# Violating the central limit theorem if using Dirac as probabilistic function?

To my (maybe wrong) understanding, the central limit theorem states that, whatever the probabilistic function that we choose, if we add the sum of a random variable that is following a given fixed chosen probabilistic function, it will eventually follow a gaussian distribution.

https://en.wikipedia.org/wiki/Central_limit_theorem

If on purpose, I choose a Dirac at x value=1, as probabilistic function (probability of 1 only for a single value), do I understand that the sum of the random variable will never follow a gaussian but will follow a Dirac distribution at a x value=number of experiments ?

Where on the definition of the Central limit theorem my counter-example is "prevented" ?

• Why do you require that a Dirac distribution have positive variance? Taking the limit as variance goes to zero is at least as valid as asserting the Dirac distribution is a probability density function (most likely obtained as a similar limit). Apr 20, 2021 at 20:53
• I don't understand well what you mean. I believe that Dirac is "formally" not a gaussian function. I believe that in the "semantic" of the central limit theorem, something (that I have missed) is preventing my example. Apr 20, 2021 at 20:55
• For example, the french version of wikipedia states that it tends the most often to a gaussian : they don't write that it always tned to a gaussian : "ce résultat affirme qu'une somme de variables aléatoires identiques et indépendantes tend (le plus souvent) vers une variable aléatoire gaussienne.". Ref : introduction of : fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_central_limite Apr 20, 2021 at 20:57
• Feel free to show what property $\delta$ has that is not reproduced by $\lim_{\sigma \rightarrow 0} N(0,\sigma^2)$. In fact, in some places, the Dirac $\delta$ is defined by this limit. Apr 20, 2021 at 20:58
• Probability density functions are functions. The Dirac delta function is not a function, it is a distribution -- the limit of functions. If you are willing to accept distributions as probability density functions, then you must permit limits of Gaussians as the result of the CLT. Apr 20, 2021 at 21:02

The Dirac distribution is just a normal distribution whose variance is zero, so in this case the CLT still holds (rather trivially, since the sample average and all the r.v.'s are constant with probability $$1$$).
• @MathieuKrisztian A constant r.v. occurs as a degenerate case of other distributions, as well. For instance, a geometric r.v. where $q = 0$ is just a constant r.v. Apr 20, 2021 at 22:51