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Limit of a multiple integral

Prove that

    \[\lim_{n \to +\infty} \idotsint \limits_{[0, 1]^n}\frac{x_1^2+x_2^2+ \cdots +x_n^2}{x_1+x_2+ \cdots +x_n} \, \mathrm{d} (x_1, x_2, \dots, x_n) = \frac{2}{3}\]

Solution

Let X_1, X_2,\dots be independent and uniform (0,1) random variables. By the law of large numbers we have

    \[\begin{matrix} \dfrac{X_1+\cdots + X_n}{ n}&\rightarrow & \mathbb{E}(X)= \dfrac{1}{2} \\\\ \dfrac{X_1^2+\cdots + X^2_n}{n}&\rightarrow & \mathbb{E}(X^2)={1\over 3} \end{matrix}\]

in probability as n \rightarrow +\infty. Therefore ,

    \[\frac{X_1^2+\cdots +X^2_n}{ X_1+\cdots +X_n}=\frac{X_1^2+\cdots +X^2_n}{ n}\cdot{n\over X_1+\cdots +X_n}\to \frac{2}{3}\]

in probability as n \rightarrow +\infty.

The ratio random variables {X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n} are bounded below by zero and above by one. This guarantees convergence of the expectations, as well.

So,

    \[\mathbb{E}\left({X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}\right)\rightarrow{2\over 3}\]

which is the required result.

Remark: In general it holds that

\displaystyle \lim_{n\rightarrow+\infty} \int_0^1 \cdots \int_0^1 f\left(\frac{x_1 + \cdots + x_n}{n}\right) \, \mathrm{d}(x_1, \dots, x_n) = \mathbb{E}\left [ f \left ( \frac{X_1+X_2+\cdots+X_n}{n} \right ) \right ]

because the distribution of (X_1,\dots,X_n) is the Lebesgue measure on [0,1]^n hence for every measurable function g,

    \[\mathbb E(g(X_1,\ldots,X_n))=\iint \limits_{[0,1]^n}g(x_1,\ldots,x_n)\mathrm dx_1\ldots\mathrm dx_n\]

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