How to investigate my proposed likelihood ratio test via a simulation study? I have developed three likelihood ratio tests (LLT) based on three different hypotheses for a certain distribution, say $X.$ I understand how to use it when dealing with real data application. I do not understand where to start for simulation study.
Say $X$ has 3 parameters, $a, b, c$ and the first hypothesis is
$a = 0$ vs. $a\neq0$.
How do I test my LLT? Should I develop two datasets, one under $H_0$ and one under $H_a,$ and proceed from there? How do I go about it?
Aim: I would like to investigate if the LLT I proposed can be used to identify whether the parameter is really $a = 0$ or $a\neq0$.
PS: I am new to this page, pardon me if my question is not correct or clear.
 A: Maybe we could give a more direct answer if we knew what test,
what parameters, and what kind of data you have in mind. Along lines of @Dave's Comment, let's look at a pooled 2-sample t test (which is
derived under the assumption that the two samples come from
populations with equal variances $\sigma_1^2 = \sigma_2^2),$
when the equal variance assumption is not true.
Let's consider testing $H_0: \mu_1 = \mu_2$ vs.
$H_a: \mu_1 \ne \mu_2.$ For a simulation, you'd have to use specific numerical values
of the parameters and sample sizes. So, suppose $\mu_1 = 50, \sigma_1 = 5, n_1 = 5$ and $\mu_2 = 50, \sigma_2 = 1, n_2 = 40.$
Thus, $H_0$ is true.
In R, we can look at one instance of such a test as follows,
where the parameter var.eq=T indicates a pooled 2-sample test is to be used. [The two samples are randomly sampled in R. In any simulation, it is a good idea to start with a set.seed statement
so that the results can be repeated, if desired.]
set.seed(111)
x1 = rnorm(5, 50, 5)
x2 = rnorm(40, 50, 1)
t.test(x1, x2, var.eq=T) 

        Two Sample t-test

data:  x1 and x2
t = -2.9049, df = 43, p-value = 0.005781
alternative hypothesis: 
 true difference in means is not equal to 0
95 percent confidence interval:
 -4.3594473 -0.7867747
sample estimates:
mean of x mean of y 
 47.11964  49.69275 

If you just want to see the P-value, you can use $-notation in R
to suppress the rest of the output.
t.test(x1, x2, var.eq=T)$p.val
[1] 0.005780852

In this particular example, the t test has given the "wrong" answer
because the small P-value indicates we should reject $H_0,$ when
in fact $\mu_1=\mu_2.$
Testing at the 5% level of significance,
we should expect such Type I errors in 5% of the tests. So the
question arises whether the example above is an "unlucky" one,
or whether something is wrong with using the pooled t test for
these particular parameters.
You can use a large-scale simulation to decide which is the case.
Below I use simulation to look at $100\,000$ such pooled t tests
to see how frequently $H_0$ is rejected.
set.seed(2022)
pv = replicate(10^5, t.test(rnorm(5, 50, 5), 
                        rnorm(40, 50, 1), var.eq=T)$p.val)
mean(pv <= 0.05)  # aprx signif. level.
[1] 0.46291

The pooled 2-sample t test gives a Type I error of about 46%
instead of 5% in this instance, where the smaller sample has
the larger population variance. That's a massive 'false discovery' rate.
Notes: (a) In the main simulation, the vector pv contains P-values of $m=100\,000$ t tests. The logical vector pv <= 0.05 contains
$m$ TRUEs and FALSEs, and its mean is the proportion of
its TRUEs.
(b) By contrast, a Welch 2-sample t test gives
reasonably good results when population variances
differ---omitting parameter var.eq=T. [Because the
same seed is used, this simulation uses exactly
the same samples as the previous one.]
set.seed(2022)
pv = replicate(10^5, t.test(rnorm(5, 50, 5), 
                        rnorm(40, 50, 1))$p.val)
mean(pv <= 0.05)  # aprx signif. level.
[1] 0.0512        #  Nearly 5%

