0 : θ = θ0 vs H1 : θ = θ1
must satisfy E0[φ(x)] = α, unless there exists a test ψ having
size E0[ψ(x)] < α and power E1[ψ(x)] = 1. This exceptional
situation occurs in the following case:
Let X1, · · · , Xn be a random sample from Unif [0, θ] for θ > 0.
We want to test H0 : θ = θ0 vs θ = θ1 < θ0 at level α. Let X(n)be the largest value in the sample and suppose that (θ1/θ0)n < α.
Then the following test is MP
ψ(x) =
1 if 0 ≤ x(n) ≤ θ10 if θ1 < x(n) ≤ θ0
and has size < α and power=1.
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Remark 3: (omitted)
28
Remark 4:
The Neyman-Pearson lemma expresses the MP level α test φ
for H0 : θ = θ0 vs θ = θ1 in terms of the likelihood ratio, or
equivalently log likelihood ratio (LLR). If X = (X1, · · · , Xn) is a
random sample from fθ(x), then the LLR is
log[f1(x)/f0(x)] =n∑i=1
log[f1(xi)/f0(xi)],
and a MP level α test for H0 vs H1 is
φ(x) =
1 if
∑ni=1 log[f1(xi)/f0(xi)] > k
γ if∑ni=1 log[f1(xi)/f0(xi)] = k
0 if∑ni=1 log[f1(xi)/f0(xi)] < k
where k and γ are determined by the condition E0[φ(x)] = α.
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Example: A Random Sample from Normal
Let X = (X1, · · · , Xn) be a random sample from N(θ, σ2), where
σ2 is known but θ is unknown.
Consider the following three hypothesis testing problems:
1. H∗0 : θ = θ0 vs H∗1 : θ = θ1
2. H0 : θ ≤ θ0 vs H∗1 : θ = θ1, θ1 > θ0
3. H0 : θ ≤ θ0 vs H1 : θ > θ0
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A Random Sample from Normal, Problem 1:H∗0 : θ = θ0 vs H∗1 : θ = θ1
Let fj be the joint pdf of X corresponding to θj. The LLR is
log[f1(x)/f0(x)] =1
2σ2[∑
(xi − θ0)2 −∑
(xi − θ1)2]
=n
2σ2(θ2
0 − θ21) +
n
σ2(θ1 − θ0)x
The MP level α test for H∗0 vs H∗1 is:
φ∗1(x) =
1 if x ≥ c10 if x < c1
if θ1 > θ0 , φ∗2(x) =
1 if x ≤ c20 if x < c2
if θ1 < θ0
where c1 = θ0 +σ/√nΦ−1(1−α) and c2 = θ0−σ/
√nΦ−1(1−α).
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A Random Sample from Normal, Problem 2:H0 : θ ≤ θ0 vs H∗1 : θ = θ1 > θ0
Suppose φ(x) is MP level α for Problem 2. Then it must satisfy
(i) supθ≤θ0Eθ[φ(x)] ≤ α and
(ii) Eθ1[φ(x)] ≥ Eθ1
[ψ(x)] ∨ ψ of level α
The power function of φ∗ obtained in Problem 1 is:
βφ∗1(θ) = Eθ[φ
∗1(x)] = Pθ[x ≥ θ0 + σ/
√nΦ−1(1− α)]
= Φ(Φ−1(α) +
√n(θ − θ0)
σ)
which is ↑ in θ, so that supθ≤θ0Eθ[φ
∗1(x)] = Eθ0
[φ∗1(x)] = α.
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A Random Sample from Normal, Problem 2:H0 : θ ≤ θ0 vs H∗1 : θ = θ1 > θ0 (continued)
Thus φ∗1 is level α.
Now if ψ satisfies (i) , then
α ≥ supθ≤θ0Eθ[ψ(x)] ≥ Eθ0
[ψ(x)]
i.e., ψ is level α for H∗0 vs H∗1.
Because φ∗1(x) is a MP for H0 : θ = θ0 vs H1 : θ = θ1(> θ0),
Eθ1[φ∗(x)] ≥ Eθ1
[ψ(x)]
Thus, φ∗ is a MP level α for H0 : θ ≤ θ0 vs H∗1 : θ = θ1.
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A Random Sample from Normal, Problem 3:H0 : θ ≤ θ0 vs H1 : θ > θ0
For this problem, we want (if possible) a uniformly most power
(UMP) level α test for H0 vs H1. In other words, we want a test
φ such that
(i) supθ≤θ0Eθ[φ(x)] ≤ α and
(ii) Eθ[φ(x)] ≥ Eθ[ψ(x)] ∨θ > θ0 whereas ψ satisfies (i).
Since φ∗1 obtained above is the same for ∨θ1 > θ0, it is UMP at
level α for H0 vs H1.
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Uniformly Most Powerful (UMP)
A test is a uniformly most powerful (UMP) test at level α for
H0 : θ ∈ Ω0 vs H1 : θ ∈ Ω1 if
(i) supθ∈Ω0Eθ[φ(x)] ≤ α and
(ii) Eθ[φ(x)] ≥ Eθ[ψ(x)] ∨θ ∈ Ω1 whereas ψ satisfies (i).
