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This post says

If the $n$ observations in the sample $$\xi_n = [x_1, \dots, x_n]$$ are the realizations of n mutually independent random variables $X_1 \dots X_n$ having the distribution function $F_X(x)$, then

$$F_n(x)\stackrel{L^2}\to F(x)$$

for any $x \in R$

What does $L^2$ on an arrow mean? Does that mean "goes to" under some condition? If yes, what condition does the $L^2$ mean?

mrtaurho
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whnlp
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2 Answers2

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Yes, it means it goes to something, i.e. it converges, and the some condition here is the $L^2$ -norm. $L^2$ is a special case of the so-called $L^p$-spaces. Functions in $L^2$ are sometimes called square integrable functions as for $f\in L^2(\Bbb R)$ we have that

$$\int_{-\infty}^\infty |f(x)|^2\mathrm dx < \infty$$

The $L^2$-norm is defined in a similiar manner, i.e.

$$\|f\|_{L^2(\Bbb R)}:=\int_{-\infty}^\infty |f(x)|^2\mathrm dx$$

mrtaurho
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  • Thanks for your answer. How does the $L^2$ space affect the distribution? Does the distribution live in the $L^2$ space? – whnlp Sep 18 '19 at 15:32
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    @whnlp Yes. The equation given in the post you linked just states that $F_n$ converges to $F$ using the norm induced by $L^2$. Are you familiar with the concept of norms in general? – mrtaurho Sep 18 '19 at 15:34
  • I know only a bit about Euclidean norm ~ I am afraid that I am not familiar with the concept of norms in general – whnlp Sep 18 '19 at 15:39
  • @whnlp No problem at all! When working within real analysis we use (without thinking about it!) the absolute value norm which enables us to, e.g. prove the convergence of a sequence. The aforementioned norm has some properties (the triangle inequalitiy for instance) which we utilize for our proofs. Here, the $L^2$-norm has for example a deep relation to integrals (by its very definition) which can be useful sometimes (e.g. in functional analysis when dealing with Fourier Transforms and similiar). – mrtaurho Sep 18 '19 at 15:52
  • @whnlp So, long story short: a norm is a way of attaching "a value" to an "object" (don't take those terms to literally), but especially a norm induces a metric, that is a way of measuring distances and thus can be used for our good old $\varepsilon$-$\delta$-proof and much more! – mrtaurho Sep 18 '19 at 15:54
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It means convergence in the $L^2$-norm, that is, $$ \| F_n(x)-F(x)\|_{ L^2(\mathbb{R}) }\to 0 $$ as $n\to \infty $.

sam wolfe
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