Question

Consider the problem of testing \( H_0: X \sim f_0 \) against \( H_1: X \sim f_1 \) based on a sample of size 1, where \[ f_0(x) = \begin{cases} 1, & 0 \le x \le 1, \\ 0, & \text{otherwise,} \end{cases} \quad f_1(x) = \begin{cases} 2x, & 0 \le x \le 1, \\ 0, & \text{otherwise.} \end{cases} \] Then, the probability of Type II error of the most powerful test of size \(\alpha = 0.1\) is equal to

• A. 0.81
• B. 0.91
• C. 0.1
• D. 1

Hint:

For simple hypotheses, the most powerful test is based on the likelihood ratio. Always compute the critical point from the size condition under \(H_0\).

Answer 1

The Correct Option is A

Step 1: Apply the Neyman–Pearson lemma.
We reject \(H_0\) for large values of the likelihood ratio \[ \Lambda(x) = \frac{f_1(x)}{f_0(x)} = 2x. \] Hence, reject \(H_0\) if \(x>c\).
Step 2: Find \(c\) using size condition.
Size \(\alpha = 0.1 \Rightarrow P_{H_0}(x>c) = 0.1\). Under \(H_0\), \(X \sim U(0,1)\), so \[ 1 - c = 0.1 \Rightarrow c = 0.9. \]
Step 3: Find probability of Type II error (\(\beta\)).
Under \(H_1\), \( f_1(x) = 2x \). \[ \beta = P_{H_1}(x \le 0.9) = \int_0^{0.9} 2x \, dx = [x^2]_0^{0.9} = (0.9)^2 = 0.81. \] Thus, Type II error = 0.81, and power = 0.19. The question asks for “probability of Type II error,” so it equals 0.81. Final Answer: \[ \boxed{0.81} \]