Discussion
Let X be a Bernoulli r.v. with parameter \(\theta\). Suppose we wish to test \(H_{0} : \theta = 1/3\) against \(H_{1} : \theta = 2/3\), based on random sample of size 1. Then :
Answer: C
There exists a unique randomized most powerful test of size 0.05.
The likelihood ratio is
\[ \begin{aligned} \frac{f_{1}(x)}{f_{0}(x)} &= \frac{(2/3)^{x}(1/3)^{1 - x}}{(1/3)^{x}(2/3)^{1 - x}} \\ &= 2^{2x - 1} \end{aligned} \]
which is 2 at x = 1 and 1/2 at x = 0. By the Neyman-Pearson lemma the most powerful test rejects for the larger ratio, that is on X = 1.
But rejecting whenever X = 1 has size \(P_{0}(X = 1) = 1/3\), far above 0.05. The only way to reach size exactly 0.05 is to reject with probability \(\gamma\) when X = 1 and never when X = 0:
\[ \gamma \cdot \tfrac{1}{3} = 0.05 \Rightarrow \gamma = 0.15 \]
That test is randomized, and the lemma makes it unique, since the rejection probability is fixed on both sample points. No non-randomized test has size 0.05 (the only non-randomized sizes are 0, 1/3, 2/3 and 1). Rejecting on X = 0 is the least powerful choice, not the most powerful.
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