I've been studying the Neyman-Pearson theorem and I want to know if my understanding about the relationship between test statistics and the significance level $\alpha$ is correct. My problem to understand it's how we can come up with different cutoffs with the same $\alpha=0.05$. What I didn't realize is the cutoffs are related to the tests and not to the p.d.f. of the distribution (right?). Since the tests $T1$ and $T_2$ are different I can have different cutoffs $C_1$ and $C_2$ with the same $\alpha$. In the context of the Neyman-Pearson theorem, I choose the most powerful test. Am I right?
Yes, of course.
$T_1$ and $T_2$ tests have their own distribution that you have to calculate. These distribution can obvioulsy be different from the pdf of the original rv
Limiting the reasoning to a system of simple hypothesis (both simple), Neymann Pearson Lemma grants you to find the Most powerful of all possible tests, say a test with a higher power with respect to any other test with the same (or lower) size $\alpha$
You will see that in many case the distribution of the test can be very hard to find thus you will have to reason on some function of the distribution in order to find the cut off bound(s)
Example
Let's have the following population density
$$f(x,\theta)=\theta x^{\theta-1}$$
where $x \in [0;1]$,$\theta>0$,
Using a simple random sample $(X_1,X_2,X_3)$ let's prove the following system
$$\begin{cases} H_0: & \theta=1 \\ H_1: & \theta=2 \end{cases}$$
First Test
I do not apply NP lemma, I choose a test by myself. I choose the following test
$T_1$: reject $H_0$ if $X_1>0.95$
(I do not mind I have 3 observation, I base my test only on the first observation $X_1$)
Let's calculate $\alpha$:
$$P(X_1>0.95|\theta=1)=\int_{0.95}^{1}dx=0.05$$
So my test has a Type I error of $5\%$...nice. Now let's calculate the power:
$$\gamma=P(X_1>0.95|\theta=2)=\int_{0.95}^{1}2x dx=0.0975\approx 10\%$$
Second Test
I apply NP lemma:
$$\frac{L(\mathbf{x}|\theta_0)}{L(\mathbf{x}|\theta_1)}\leq k$$
$$\frac{1}{8X_1X_2X_3}\leq k$$
taking the logarithm,
$$\Sigma_i(-log X_i)\leq c$$
Observing that $-log X_i\sim Exp(\theta)$
We have that the critical region is
$$Y\leq c$$
where $Y\sim Gamma(3;\theta)$ that is $2\theta Y\sim \chi_{(6)}^2$
Now to compare the two test, let's fix $\alpha=0.05$ also fot $T_2$
$$0.05=P(\chi_{(6)}^2<c) \rightarrow c=1.64$$
that is $Y=0.82$
Finally let's calculate $T_2$ power
$$\gamma=P(Y<0.82|\theta=2)=P(\chi_{(6)}^2<3.27)\approx 22.6\%$$
whichi is greater than $T_1$ power...