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Show $$\lim_{n \to\infty} \int_0^1 \cdots \int_0^1 \int_0^1 \frac{ x_1^2 + \cdots + x_n^2}{x_1 + \cdots + x_n} \, dx_1 \cdots dx_n = \frac 2 3.$$

Not sure how to start off this iterated integral question, any help would be appreciated.

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    $\begingroup$ On the assumption that you meant $2/3$ rather than $2/34,$ this amounts to $$ \lim_{n \to\infty} \int_0^1 \cdots \int_0^1 \frac{ x_1^2 + \cdots + x_n^2}{x_1 + \cdots + x_n} \, dx_1 \cdots dx_n = \frac{\int_0^1 \cdots \int_0^1 (x_1^2+\cdots + x_n^2) \,dx_1 \cdots dx_n}{\int_0^1 \cdots \int_0^1 (x_1 + \cdots + x_n) \, dx_1 \cdots dx_n }. $$ $\endgroup$ Mar 27, 2017 at 18:10
  • $\begingroup$ In the comment above, note that the expression on the right side does not actually depend on $n. \qquad$ $\endgroup$ Mar 27, 2017 at 18:19
  • $\begingroup$ My first idea involved traces of $M$ and $M^2$ which are sums of eigenvalues and eigenvalues squared, but then I did not manage to connect it to integration in any elegant easy-to-explain way. Maybe someone else could think of how. $\endgroup$ Apr 15, 2017 at 13:19

4 Answers 4

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Let $X_1,\ldots,X_n$ be independent random variables, each distributed uniformly on the interval $[0,1]$. Your question is then equivalent to $$ \lim_{n\to\infty}\mathbb E\frac{X_1^2+\cdots+X_n^2}{X_1+\cdots+X_n}=\frac{2}{3}. $$

We will deduce this from the Strong Law of Large Numbers and the Bounded Convergence Theorem. Consider an infinite iid sequence $(X_i)_{i=1}^{\infty}$ of uniform $[0,1]$ random variables on a probability space $\Omega$. By the Strong Law of Large Numbers, each of the events $$ \left\{\omega\in\Omega\colon\lim_{n\to\infty}\frac{X_1(\omega)+\cdots+X_n(\omega)}{n}=\frac{1}{2}\right\} $$ and $$\left\{\omega\in\Omega\colon\lim_{n\to\infty}\frac{X_1(\omega)^2+\cdots+X_n(\omega)^2}{n}=\frac{1}{3}\right\} $$ occurs with probability 1. Therefore, it holds with probability 1 that $$ \lim_{n\to\infty}\frac{X_1^2+\cdots+X_n^2}{X_1+\cdots+X_n}=\frac{2}{3}. $$ Taking expectations gives that $$ \mathbb E\lim_{n\to\infty}\frac{X_1^2+\cdots+X_n^2}{X_1+\cdots+X_n}=\frac{2}{3}. $$ Since $X_i^2\leq X_i$ for all $i$, the quantity inside the limit is bounded above by $1$. Thus by the Bounded Convergence Theorem we may interchange the expectation with the limit, and therefore $$ \lim_{n\to\infty}\mathbb E\frac{X_1^2+\cdots+X_n^2}{X_1+\cdots+X_n}=\frac{2}{3}. $$

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Suppose $X_1,X_2,X_3,\ldots$ are independent random variables, each uniformly distributed on the interval $[0,1].$ Then for each value of $n$ we have $\operatorname{E}(X_n^2) = 1/3.$ The weak law of large numbers says $$ \operatorname*{l.i.p.}_{n\to\infty} \frac{X_1^2 + \cdots + X_n^2} n = \frac 1 3 $$ where $\operatorname{l.i.p.}$ means "limit in probability", and that is defined by saying $$ \text{for every } \varepsilon>0\ \lim_{n\to\infty} \Pr\left( \left| \frac{X_1^2+\cdots + X_n^2} n - \frac 1 3 \right| < \varepsilon \right) = 1. $$ Similarly $$ \operatorname*{l.i.p.}_{n\to\infty} \frac{X_1+\cdots + X_n} n = \frac 1 2. $$ In general, $\Pr(A\cap B) \ge \Pr(A) + \Pr(B) - 1.$ Thus \begin{align} & \Pr\left( \left| \frac{X_1^2+\cdots + X_n^2} n - \frac 1 3 \right| < \varepsilon \text{ and } \left| \frac{X_1+\cdots + X_n} n - \frac 1 2 \right| < \varepsilon \right) \\[10pt] \ge {} & \Pr\left( \left| \frac{X_1^2+\cdots + X_n^2} n - \frac 2 3 \right| < \varepsilon\right) + \Pr\left( \left| \frac{X_1+\cdots + X_n} n - \frac 1 2 \right| \right) - 1. \end{align} Next you need to say that if one number is near $1/3$ and another near $1/2$, then the quotient is near $2/3.$

This sketch of an argument leaves a lot of details to be filled in.

