My understanding of a probability distribution with compact support, means that the set of valid inputs to query the probability for much be compact.

And that in this basically mean that that that set must be bounded, and closed.

The support for a $n$ dimensional Gaussian distribution is all of $\mathbb{R}^n$. It is not bounded, as far as I can tell -- you give me any point, and I can find a point that is further from the origin. (Or more generally, give me an open ball and I can find an open ball that encloses it). And (importantly perhaps) all mentioned points (in the open balls) are all points that have a chance that they could be sampled from a Gausssian distribution -- so they are actually in the support

Thus the Gaussian distribution does not have a compact support.

Is my reasoning correct?

I ask because I am looking at a paper describing a non-parametric estimator, that only works for estimating distributions with a compact support, and their first example is estimating a Gaussian.

  • $\begingroup$ You do not need this complicated reasoning. The pdf of a 2D gaussian for example is non zero on the whole of $\mathbb{R^2}$ because $exp(-(ax^2+by^2+2cxy))>0$ for all $x,y$. $\endgroup$
    – Jean Marie
    Sep 5, 2017 at 8:22

1 Answer 1


You are right in confirming that Gaussian do not have a compact support.

With regards to your paper, it is not rare in statistical application to find cases where methods are applied outside their domain of rigorous validity. This does not of course necessarily yield incorrect, from the practical point of view, results.

Imagine to take a Gaussian, and truncate it after some threshold. What would the mistake be, induced by such operation? Under suitable conditions, it might well be fully negligible.

  • $\begingroup$ Yes, indeed. I believe they are effectively bounding it about 7 std deviations from its mean. Where the mistake would indeed be negligible. $\endgroup$ Sep 5, 2017 at 8:27

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