Question

Let \(X_1,X_2\) be a random sample of size \(2\) from a distribution with the probability density function

Hint:

For simple versus simple testing, the Neyman-Pearson lemma says to reject \(H_0\) for large values of the likelihood ratio \(\frac{L_1}{L_0}\).

Answer 1

Correct Answer: 0.615

Step 1: Write the likelihood ratio.
For the sample \(X_1,X_2\), the likelihood is
\[ L(\alpha)=\prod_{i=1}^{2}\alpha X_i^{\alpha-1} \] \[ L(\alpha)=\alpha^2(X_1X_2)^{\alpha-1} \] For testing
\[ H_0:\alpha=1 \] against
\[ H_1:\alpha=2, \] the likelihood ratio is
\[ \frac{L(2)}{L(1)} = \frac{2^2(X_1X_2)}{1} \] \[ =4X_1X_2 \] Thus, the most powerful test rejects \(H_0\) for large values of
\[ X_1X_2 \]

Step 2: Compute the observed value of the test statistic.
Given observations are
\[ X_1=\frac14,\qquad X_2=\frac12 \] So,
\[ X_1X_2=\frac14\cdot \frac12 \] \[ =\frac18 \]

Step 3: Define the \(p\)-value.
The \(p\)-value is the probability under \(H_0\) of getting a value at least as extreme as the observed value.
Since rejection is for large \(X_1X_2\),
\[ p\text{-value}=P_{H_0}\left(X_1X_2\geq \frac18\right) \] Under \(H_0:\alpha=1\),
\[ X_1,X_2\sim U(0,1) \] independently.

Step 4: Calculate the probability.
Let
\[ U=X_1,\qquad V=X_2 \] where \(U,V\) are independent \(U(0,1)\) random variables.
We need
\[ P\left(UV\geq \frac18\right) \] First calculate the complement:
\[ P\left(UV<\frac18\right) \] For \(0<c<1\),
\[ P(UV<c)=c(1-\log c) \] Here,
\[ c=\frac18 \] Therefore,
\[ P\left(UV<\frac18\right) = \frac18\left(1-\log\frac18\right) \] Since
\[ \log\frac18=-\log 8, \] we get
\[ P\left(UV<\frac18\right) = \frac18(1+\log 8) \] Thus,
\[ P\left(UV\geq \frac18\right) = 1-\frac18(1+\log 8) \] \[ = 1-\frac18(1+2.0794) \] \[ = 1-0.3849 \] \[ = 0.6151 \] Rounded off to three decimal places,
\[ 0.615 \]

Step 5: Final conclusion.
Hence, the \(p\)-value is
\[ \boxed{0.615} \]