Discussion
Suppose X has a distribution with probability mass function that of discrete uniform on \(\{-1, 0, 1\}\) under \(H_{0}\) and \(U(-1, 1)\) under \(H_{1}\). Then the test function
\[ \phi(x) = \begin{cases} 1 & \text{if } x \lt 0 \\ 0 & \text{if } x \ge 0 \end{cases} \]
has size :
Answer: B
The size of the test is 1/3.
The size is the probability of rejecting \(H_{0}\) when \(H_{0}\) is true. The test rejects when \(x \lt 0\), and under \(H_{0}\) the only negative value is \(-1\), with probability 1/3:
\[ E_{H_{0}}[\phi(X)] = P_{H_{0}}(X = -1) = \tfrac{1}{3} \]
Size is computed under \(H_{0}\) alone, so it makes no difference that \(H_{1}\) is continuous; the test is perfectly well defined. 1/2 is \(P_{H_{1}}(X \lt 0)\), which is the power, not the size.
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