| $s=1$ | $s=2$ | $s=3$ | $s=4$ | |
|---|---|---|---|---|
| $f_a(s)$ | $1/3$ | $1/6$ | $1/12$ | $5/12$ |
| $f_b(s)$ | $1/2$ | $1/4$ | $1/6$ | $1/12$ |
This exercise asks you to find the UMP size 0.10 test for $H_0:\theta = a$ versus $H_a: \theta = b$
I know the solution for this test gives a UMP test with $c_0=3/2$ and $\gamma=1/30$.
I for the life of me do not know how they computed these numbers.
My attempt so far is this.
We know by Neyman-Pearson that a UMP test is given by, where $H_0:\theta = \theta_0$ versus $H_a: \theta = \theta_1$
$$
\phi_0(x)= \Bigg\{
\begin{matrix}
1 & \frac{f_{\theta_1}}{f_{\theta_0}} > c_0\\
\gamma & \frac{f_{\theta_1}}{f_{\theta_0}} = c_0\\
0 & \frac{f_{\theta_1}}{f_{\theta_0}} < c_0
\end{matrix}
$$
We also know that $P(\frac{f_{\theta_1}}{f_{\theta_0}} > c_0) \leq \alpha$.
So I need to compute $\frac{f_{\theta_1}}{f_{\theta_0}} = \frac{f_{\theta_b}}{f_{\theta_a}}$
So to find the Likelihood for each hypothesis I take the values of the pmf at each point and multiply them. Then I take the the ration of the two likelihoods ending up with $9/2$.
So now with $P(9/2>c_0)\leq 0.1$ I try to compute $c_0$. I find the mean and variance under the null hypothesis, and try to standardize the above to use the t-distribution.
I end up with $P(\frac{9/2-31/12}{\sqrt{251}/12} > \frac{c_0-31/12}{\sqrt{251}/12}) \leq 0.1$. Using the t distribution I get $1-G_3(\frac{c_0-31/12}{\sqrt{251}/12}) \leq 0.1$. Moving stuff around I get.
$-G_3(\frac{c_0-31/12}{\sqrt{251}/12}) \leq -0.9$, and dividing by $-1$ I get: $G_3(\frac{c_0-31/12}{\sqrt{251}/12}) \geq 0.9$. Then using the t-table I find that I get $\frac{c_0-31/12}{\sqrt{251}/12} = 1.638$. Isolating for $c_0$ I get $c_0 = 4.76733$
Which is obviously wrong. How am I supposed to go about these kinds of questions?
Best Answer
NP lemma tells you to reject $H_0:\theta=a$ in favour of $H_1:\theta=b$ for large values of the ratio $r(x)=f_b(x)/f_a(x)$. So compute $r(x)$ for every $x$:
\begin{array}{|c|c|c|c|c|} \hline x&1&2&3&4\\ \hline r(x)&3/2& 3/2 & 2 & 1/5\\ \hline \end{array}
As you can see, $$r(3)>r(1)=r(2)>r(4) \tag{$\star$}$$
By NP lemma, an MP (or UMP) size $0.1$ test takes the form
$$\phi(x)=\begin{cases} 1&, \text{ if }x\in R_1 \\ \gamma &, \text{ if }x\in R_2 \\ 0&, \text{ otherwise } \end{cases} $$
where $\gamma \in[0,1]$ and the regions $R_1,R_2$ are such that
$$E_{H_0}\phi(X)=P_{H_0}(X\in R_1)+\gamma P_{H_0}(X\in R_2)=0.1$$
Further, $R_1,R_2$ are decided keeping in mind the order in $(\star)$. That is to say, the sample point $3$ will be the first to enter the rejection region, followed by $1$ and/or $2$, and finally $4$.
Here you can only choose $R_1=\{3\}$, because $P_{H_0}(X=3)=\frac1{12}=0.0833<0.1$ and the probabilities of $X$ taking the other values under $H_0$ already exceed $0.1$.
As for $R_2$, you can take $R_2=\{1\}$ or $R_2=\{2\}$ or $R_2=\{1,2\}$. This gives you 3 UMP tests, each with a different $\gamma$ found subject to the size $0.1$ restriction. You can check that all three tests have the same power $E_{H_1}\phi(X)$. The solution you mention takes $R_2=\{1,2\}$.
Lastly, 't' distribution is completely off-topic here and your attempt to bring this into the solution makes no sense.