Question

Let \(Y_1,Y_2\) be a random sample of size \(2\) from a distribution with the probability mass function

• A. The test \(\psi\) rejects \(H_0\) with probability \(1\) if the observed value of \(Y_1+Y_2\) is \(0\)
• B. The test \(\psi\) rejects \(H_0\) with probability \(1\) if the observed value of \(Y_1+Y_2\) is \(1\)
• C. The test \(\psi\) rejects \(H_0\) with probability \(\dfrac{5}{16}\) if the observed value of \(Y_1Y_2\) is \(1\)
• D. The power of the test \(\psi\) is \(\dfrac{621}{1024}\)

Hint:

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.

Answer 1

The Correct Option is A, D

Step 1: Shape of the test.
The likelihood ratio for $\theta=0.75$ over $\theta=0.20$ decreases in $S=Y_1+Y_2$, so the most powerful test rejects for small $S$.

Step 2: Null probabilities.
Under $\theta=0.20$, $P(S=0)=0.04$ and $P(S=1)=0.064$.

Step 3: Build the level $0.05$ test.
Since $0.04<0.05$, reject outright at $S=0$ (so (A) holds), then randomise at $S=1$ with $\gamma=\frac{0.01}{0.064}=\frac5{32}$, not $1$, so (B) fails. If $Y_1Y_2=1$ then $S=2$, never in the rejection scheme, so (C) fails.

Step 4: Power (D).
Under $\theta=0.75$, $P(S=0)=\frac9{16}$ and $P(S=1)=\frac9{32}$, so power $=\frac9{16}+\frac5{32}\cdot\frac9{32}=\frac{576+45}{1024}=\frac{621}{1024}$. (D) holds.

Step 5: Collect.
\[ \boxed{(A),(D)} \]