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  • $\begingroup$ (+1) Some details need to be filled, but the idea is crystal clear. $\endgroup$ Mar 27, 2017 at 18:48
  • $\begingroup$ The disadvantage of this argument is that it can be understood only by those who know the weak law of large numbers. Here's a sketch of the proof of that law: Show that $\displaystyle \operatorname{var} \frac{X_1 + \cdots + X_n} n = \frac{\operatorname{var} (X_1)} n$ and $\displaystyle \operatorname{E} \frac{X_1 + \cdots + X_n} n = \operatorname{E} (X_1),$ and use Chebyshev's inequality. $\qquad$ $\endgroup$ Mar 27, 2017 at 18:56
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    $\begingroup$ Even if $$\dfrac{X_1^2+\ldots+X_n^2}{X_1+\ldots+X_n} \to \dfrac23$$ in probability, it does not follow that the expectations converges. $\endgroup$
    – NCh
    Mar 28, 2017 at 3:16
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    $\begingroup$ @MichaelHardy Fortunately, for values inside $[0,1]$ the numerator is less then denominator, so the dominated convergence theorem implies the required fact. $\endgroup$
    – NCh
    Apr 12, 2017 at 17:20
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    $\begingroup$ Just a remark on @NCh 's comment, so we may use the $\text{strong}$ law of large numbers to show $\frac{X_1^2+\cdots+X_n^2}{X_1+\cdots+X_n} \to \frac{2}{3}$, and then use dominated convergence theorem to conclude the desired result. $\endgroup$ Apr 13, 2017 at 5:00
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My proof based on calculation of asymptotic on integral. Let us consider the integral at lagre fixed $n$ $$ \int_0^1 \cdots \int_0^1\, dx_1 \cdots dx_n \frac{ x_1^2 + \cdots + x_n^2}{x_1 + \cdots + x_n} = n\int_0^1 \cdots \int_0^1 \, dx_1 \cdots dx_n \frac{ x_1^2}{x_1 + \cdots + x_n} = $$ $$ n\int_0^1 \cdots \int_0^1 \, dx_1 \cdots dx_n \int_0^\infty d\lambda\,\, x_1^2 e^{-\lambda(x_1 + \cdots + x_n)} $$ After integration over $x_2\cdots x_n$ we obtain $$ n\int_0^1 dx_1 x_1^2 \int_0^\infty d\lambda\,\Bigr(\frac{1-e^{-\lambda}}{\lambda}\Bigr)^{n-1} e^{-\lambda x_1} $$ After that we can calculate asymptotic of integral over $\lambda$. We can present integral over $\lambda$ in the following form $$ \int_0^\infty d\lambda e^{-n f(\lambda)} G(\lambda) $$ where $f(\lambda)=\log \lambda-\log(1-e^{-\lambda})$ and $G(\lambda)=\dfrac{\lambda e^{-\lambda x_1}}{1-e^{-\lambda}} $ Since the function $f$ is a monotonically decreasing function we can obtain asymptotic of its integral using integral by parts. $$ \int_0^\infty d\lambda e^{-n f(\lambda)} G(\lambda)=\frac{G(0)}{n f'(0)}+o\Bigr(\frac{1}{n}\Bigr) $$ And we obtain $$ 2\int_0^1 dx_1 x_1^2=\frac{2}{3} $$

