hypothesis testingneyman-pearson-lemma
I'm reading an online course and I'm really confused about the Neyman-Pearson lemma. It states
From my understanding, does it mean that critical region could be any region like (a,b) and (c,+00)?
And I'm very confused with most powerful critical region according to its definition
what does K mean? Can K be any constant number?
Anyway you look at it, the answer looks good.
You have discovered that the log likelihood ratio is linear:
$$\log\left(\frac{f(x;\theta_1)}{f(x;\theta_0)}\right) = 25\log(2) - x\log(3).$$
The left hand panel of the figure plots this. (Since only the shape of the log likelihood ratio matters, no value axis is shown.)
This reflects the fact that the relative probability that $X$ came from $\theta_1$ compared to $\theta_0$ decreases as $X$ grows larger. Those probabilities appear in the middle panel of the figure: red, peaking near $25\theta_1=12.5$, for $\theta_1$, and blue, peaking near $25\theta_0=18.75$, for $\theta_0$. That suffices to show the UMP critical region must be some interval starting at $X=0$.
A good quantitative check of the result is afforded by the cumulative distribution functions, shown in the right panel with comparable colors. The dotted horizontal line is at a value of $\alpha=0.05$. The CDF corresponding to $\theta_0$ rises above $\alpha$ between $X=14$ and $X=15$, showing that the false positive rate must be less than $\alpha$ for the critical region $\{0,1,\ldots, 14\}$ but greater than $\alpha$ if $15$ is also included. Consequently $\{0,1,\ldots, 14\}$ is the unique solution.
I think you understood the lemma well.
Why it does not work for a composite alternative? As you can see in the likelihood ratio, we need to plug in the parameter(s) for the alternative hypothesis. If the alternative is composite, which parameter are you going to plug in?
Best Answer
Both of the regions you shaded correspond to $\alpha=0.05$, assuming that your plot represents the null distribution.
The best critical region would be the one of these that would be most likely under the alternative hypothesis. If the alternative hypothesis was $H_a:\mu = C$ for any $C>0$, then the tail critical region would be the best critical region because it would be the critical region with the largest probability under the alternative.
K is a unique constant so that the test has the specified significant level, as the commentator JohnK noted.