Although requirement (ii) is very stringent, UMP tests do exist
in a certain type of situations. We have seen such situation in
the normal example.
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Monotone Likelihood Ratio (MLR) Family
• A family of pdfs or pmfs for a univariate random variable X is
said to be a monotone likelihood ratio (MLR) family if there
exists a real-valued T (x) such that for any θ1 < θ2 in Ω,
fθ2(x)/fθ1
(x)
is a nondecreasing function of T (x).
• If fθ1(x) = 0 and fθ2
(x) > 0, define fθ2(x)/fθ1
(x) = +∞
36
Some Examples of MLR
• Suppose X ∼ Binomial(n, θ). Then
f(x|θ2)/f(x|θ1) = (θ2/θ1)x(1− θ2
1− θ1)n−x = (
θ2(1− θ1)
θ1(1− θ2))x(
1− θ2
1− θ1)n
Thus, the family is MLR.
• Any regular exponential family with f(x|θ) = h(x)g(θ)expη(θ)T (x)has an MLR.
– N(θ, σ2) with known σ2.
– N(µ, θ = σ2) with known µ and θ > 0.
– Poisson(θ) for θ > 0
37
Some Examples of MLR
• The hypergeometric distribution H(N,n, θ) with
pθ(x) =
(θx
)(N−θn−x
)(Nn
) ,
where x = max(0, θ+n−N), · · · ,min(n, θ). [θ white and N−θblack balls in a box from which n balls are drawn at random,Pθ(x) is the probability of x white balls in the sample.]
• Unif(0, θ). [We define a/0 =∞ for ∨a > 0.]
• The Cauchy distributions C(θ,1) with Pθ(x) = 1/[π(1 + (x−θ)2)], x ∈ R, θ ∈ R is NOT a MLR family.
38
Theorem of UMP
Suppose that the family of pdfs/pmfs Pθ, θ ∈ R has a MLR inT (X). Then
1. There exists a UMP level α test for H0 : θ ≤ θ0 vs H1 : θ > θ0given by
φ(x) =
1 if T (x) > cγ if T (x) = c0 if T (x) < c
where c and 0 ≤ γ ≤ 1 are determined by Eθ0[φ(x)] = α
2. The power function β(θ) = Eθ[φ(x)] of this φ is strictly in-creasing at all θ for which β(θ) < 1.
39
Proof
First consider H∗0 : θ = θ0 vs H∗1 : θ = θ1 where θ1 > θ0 isfixed. Then a MP level α test for H∗0 vs H∗1 rejects H∗0 forlarge values of fθ1
(x)/fθ0(x), i.e., for large values of T (x) (by
the MLR property). Moreover, by the existence part of the N-Plemma, there exist c and 0 ≤ γ ≤ 1 such that the test φ(x) =I(c,∞)(T (x)) + γIc(T (x)) satisfies Eθ0
[φ(x)] = α.
Since the forms:
φ(x) =
1 T (x) > c0 T (x) < c
and φ(x) =
1 f
θ′′(x) > kf
θ′(x)
0 fθ′′(x) < kf
θ′(x)
are equivalent for any θ′< θ
′′, this test is MP at level α
′= β(θ
′)
for testing H∗∗0 : θ = θ′
vs H∗∗1 : θ = θ′′
whenever θ′< θ
′′(by the
sufficiency part of the N-P lemma).
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Proof (continued)
Next note that by the Corollary to the N-P lemma, β(θ′′) > α
′=
β(θ′
if α′< 1, which proves that β(θ) is strictly increasing, so
long as it is < 1.
Now note that for this test, β(θ) = Eθ[φ(x)] ≤ α ∨θ ≤ θ0, whichmakes φ a level α test for H0 : θ ≤ θ0. Let
Ψα = all tests ψ such that supθ≤θ0Eθ[ψ(x)] ≤ α
and
Ψ∗α = all tests ψ such that Eθ0[ψ(x)] ≤ α
Then Ψα ⊂ Ψ∗α. We have shown that Eθ1[φ(x)] ≥ Eθ1
[ψ(x)]∨ψ ∈ Ψ∗α. Hence Eθ1
[φ(x)] ≥ Eθ1[ψ(x)] ∨ψ ∈ Ψα (by the N-P
Lemma). This makes φ a MP test at level α for H0 : θ ≤ θ0 vsH∗1 : θ = θ1 > θ0.
41
Proof (continued)
Finally, since φ is independent of θ1 > θ0, it is a UMP test at
level α for H0 : θ ≤ θ0 vs H1 : θ > θ0.
Note, the theory for H0 : θ ≥ θ0 vs H1 : θ < θ can be stated and
proved analogously.
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Example: Uniform Distribution
Let X = (X1, · · · , Xn) be a random sample from Unif(0, θ).
(i) Find a MP test for H0 : θ = θ0 vs H1 : θ = θ1, θ1 > θ0.
(ii) Show that any level α test with the rejection region X(n) >
θ0 is a UMP level α test for H0 : θ ≤ θ0 vs H1 : θ > θ0.
