Let $X_{1}, \ldots, X_{n}$ be a sample from a discrete distribution. Let the null hypothesis $H_{0}$ state that the distribution is uniform on $\{0,1,2\},$ i.e., $P\left(X_{i}=0\right)=P\left(X_{i}=1\right)=P\left(X_{i}=\right.$
2)$=\frac{1}{3}$.
Let the alternative $H_{1}$ be that the distribution is given by $P\left(X_{i}=0\right)=\frac{1}{5}=P\left(X_{i}=1\right), P\left(X_{i}=2\right)=\frac{3}{5}$. Consider testing $H_{0}$ vs $H_{1}$.
I want to construct a Neyman-Pearson most powerful test among tests with level at most $\alpha .$
But I have no idea how to construct such a test from given information. Please help me with this. Thankyou.
Let $m$ denote the number of $2$’s in the sample. The Neyman–Pearson lemma tells us that we should reject the null hypothesis whenever $$ \frac{L_1(m)}{L_0(m)} \ge k_\alpha\text, $$ where $L_0$ is the likelihood under $H_0$, $L_1$ the likelihood under $H_1$, and $k_\alpha$ is chosen to have the appropriate power. The likelihood ratio here is of the form $$ \frac{\left(\frac15\right)^{n - m} \left(\frac35\right)^m}{\left(\frac13\right)^n} = \frac{3^{n + m}}{5^n}\text. $$
You should therefore reject the null hypothesis whenever $m$ is high enough, i.e., $$ \frac{3^{n + m}}{5^n} \ge k_\alpha \iff m \ge m_\alpha\text. $$
Intuitively, this makes sense because the $2$’s are more likely under $H_1$, so if we have many of them, we should prefer $H_1$! How do we choose $m_\alpha$? For any given critical value $m_c$, $$ \alpha = \left(\frac{1}{3}\right)^n \sum_{j = m_c}^{n} \binom{n}{j}\text. $$
Therefore, you can try different $m_c$ from $0$ until $n$ and see which one gives you the required power $\alpha$.