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The most powerful test for the variance of a normal distribution
Let X1 , . . . , Xn be a random sample from a normal distribution with known mean µ and
unknown variance σ 2 . Suggested are two hypotheses: σ = σ0 and σ = σ1 .
Let us derive the likelihood ratio criterion at significance level α, for each 0 < α < 1.
Form the likelihood quotient,
P (X1 ≈ x1 , . . . , Xn ≈ xn | σ = σ0 )
P (X1 ≈ x1 , . . . , Xn ≈ xn | σ = σ1 )
Qn
−(xi −µ)2 /2σ02 dx
√ 1
i
i=1 2πσ0 e
= Qn
2 /2σ 2
1
−(x
−µ)
i
√
1 dxi
i=1 2πσ1 e
)
(
n
n
σ1
σ02 − σ12 X
=
(xi − µ)2 .
exp
σ0
2σ02 σ12 i=1
l(x1 , . . . , xn ) =
n
X
The condition l(x1 , . . . , xn ) < c is equivalent to
(xi − µ)2 > c̃, if σ0 < σ1 , and to
i=1
2σ2
2σ
σ0
2
0 1
+ ln c .
(xi − µ) < c̃, if σ0 > σ1 , where c̃ = 2
n ln
σ1
σ0 − σ12
i=1
n
X
Thus
(Xi − µ)2 is the statistic of the test. The critical value c̃ is determined by α.
n
X
i=1
Assume that σ0 < σ1 .
Under the hypothesis σ = σ0 , we have
Xi − µ
σ0
2
∼
χ21
and
n X
Xi − µ 2
i=1
σ0
∼ χ2n .
Therefore the condition
α = P (l(X1 , . . . , Xn ) < c σ = σ0 ) = P
n X
Xi − µ 2
i=1
σ0
!
c̃ > 2 σ = σ0
σ0
gives c̃ = σ02 χ2n (α), where χ2n (α) is the α–upper quantile for the chi-square distribution
with n degrees of freedom. Note that c̃ does not depend on σ1 .
Under the hypothesis σ = σ1 , we have
n X
Xi − µ 2
i=1
σ1
β = P (l(X1 , . . . , Xn ) ≥ c σ = σ1 ) = P
Hence β = F
∼ χ2n , and therefore
n X
Xi − µ 2
!
c̃ ≤ 2 σ = σ1 .
σ1
σ1
i=1
σ02 2
χ (α) , where F (x) is the cumulative distribution function of χ2n .
σ12 n
Observe that β ≤ 1 − α is monotone decreasing in α and σ1 , and monotone increasing in σ0 ,
σ0
and that χn (1 − β) =
χn (α).
σ1
EXAMPLE Examined are size–25 random samples from a normal distribution with mean
µ = 0 and unknown variance σ 2 . Consider the null hypothesis σ = 1 versus the alternative
hypothesis σ = 2. What is the likelihood ratio (rejection) criterion at the significance level
α = 0.05? Find the probability of a type II error.
Solution.
If the null hypothesis σ = 1 is true, then
25
X
Xi2 ∼ χ225 .
i=1
Since 0.05 = P
25
X
i=1
!
Xi2 > c̃ σ = 1 , we have c̃ = χ225 (0.05) = 37.65.
So the null hypothesis is to be rejected whenever
25
X
Xi2 > 37.65.
i=1
The type II error probability is β = P
25
X
i=1
!
Xi2 /4 ≤ 37.65/4 σ = 2 = P (χ225 ≤ 9.4125) = 0.002.