How could Lilliefors use Monte Carlo if the estimand is not distribution-free?

Question

Lilliefors test is a well-known statistical test for normality. Its idea is based on the Kolmogorov-Smirnov test, except the CDF is replaced by the CDF of the normal distribution with $\mu, \sigma^2$ chosen as the estimated ones from the data.

In other words given data $\{X_1, \ldots, X_n\}$ we form its empirical CDF $F_n$, and consider the estimand $$E := \sup_x \left|F_n(x) - \text{cdf}_{N(\hat{\mu}, \hat{\sigma^2})}(x)\right|.$$ The idea is that if the data does come from a normal distribution, then $E$ must be small. In [1], Lilliefors claimed to use Monto Carlo method to compute numbers that defines the smallness.

Question

Since $E$ is not distribution-free, shouldn't have Lilliefors done a Monto Carlo numerical estimate for all kinds of $\mu$ and $\sigma^2$? Why would a single table be enough?

Reference

• [1] On the Kolmogorov-Smirnov Test for Normality with Mean and Variance Unknown-[Hubert W. Lilliefors]
• [2] The Probability Integral Transformation When Parameters are Estimated from the Sample-[F. N. David, N. L. Johnson]

Related

• How can one compute Lilliefors' test for arbitrary distributions?

Answer 1

shouldn't have Lilliefors done a Monto Carlo numerical estimate for all kinds of $\mu$ and $\sigma^2$?

If the $X_i$ are iid Gaussian variables, then $E$ is independent from distribution parameters $\sigma$ and $\mu$, but depends only on the sample size $n$.

Consider generating samples from a distribution $N(\mu,\sigma^2)$ by first generating samples from $N(0,1)$ then multiplying by $\sigma$ and adding $\mu$. That multiplication and addition will make the estimates $\hat{\mu}$ and $\hat\sigma$ shifted and scaled with similar factors. The two distributions $F_n(x)$ and $\text{cdf}_{N(\hat{\mu}, \hat{\sigma^2})}(x)$ will be shifted and stretched in the same way and the parameters $\mu$ and $\sigma$ have no effect.

An example is in the following image:

The image above is a sample of size $n = 20$ generated with $\mu = 0$ and $\sigma = 1$, the sample below is a sample of size $n = 20$ generated with $\mu = 1$ and $\sigma = 0.5$ (effectively the same samples with just a different position and scale).

The only difference is that the curves are shifted and squeezed in the horizontal direction, but because this happens in the same way for the empirical distribution function and the estimated cumulative distribution function, the distance in the vertical direction remains the same.

Answer 2

It might not be distribution-free but for a given sample size it is location and scale free.

The internally standardized joint distribution of the observations does depend on the form of the distribution, but at the normal it does not depend on the specific values of $\mu$ and $\sigma$.

Specifically, for i.i.d. observations from a normal distribution, the vector of internally standardized values is a pivotal quantity; its joint distribution doesn't depend on any unknown parameters.

Even if you didn't wish to/know how to show it algebraically, it's easy enough to look at simulation and see that it doesn't matter what $\mu$ and $\sigma$ are.

As such, the distribution of the test statistic, which depends on those standardized values in turn isn't affected by what values $\mu$ and $\sigma$ take.

Consequently he could just simulate standard normals. I don't know that he did - it's a long time since I read the paper - but it would make sense. Nothing would change if he used any other normal, though, he could please himself.

Similar comments apply to the other two cases Lilliefors looked at in that paper (i.e. just $\mu$ or just $\sigma$ unknown) and to the exponential cases that were covered in his other paper.