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416 8. Tests of Hypotheses

since Cauchy(0, 1) pdf is symmetric around x = 0. Hence, the test’s power is
given by 1 – 2/π arctan(k) with When α = .05, the power turns
out to be .39791. !
Look at the Exercises 8.3.8-8.3.10 and the Exercise 8.3.20.

8.3.3 Applications: Observations Are Non-IID
In the formulation of the Neyman-Pearson Lemma, it is not crucial that the
X’s be identically distributed or they be independent. In the general situation,
all one has to work with is the explicit forms of the likelihood function under
both simple hypotheses H and H . We give two examples.
1
0
Example 8.3.11 Suppose that X and X are independent random variables
2
1
respectively distributed as N(µ, σ ) and N(µ, 4σ ) where µ ∈ ℜ is unknown
2
2
and σ ∈ ℜ + is assumed known. Here, we have a situation where the X’s are
independent but they are not identically distributed. We wish to derive the MP
level a test for H : µ = µ versus H : µ = µ (> µ ). The likelihood function is
0
1
0
0
1
given by
where x = (x ,x ). Thus, one has
1 2
In view of the Neyman-Pearson Lemma, we would reject H if and only if
0
L(x; µ )/L(x; µ ) is “large”, that is when 4X + X > k. This will be the MP
2
1
0
1
level α test when k is chosen so that P {4X + X > k} = α. But, under H ,
1
µ0
2
0
the statistic 4X + X is distributed as N(5µ , 20σ ). So, we can rephrase the
2
0
2
1
MP level a test as follows:
We leave out the power calculation as Exercise 8.3.15. !
Example 8.3.12 Let us denote X’ = (X , X ) where X is assumed to have
1 2
the bivariate normal distribution, with the unknown param-
eter µ ∈ ℜ. Refer to Section 3.6 for a review of the bivariate normal distribu-
tion. We wish to derive the MP level a test for H : µ = µ versus H : µ = µ (>
1
0
0
1
µ ). The likelihood function is given by
0
Thus, one has