How to find MP test and UMP test

Lemma) Neyman-Pearson Lemma.
Suppose $f(\underline{x};\theta)$ is joint pdf of $X_1,...,X_n$ and
$\phi(\underline{x})=\left\{\begin{matrix} 1 & f(\underline{x};\theta_1) > kf(\underline{x};\theta_0)\\ 0 & f(\underline{x};\theta_1) < kf(\underline{x};\theta_0) \end{matrix}\right.$ satisfying $E_{\theta_0}(\phi(\underline{x}))=P_{\theta_0}(\phi(\underline{x})=1)=\alpha$ for $k>0$
Also, consider $H_0:\theta=\theta_0$ vs $H_1: \theta=\theta_1$

Then,
(a) $\phi(\underline{x})$ is MP test of level $\alpha$
(b) Every level $\alpha$ test is size $\alpha$ test
(c) If $\phi'(\underline{x})$ is another level $\alpha$ MP test, then $\beta_{\phi}(\theta_0)=\beta_{\phi'}(\theta_0)$ and $\beta_{\phi}(\theta_1)=\beta_{\phi'}(\theta_1)$
pf)

Corollary)
Suppose $T$: sufficient statistic for $\theta$.
If $g(T;\theta)$ is pdf of $T,$
for $H_0:\theta=\theta_0$ vs $H_1:\theta=\theta_1,$
MP test is $\phi(\underline{x})=\left\{\begin{matrix} 1 & g(T;\theta_1) > k g(T;\theta_0)\\ 0 & g(T;\theta_1) < k g(T;\theta_0) \end{matrix}\right.$ satisfying $E_{\theta_0}(\phi(\underline{x}))=P_{\theta_0}(\phi(\underline{x})=1)=\alpha$ for $k>0$
pf)

Corollary)
For $H_0:\theta=\theta_0$ vs $H_1:\theta=\theta_1,$
if $\phi(\underline{x})$ is size $\alpha$ MP test, $E_{\theta_1}(\phi(\underline{x})) \geq \alpha=E_{\theta_0}(\phi(\underline{x}))$
pf)

ex1) $X \sim b(n,\theta), \space H_0:\theta=\theta_0$ vs $H_1:\theta=\theta_1(\theta_1>\theta_0)$

✔︎ If $\phi^{MP}$ does not depend on $\theta_1$, $\phi^{MP}$ is a uniformly most powerful test of size $\alpha$ and denote $\phi^{UMP}$

ex2) $X_1,...,X_n \overset{iid}{\sim} N(\theta,1), \space H_0: \theta=0$ vs $H_1:\theta>0$

ex3) $X_1,...,X_n \overset{iid}{\sim} N(\theta,1), \space H_0: \theta=0,-1$ vs $H_1:\theta>0$

✔︎ If $\beta_{\phi}(\theta)$ is nondecreasing, the UMP size $\alpha$ test for $H_0:\theta=\theta_0$ vs $H_1:\theta>\theta_0$ becomes UMP size $\alpha$ test for $H_0:\theta \leq \theta_0$ vs $H_1:\theta>\theta_0$

ex4) $X_1,...,X_n \overset{iid}{\sim} N(\theta,1), \space H_0: \theta \leq 0$ vs $H_1:\theta > 0$

Def)
$T$ has a monotone likelihood ratio(MLR) if
for every $\theta_2>\theta_1, \space g(t;\theta_2)/g(t;\theta_1)$ is a monotone function on $t$

$g(t;\theta)$: pdf or pmf of $t$

ex1) $T \sim Exp(\theta), \theta_2 > \theta_1$

Theorem)
For $H_0:\theta \leq \theta_0$ vs $H_1:\theta > \theta_0$,
if $T$ is sufficient statistic for $\theta$ and has nondecreasing(nonincreasing) MLR,
then $T>(<)t_0$ is UMP size $\alpha$ test where $\alpha=P_{\theta_0}(T>t_0)$
pf)

ex1) $X_1,...,X_n \overset{iid}{\sim} N(\theta,1), \space H_0: \theta \leq 0$ vs $H_1: \theta>0$

ex2) $X_1,...,X_n \overset{iid}{\sim} N(0,\theta), \space H_0: \theta \leq \theta_0$ vs $H_1: \theta > \theta_0$