Solution :(i) The likelihood ratio is
λ(x) =fθ1
(x)
fθ0(x)
=
(θ0
θ1
)n I(x(n) ≤ θ1)
I(x(n) ≤ θ0)=
∞ x(n) > θ0
(θ0/θ1)n 0 < x(n) ≤ θ0
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Solution (continued)
By the Neyman-Pearson Lemma, a UMP level α satisfies
φ(x) =
1 λ(x) > (θ0/θ1)n
γ λ(x) = (θ0/θ1)n
0 λ(x) < (θ0/θ1)n this cannot happen
which is equivalent to
φ(x) =
1 x(n) > θ0γ 0 < x(n) ≤ θ0
Because it is a level α test, γ satisfies α = Eθ0[φ(x)] = Pθ0
[X(n) >
θ0] = γ.
Note, because the ratio λ(x) is nondecreasing in X(n), the test
is also a UMP for H0 : θ ≤ θ0 vs H1 : θ > θ0.
44
Solution (continued) Now, let’s determine the rejectionregion using the theorem to construct UMP for MLR family. Thefact that the family is a MLR family in X(n) implies that a UMPlevel α test for H0 : θ < θ0 vs H1 : θ > θ0 is:
φ2(x) =
1 x(n) > k
0 0 < x(n) < k
where k satisfies α = Pθ0(X(n) > k) = (1− k/θ0)n = 1− (k/θ0)n,
which gives k = θ0(1− α)1/n.
Since both φ and φ2 are UMP level α test, UMP is not uniquein this situation. This happens because the likelihood ratio λ(x)is constant on [0, θ0]. In fact we can show that they have thesame power function, i.e., for any θ > θ0,
β(θ) = β1(θ) = 1− (1− α)(θ0
θ
)n45
UMP Does Not Always Exist
• In many situations, because the class of level α tests is so
large, no one dominates all the others in power.
• Example. Let (X1, · · · , Xn) be a random sample from ∼N(θ, σ2) with σ2 known, H0 : θ = θ0 and H1 : θ 6= θ0. Here
the tests
φ1(x) =
1 x > θ0 + σ√
nΦ−1(1− α)
0 otherwiseand φ1(x) =
1 x < θ0 − σ√
nΦ−1(1− α)
0 otherwise
are the UMP level α tests for H0 : θ = θ0 vs K1 : θ > θ0 and
K2 : θ < θ0 respectively.
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• Consider an alternative parameter θ1. Eθ1[φ1(x)] < α ∨θ1 <
θ0 and Eθ1[φ2(x)] < α ∨θ1 > θ0.
• φ1 is UMP level α when θ1 > θ0; φ2 is UMP level α whenθ1 < θ0.
• φ1 is more powerful than φ2 when θ1 > θ0, but φ2 is morepowerful when θ1 < θ0.
• As a result, a UMP test at level α for H:θ = θ0 vs H1 : θ 6= θ0does not exist.
Draw the power functions here (p394 of Casella andBerger)
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Unbiased Test
• In the above example, UMP does not exist. To find/definea “goog” test in the situation, we then narrow the class oflevel α test to the class of “unbiased” level α tests.
• Definition. A test φ is said to be an unbiased test at levelα for H0 : θ ∈ Ω0 vs H1 : θ ∈ Ω1 if
(i) Eθ[φ(x)] ≤ α ∨θ ∈ Ω0
(ii) Eθ[φ(x)] ≥ α ∨θ ∈ Ω1
• An unbiased test rejects the null hypothesis H0 with at leastas much probability when it is false as when it is true.
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Uniformly Most Powerful Unbiased (UMPU)Test
• A test φ is said to be a uniformly most powerful unbiased
(UMPU) test at level α for H0 : θ ∈ Ω0 vs H1 : θ ∈ Ω1 , if
(i) φ is an unbiased test at level α for H0 vs H1
(ii) if Eθ[φ(x)] ≥ Eθ[ψ(x)] ∨θ ∈ Ω1, where ψ is also an unbi-
ased level α test for H0 vs H1.
• UMPU tests usually exist for several types of hypothesis tests
for a natural exponential family.
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UMPU (continued)
• For testing H:θ = θ0 vs H1 : θ 6= θ0 at level α, an unbiasedtest must satisfy Eθ0
[φ(x)] ≤ α and Eθ[φ(x)] ≥ α ∨θ 6= θ0.
• Neither φ1 nor φ2 in the above example is an unbiased test.
• If we restrict to the class of unbiased level α tests, then testssuch as φ1 and φ2 would not quality, and in the restrictedclass, a UMP test does exist.
• The following test is an UMPU:
φ3(x) =
1 |x− θ0| > σ√
nΦ−1(1− α/2)
0 otherwise50
• Note. φ3(x) = I[0,∞)(|x− θ0| > σ√n
Φ−1(1− α/2))
• In practice, MP, UMP, or UMPU test often does not exist.
• We often seek for intuitively reasonable solutions. These
solutions frequently coincide with optimal tests.
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