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I found this on this website:

Convergence in probability: $\lim_{n\to\infty}\int_0^1\cdots\int_0^1\frac{x_1^2+x_2^2+\cdots+x_n^2}{x_1+x_2+\cdots +x_n}dx_1\cdots dx_n=\frac23$

$\lim_{n\to\infty} \underbrace{\int_{0}^{1}\cdots \int_{0}^{1}}_{n}\frac{x_1^{505}+\cdots +x_n^{505}}{x_1^{2020}+\cdots +x_n^{2020}}dx_1\cdots dx_n$

I try to generalize its conclusions:

I believe:

with $q>p>0$,

$$ \lim_{n\to+\infty}\int_0^1\cdots\int_0^1\frac{x_1^q+x_2^q+\cdots+x_n^q}{x_1^p+x_2^p+\cdots+x_n^p}dx_1dx_2\cdots dx_n=\frac{p+1}{q+1} $$

The answers below this page give perfect proof.

We can continue to promote it.

With $f\in[0,1]$,$g\in[0,1]$,$\exists C>0 ,\forall x\in[0,1]$,$0<f(x)<Cg(x)<\infty$ and $\int_0^1g(x)dx<\infty$

$$ \lim_{n\to+\infty}\int_0^1\cdots\int_0^1\frac{\sum_{i=1}^{n}f(x_i)}{\sum_{i=1}^{n}g(x_i)}dx_1dx_2\cdots dx_n=\frac{\int_0^1f(x)dx}{\int_0^1g(x)dx} $$

Proof of this conclusion I find it here today.

The proof about $\lim_{n\rightarrow\infty}\int _0^1\cdots\int _0^1 \frac{\sum_{i=1}^n f(x_i)}{\sum_{i=1}^n g(x_i)}dx_1\cdots dx_n$

Well, this is over, and you can use this as a navigation transit page.

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1 Answer 1

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The conclusion is indeed true. Let $X_1, X_2, \ldots$ be i.i.d. random variables uniformly distributed over $[0, 1]$. Since $0 < p < q$,

$$ 0 \leq \frac{X_1^q + \cdots + X_n^q}{X_1^p + \cdots + X_n^p} \leq 1 $$

with probability one. Moreover, by the strong law of large numbers (SLLN),

$$ \lim_{n\to\infty} \frac{X_1^q + \cdots + X_n^q}{X_1^p + \cdots + X_n^p} = \frac{\lim_{n\to\infty} \frac{1}{n}(X_1^q + \cdots + X_n^q)}{\lim_{n\to\infty} \frac{1}{n}(X_1^p + \cdots + X_n^p)} = \frac{\mathbf{E}[X_1^q]}{\mathbf{E}[X_1^p]} = \frac{p+1}{q+1} $$

holds with probability one. So by the dominated convergence theorem,

$$ \int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1^q + \cdots + x_n^q}{x_1^p + \cdots + x_n^p} \, \mathrm{d}x_1 \cdots \mathrm{d}x_n = \mathbf{E}\left[ \frac{X_1^q + \cdots + X_n^q}{X_1^p + \cdots + X_n^p} \right] \to \mathbf{E}\left[\frac{\mathbf{E}[X_1^q]}{\mathbf{E}[X_1^p]}\right] = \frac{p+1}{q+1}. $$

The same argument applies to the proposed generalization as well.

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    $\begingroup$ +1 Very nice. ${}$ $\endgroup$
    – copper.hat
    Nov 6, 2023 at 4:47
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    $\begingroup$ I wonder if there is a way to prove the claim without such advanced math… $\endgroup$ Nov 6, 2023 at 4:52

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