Consider a Bernoulli random variable:
$$X_i= \begin{cases} 1, & \text{with probability }p \\ 0, & \text{with probability }1-p \end{cases}$$
You observe the outcomes of two Bernoulli trials and want to test $H_o : p=0$ against $H_1 : p=0.5$. Use Neyman-Pearson lemma to determine the most powerful test of $H_o$ versus $H_1$. What are the Type 1 $(\alpha)$ and Type 2 $(\beta)$ errors for your test?
Here's how I tried. Please point out my mistakes and correct me:
$$f(x_i)= \begin{cases} (1-p)^{1-x_i} p^{x_i}, & x_i =0,1 \\ 0, & \text{elsewhere} \\ \end{cases}$$
$H_o: p=p'=0$
$H_1: p=p''=0.5$
$\frac {L(p')}{L(p'')}$= $\frac {0}{1/4}$ = $0 \le k$ where $k$ is a positive number.
$\implies \frac {L(p')}{L(p'')} \lt k$ (for all $k \gt 0)$ $\implies \frac {L(p')}{L(p'')} \lt k$ for all $X_i$s (here for $X_1$ and $X_2$)
So here my critical region is all values of $X_i$ i.e., $X_i=0,1$ or to say $[X_1=0,1$ and $X_2=0,1]$
$\alpha = P_{H_o}(X_1=0,1$ and $X_2=0,1) = 0$
$\beta =1- P_{H_1}(X_1=0,1$ and $X_2=0,1)$
But, $P_{H_1}(X_1=0,1$ and $X_2=0,1)$
$=P_{H_1}(X_1=0,X_2=0) + P_{H_1}(X_1=0,X_2=1) + P_{H_1}(X_1=1,X_2=0) + P_{H_1}(X_1=1,X_2=1)$
$=(\frac12)(\frac12)4 =1$
So, $\beta = 1-1=0$
But both $\alpha$ and $\beta$ cant be zero? someone please tell me how to apply neyman pearson in such cases.
$$ \frac{L(p')}{L(p'')} = \begin{cases} \frac{0}{1/4} & \text{if }X_1=X_2=1, \\[10pt] \frac{0}{1/4} & \text{if }X_1=1\ \&\ X_2=0, \\[10pt] \frac{0}{1/4} & \text{if }X_1=0\ \&\ X_2=1, \\[10pt] \frac{1}{1/4} & \text{if }X_1=X_2=0. \end{cases} $$ The probability of a false positive (Type I error, falsely rejecting $H_0$) is $0$, since $X_1+X_2$ cannot be positive if $p=0$. But the probability of failing to reject $H_0$ if the alternative is true is $1/4$, since $\Pr(X_1=X_2=0\mid p=0.5)=1/4$.