# Testing for membership in a distribution with one single value

I came across an interesting little quirk when I was coding up a hypothesis test in Python.

I was wondering why very time I tested some values for normalcy using the Anderson–Darling test, I was getting a positive.

Turns out my loop was just testing over one single value (oops)! But then I got to thinking, I wonder if this is generally the case -- can a single real-valued number be shown to belong to every distribution that has real-valued numbers as it's support? Is there a trivial result that states this somewhere in the statistical literature?

• What do you mean by "getting a positive"? If you simply mean "failed to reject the null hypothesis", then you're gravely mistaken about what it implies. It certainly doesn't mean that the sample is "shown to belong" to the distribution being tested. Commented May 10, 2014 at 9:12
• Are you estimating parameters, or is this for fully specified distributions? Commented May 10, 2014 at 9:19
• No, a single real-valued number cannot "belong to every distribution that has real-valued numbers as it's support". For example, the continous uniform distribution on $[0,1]$ has real valued numbers as its support, though not all real valued numbers, but $x = 2$ cannot be a realization of this uniform distribution. For a number to be a (potential) realisation from a distribution it needs to belong to its support. Commented May 10, 2014 at 11:12

I see three* main possibilities:

1) You're dealing with fully specified distributions.

In this case, your assertion is plainly false, since a value like 3 is immediately rejected at typical significance levels

> ADGofTest::ad.test(3,pnorm)

Anderson-Darling GoF Test

data:  3  and  pnorm
AD = 5.6091, p-value = 0.002753
alternative hypothesis: NA


2) you're fitting a mean, but specifying a standard deviation

In this case, the estimated mean is the single observation; it's impossible for an observation in the center of the estimated distribution to be "in the tail" of the distribution. The p-value will always be 1.

3) You're estimating both parameters. In that case you have some 'splainin' to do, because you can't estimate two parameters from one data point.

* (I discount the possibility that you're specifying a mean and estimating the variance. You'd have mentioned that specifically.)