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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" ?

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  • $\begingroup$ 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). $\endgroup$
    – Eric Towers
    Apr 20, 2021 at 20:53
  • $\begingroup$ 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. $\endgroup$
    – Mathieu Krisztian
    Apr 20, 2021 at 20:55
  • $\begingroup$ 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 $\endgroup$
    – Mathieu Krisztian
    Apr 20, 2021 at 20:57
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    $\begingroup$ 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. $\endgroup$
    – Eric Towers
    Apr 20, 2021 at 20:58
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    $\begingroup$ 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. $\endgroup$
    – Eric Towers
    Apr 20, 2021 at 21:02

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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 $1$).

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  • $\begingroup$ This is very interesting. Although, I'm not sure, academically speaking, that the Dirac distribution could be classified as a gaussian. Let's see other possible answers. I believe that the central limit theorem expresses somewhere that I don't have the right to use the Dirac, because it is not a "function", or something like this. $\endgroup$
    – Mathieu Krisztian
    Apr 20, 2021 at 20:52
  • $\begingroup$ 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 $\endgroup$
    – Mathieu Krisztian
    Apr 20, 2021 at 20:57
  • $\begingroup$ @MathieuKrisztian "Academically speaking" a constant r.v. is a degenerate case of a normal r.v., the same way three points in a line are a degenerate case of a triangle. Just like three points in a line would be a triangle with zero area, a constant r.v. would be a normal r.v. with zero variance. $\endgroup$
    – Rivers McForge
    Apr 20, 2021 at 22:47
  • $\begingroup$ @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. $\endgroup$
    – Rivers McForge
    Apr 20, 2021 at 22:51

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