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    $\begingroup$ Well done! Up to the last step of the asymptotics, everything is easily believable by a student who has taken calculus. A small thing: Do you perhaps have the wrong sign for $f$, i.e. should it perhaps be $f(\lambda)=\log\lambda-\log(1-e^{-\lambda})$? I will wait a while with the bounty, but unless someone shows something even more elementary, the bounty is yours. Again, well done, and thank you! $\endgroup$
    – mickep
    Apr 15, 2017 at 8:03
  • $\begingroup$ In my point of wiev the simplest proof of it can be done using coment on the question, which can be turn into a rigorous proof. $\endgroup$
    – Peter
    Apr 15, 2017 at 13:43
  • $\begingroup$ The function $\dfrac{1-e^{-\lambda}}{\lambda}=1-\dfrac{\lambda}{2}+\dfrac{\lambda^2}{6}\pm\dots$ will add to the integral only in some interval $[0,B/(n-1)]$ for some $B\gt0$ and with growing n we can write $\left(\dfrac{1-e^{-\lambda}}{\lambda}\right)^{n-1}\to\left(1-\dfrac{\lambda}{2}\right)^{n-1}$ on this interval. Hence $$\int_0^\infty\left(\dfrac{1-e^{-\lambda}}{\lambda}\right)^{n-1}d\lambda\to\int_0^{B/(n-1)}\left(1-\dfrac{\lambda}{2}\right)^{n-1}d\lambda\to\dfrac{1}{n-1}\int_0^B \left(1-\dfrac{t}{2(n-1)}\right)^{n-1}dt\to\dfrac{1}{n-1}\int_0^B e^{-t/2}dt\to\dfrac{2}{n-1}$$ $\endgroup$ Apr 15, 2017 at 14:22
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We follow the idea of @peter, and also prepare for an integration by parts. We try, however, to keep the argument as elementary as possible. $$\begin{aligned} I_n & =\int_{[0,1]^n}\frac{x^2+x_2^2+\cdots+x_n^2}{x+x_2+\cdots+x_n}\,dx\,dx_2\,\ldots\,dx_n\\ & = n\int_0^{+\infty}\int_0^{1}x^2e^{-\lambda x}\,dx \biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^{n-1}\,d\lambda\\ & = n\int_0^{+\infty}\biggl(\frac{2}{\lambda^3}-\frac{e^{-\lambda}(2+2\lambda+\lambda^2)}{\lambda^3}\biggr)\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^{n-1}\,d\lambda\\ & = \int_0^{+\infty}\biggl(\frac{2}{\lambda^3}-\frac{e^{-\lambda}(2+2\lambda+\lambda^2)}{\lambda^3}\biggr) \biggl(\frac{e^{-\lambda}}{\lambda}-\frac{1-e^{-\lambda}}{\lambda^2}\biggr)^{-1} \frac{d}{d\lambda}\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n\,d\lambda\\ \end{aligned} $$ It happens that $$ \biggl(\frac{2}{\lambda^3}-\frac{e^{-\lambda}(2+2\lambda+\lambda^2)}{\lambda^3}\biggr) \biggl(\frac{e^{-\lambda}}{\lambda}-\frac{1-e^{-\lambda}}{\lambda^2}\biggr)^{-1} =\frac{\lambda}{e^\lambda-1-\lambda}-\frac{2}{\lambda}. $$ Thus, integrating by parts, $$ \begin{aligned} I_n & =\biggl[\biggl(\frac{\lambda}{e^\lambda-1-\lambda}-\frac{2}{\lambda}\biggr) \biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n\biggr]_0^{+\infty}\\ & \quad- \int_0^{+\infty}\biggl(\frac{2}{\lambda^2}-\frac{\lambda^2}{(e^\lambda-1-\lambda)^2}-\frac{\lambda-1}{e^{\lambda}-1-\lambda}\biggr) \biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n \,d\lambda\\ & =\frac{2}{3}- \int_0^{+\infty}\biggl(\frac{2}{\lambda^2}-\frac{\lambda^2}{(e^\lambda-1-\lambda)^2}-\frac{\lambda-1}{e^{\lambda}-1-\lambda}\biggr) \biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n \,d\lambda. \end{aligned} $$ We need to show that the last integral tends to $0$ as $n\to+\infty$. The absolute value of the first factor is bounded by some constant $C$. Indeed, close to $0$ a Maclaurin expansion shows that it is of size $1/18+O(\lambda)$. Moreover, it is continuous for $0<\lambda<+\infty$ and tends to $0$ as $\lambda\to +\infty$. For the other factor, one can use that $$ 0<\frac{1-e^{-\lambda}}{\lambda}\leq \frac{2}{2+\lambda}\quad\text{for $\lambda>0$.} $$ Thus, $$ \begin{aligned} 0 & < \biggl|\int_0^{+\infty}\biggl(\frac{2}{\lambda^2} -\frac{\lambda^2}{(e^\lambda-1-\lambda)^2}-\frac{\lambda-1}{e^{\lambda}-1-\lambda}\biggr) \biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n\,d\lambda\biggr| \\ & \leq \int_0^{+\infty} M\biggl(\frac{2}{2+\lambda}\biggr)^n\,d\lambda =\frac{2M}{n-1}\to 0\quad\text{as $n\to+\infty$.} \end{aligned} $$ We conclude that $\lim_{n\to+\infty}I_n=2/3$. Of course, Peter will receive the bounty.

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