Question

Consider the two probability density functions (pdfs): \[ f_0(x) = \begin{cases} 2x, & 0 \le x \le 1, \\ 0, & \text{otherwise}, \end{cases} f_1(x) = \begin{cases} 1, & 0 \le x \le 1, \\ 0, & \text{otherwise.} \end{cases} \] 

Let $X$ be a random variable with pdf $f \in \{f_0, f_1\}$. Consider testing  $H_0: f = f_0(x)$ against $H_1: f = f_1(x)$ at $\alpha = 0.05$ level of significance. For which observed value of $X$, the most powerful test would reject $H_0$? 
 

• A. 0.19
• B. 0.22
• C. 0.25
• D. 0.28

Hint:

In the Neyman–Pearson test, reject $H_0$ for values that make $\frac{f_1(x)}{f_0(x)}$ largest — that is, where the likelihood ratio is maximized.

Answer 1

The Correct Option is A, B

Step 1: Neyman–Pearson lemma.
The most powerful (MP) test rejects $H_0$ for large values of the likelihood ratio \[ \Lambda(x) = \frac{f_1(x)}{f_0(x)}. \]

Step 2: Compute the ratio.
\[ \Lambda(x) = \frac{1}{2x}, 0 < x \le 1. \] This ratio is largest when $x$ is small.

Step 3: Determine critical region.
Reject $H_0$ for $x < c$, such that $P_{H_0}(X < c) = \alpha = 0.05$. Under $H_0$, $F_0(x) = x^2$. So, $x^2 = 0.05 $\Rightarrow$ x = \sqrt{0.05} \approx 0.2236$.

Step 4: Conclusion.
Reject $H_0$ when $x < 0.2236$. Among given choices, $x = 0.19$ satisfies this. \[ \boxed{x = 0.19.} \]