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STAT 533 Mathematical Statistics II
Lecture 5
Dajiang Liu
September 30, 2007
Example 1: Consider a mixture of 2 distributions, H(x) = θF (x) + (1 − θ)G(x), and we want to test the hypothesis that
H 0 : θ < θ 0 vs.H 1 : θ > θ 0
For the simplicity of maximization, we just assume we collected one data point x.
Now we rewrite the mixture distribution into
H(x; θ) = θ(F (x) − G(x)) + G(x)
and its density h(x; θ) = θ(f (x) − g(x)) + g(x)
So we could see if f (x) > g(x) at the data point, we have h(x; θ) is an increasing function of θ. h(x; θ) is maximized at θ = θ 0 under the NULL, and if f (x) < g(x), h(x; θ) is maximized at θ = 0, so
L∗ 0 = [θ 0 (f (x) − g(x)) + g(x)]I(f (x) > g(x)) + g(x)I(f (x) < g(x))
By similar argument, we have
L∗^ = f (x)I(f (x) > g(x)) + g(x)I(f (x) < g(x))
So the likelihood ratio is given by
λ(x) =
L∗ 0
L∗^
θ 0 (f (x) − g(x)) + g(x) f (x)
I(f (x) > g(x)) + 1I(f (x) < g(x))
We will reject the null if the the likelihood ratio is small. Consider the size constraint
Pθ∈Θ 0 [λ < kα] = Pθ∈Θ 0 [I[
g(x) f (x)
kα − θ 0 1 − θ 0
]]
We need to figure out the value kα to make the size constraint satisfied.
However, this is usually handled when f (x) and g(x) has monotone likeli- hood ratios, which is like the case in UMP test.
Example 2: Assume we have samples from 2 distributions: X 11 ,... , Xn 11 from U (0, θ 1 ) and X 12 ,... , Xn 22 from U (0, θ 2 ). And now we want to test the hypothesis H 0 : θ 1 = θ 2 vs. H 1 : θ 1 6 = θ 2 So we need to estimate the MLE under the null constraint and uncon- strained MLE, which is very standard calculation. We know θˆ 1 = max x 1 i and θˆ 2 = max x 2 i and for the constrained case, we can just pool the two sam- ples together,and obtain θˆ = max xij. To simplify the notations, we define Yi = maxj xij for i = 1, 2, and Y = max Xij , so the likelihood ratio is then given by
λ = Y 1 n 1 Y 2 n^2 Y n
and to make latter compuations feasible, we rewrite
λ = (
Y 1
Y 2
)n^1 I(Y 2 > Y 1 ) + (
Y 2
Y 1
)n^2 I(Y 1 > Y 2 )
So in order to find out the rejection region of size α we need to find kα such that Pθ 1 =θ 2 [λ < kα] = α
we have the following calculation
P [(
Y 1
Y 2
)n^1 I(Y 2 > Y 1 ) + (
Y 2
Y 1
)n^2 I(Y 1 > Y 2 ) < kα] = P [
Y 1
Y 2
< k n^1 α^1 , Y^1 Y 2
< 1]
+P [
Y 2
Y 1
< k n^1 α 2 , Y^2 Y 1
< 1]
∫ (^) θ
0
n 2 θn 2
Y 2 n 2 −^1 dY 2
∫ (^) Y 2 k n^1 α^1
0
n 1 θn 1
Y 1 n 1 −^1 dY 1
∫ (^) θ
0
n 1 θ 1 n
Y 1 n 1 −^1 dY 1
∫ (^) Y 1 k n^1 α^2
0
n 2 θn 2
Y 2 n 2 −^1 dY 2
= kα
Remark: (brought up by Dr. Ensor) The likelihood ratio now is not asymp- totically a χ^2 distributed statistic, and the support of the distribution now de- pend on the parameter, which violates one of the key regularity conditions for the asymptotic result to hold.
