Question:
Let \(Y_1,Y_2\) be a random sample of size \(2\) from a distribution with the probability mass function
Let \(Y_1,Y_2\) be a random sample of size \(2\) from a distribution with the probability mass function
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For simple versus simple hypothesis testing, use the Neyman-Pearson lemma. Reject \(H_0\) where the likelihood ratio \(\frac{L_1}{L_0}\) is large.
Updated On: Jun 4, 2026
- The test \(\psi\) rejects \(H_0\) with probability \(1\) if the observed value of \(Y_1+Y_2\) is \(0\)
- The test \(\psi\) rejects \(H_0\) with probability \(1\) if the observed value of \(Y_1+Y_2\) is \(1\)
- The test \(\psi\) rejects \(H_0\) with probability \(\dfrac{5}{16}\) if the observed value of \(Y_1Y_2\) is \(1\)
- The power of the test \(\psi\) is \(\dfrac{621}{1024}\)
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The Correct Option is A, D
Solution and Explanation
Step 1: Write the likelihood ratio.
For the sample \(Y_1,Y_2\), the likelihood is
\[ L(\theta)=\theta^2(1-\theta)^{Y_1+Y_2} \] Let
\[ S=Y_1+Y_2 \] The likelihood ratio is
\[ \frac{L(0.75)}{L(0.20)} = \left(\frac{0.75}{0.20}\right)^2 \left(\frac{0.25}{0.80}\right)^S \] Since
\[ \frac{0.25}{0.80}<1, \] the likelihood ratio decreases as \(S\) increases.
Therefore, the most powerful test rejects \(H_0\) for small values of \(S=Y_1+Y_2\).
Step 2: Find the rejection probabilities under \(H_0\).
Under \(H_0:\theta=0.20\),
\[ P(Y=y)=0.20(0.80)^y \] Now,
\[ P_0(S=0)=P_0(Y_1=0,Y_2=0) \] \[ =(0.20)^2=\frac{1}{25}=0.04 \] Also,
\[ P_0(S=1)=P_0((Y_1,Y_2)=(1,0)\text{ or }(0,1)) \] \[ =2(0.20)^2(0.80) \] \[ =0.064 \]
Step 3: Determine the randomized test.
The level of the test is \(0.05\).
Since
\[ P_0(S=0)=0.04<0.05, \] we reject \(H_0\) with probability \(1\) when
\[ S=0 \] Then we randomize when
\[ S=1 \] Let the randomization probability be \(\gamma\). Then
\[ 0.04+\gamma(0.064)=0.05 \] \[ \gamma=\frac{0.01}{0.064} \] \[ \gamma=\frac{5}{32} \] Thus, the test rejects \(H_0\) with probability \(1\) if \(Y_1+Y_2=0\), and with probability \(\frac{5}{32}\) if \(Y_1+Y_2=1\).
Hence, option (A) is true and option (B) is false.
Step 4: Check option (C).
If
\[ Y_1Y_2=1, \] then since \(Y_1,Y_2\) are non-negative integers, we must have
\[ Y_1=1,\qquad Y_2=1 \] So,
\[ Y_1+Y_2=2 \] But the test rejects only for \(S=0\), and randomizes only for \(S=1\).
Therefore, the rejection probability when \(Y_1Y_2=1\) is \(0\), not \(\frac{5}{16}\).
Hence, option (C) is false.
Step 5: Find the power of the test.
Under \(H_1:\theta=0.75\),
\[ P_1(S=0)=(0.75)^2=\frac{9}{16} \] Also,
\[ P_1(S=1)=2(0.75)^2(0.25) \] \[ =2\cdot \frac{9}{16}\cdot \frac14 \] \[ =\frac{9}{32} \] Therefore, the power is
\[ P_1(S=0)+\frac{5}{32}P_1(S=1) \] \[ =\frac{9}{16}+\frac{5}{32}\cdot \frac{9}{32} \] \[ =\frac{576}{1024}+\frac{45}{1024} \] \[ =\frac{621}{1024} \] Hence, option (D) is true.
Step 6: Final conclusion.
The true statements are
\[ \boxed{(A)\text{ and }(D)} \]
For the sample \(Y_1,Y_2\), the likelihood is
\[ L(\theta)=\theta^2(1-\theta)^{Y_1+Y_2} \] Let
\[ S=Y_1+Y_2 \] The likelihood ratio is
\[ \frac{L(0.75)}{L(0.20)} = \left(\frac{0.75}{0.20}\right)^2 \left(\frac{0.25}{0.80}\right)^S \] Since
\[ \frac{0.25}{0.80}<1, \] the likelihood ratio decreases as \(S\) increases.
Therefore, the most powerful test rejects \(H_0\) for small values of \(S=Y_1+Y_2\).
Step 2: Find the rejection probabilities under \(H_0\).
Under \(H_0:\theta=0.20\),
\[ P(Y=y)=0.20(0.80)^y \] Now,
\[ P_0(S=0)=P_0(Y_1=0,Y_2=0) \] \[ =(0.20)^2=\frac{1}{25}=0.04 \] Also,
\[ P_0(S=1)=P_0((Y_1,Y_2)=(1,0)\text{ or }(0,1)) \] \[ =2(0.20)^2(0.80) \] \[ =0.064 \]
Step 3: Determine the randomized test.
The level of the test is \(0.05\).
Since
\[ P_0(S=0)=0.04<0.05, \] we reject \(H_0\) with probability \(1\) when
\[ S=0 \] Then we randomize when
\[ S=1 \] Let the randomization probability be \(\gamma\). Then
\[ 0.04+\gamma(0.064)=0.05 \] \[ \gamma=\frac{0.01}{0.064} \] \[ \gamma=\frac{5}{32} \] Thus, the test rejects \(H_0\) with probability \(1\) if \(Y_1+Y_2=0\), and with probability \(\frac{5}{32}\) if \(Y_1+Y_2=1\).
Hence, option (A) is true and option (B) is false.
Step 4: Check option (C).
If
\[ Y_1Y_2=1, \] then since \(Y_1,Y_2\) are non-negative integers, we must have
\[ Y_1=1,\qquad Y_2=1 \] So,
\[ Y_1+Y_2=2 \] But the test rejects only for \(S=0\), and randomizes only for \(S=1\).
Therefore, the rejection probability when \(Y_1Y_2=1\) is \(0\), not \(\frac{5}{16}\).
Hence, option (C) is false.
Step 5: Find the power of the test.
Under \(H_1:\theta=0.75\),
\[ P_1(S=0)=(0.75)^2=\frac{9}{16} \] Also,
\[ P_1(S=1)=2(0.75)^2(0.25) \] \[ =2\cdot \frac{9}{16}\cdot \frac14 \] \[ =\frac{9}{32} \] Therefore, the power is
\[ P_1(S=0)+\frac{5}{32}P_1(S=1) \] \[ =\frac{9}{16}+\frac{5}{32}\cdot \frac{9}{32} \] \[ =\frac{576}{1024}+\frac{45}{1024} \] \[ =\frac{621}{1024} \] Hence, option (D) is true.
Step 6: Final conclusion.
The true statements are
\[ \boxed{(A)\text{ and }(D)} \]
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