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I found an interesting problem on the internet:
$\lim_{n\to \infty} \frac{1}{n}\int_0^n \max(\{x\},\{\sqrt{2}x\},\{\sqrt{3}x\})dx$, where $\{x\}$ is the fractional part of $x$.

I have known that this problem can be regarded as calculating the expectation of the maximum of three independent uniform random variables $X_i \overset{i.i.d}{\sim} \mathrm{Unif}(0,1)$. However, I do not understand the reasoning behind why this is the case.

amWhy
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Zhou_Key_Error
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1 Answers1

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Step 1. Note that for any integers $p, q$ with $(p, q) \neq (0, 0)$, the number $\alpha = p\sqrt{2} + q\sqrt{3}$ is irrational. Hence,

\begin{align*} \frac{1}{N} \sum_{n=0}^{N-1} e^{2\pi i (p\sqrt{2}(x+n) + q\sqrt{3}(x+n))} &= \frac{e^{2\pi i \alpha x}}{N} \cdot \frac{1 - e^{2\pi i \alpha N}}{1 - e^{2\pi i \alpha}} \to 0 \end{align*}

as $N \to \infty$. Then by the multi-dimensional Weyl’s Criterion, the empirical measure

$$ \mu_{x,N}(\mathrm{d}x, \mathrm{d}y) = \frac{1}{N} \sum_{n=0}^{N-1} \delta_{(\{ \sqrt{2}(x+n) \}, \{ \sqrt{3}(x+n) \})} $$

converges in distribution to the uniform distribution on $[0, 1]^2$.

Step 2. Now we return to OP's integral. Simplifying,

\begin{align*} &\frac{1}{N} \int_{0}^{N} \max(\{x\}, \{\sqrt{2}x\}, \{\sqrt{3}x\}) \, \mathrm{d}x \\ &= \frac{1}{N} \sum_{n=0}^{N-1} \int_{0}^{1} \max(x, \{\sqrt{2}(x+n)\}, \{\sqrt{3}(x+n)\}) \, \mathrm{d}x \\ &= \int_{0}^{1} \biggl( \int_{[0,1]^2} \max(x, y, z) \, \mu_{x,N}(\mathrm{d}y, \mathrm{d}z) \biggr) \, \mathrm{d}x. \end{align*}

By Step 1, for each $x \in [0, 1]$ we know that

$$ \int_{[0,1]^2} \max(x, y, z) \, \mu_{x,N}(\mathrm{d}y, \mathrm{d}z) \to \int_{[0,1]^2} \max(x, y, z) \, \mathrm{d}y\mathrm{d}z $$

as $N \to \infty$. So by the dominated convergence theorem,

\begin{align*} &\frac{1}{N} \int_{0}^{N} \max(\{x\}, \{\sqrt{2}x\}, \{\sqrt{3}x\}) \, \mathrm{d}x \\ &\to \int_{0}^{1} \biggl( \int_{[0,1]^2} \max(x, y, z) \, \mathrm{d}y\mathrm{d}z \biggr) \, \mathrm{d}x \\ &= \int_{[0,1]^3} \max(x, y, z) \, \mathrm{d}x\mathrm{d}y\mathrm{d}z. \end{align*}

This is equal to $\mathbb{E}[\max\{U_1, U_2, U_3\}]$ for IID uniform random variables $U_1, U_2, U_3$ on $[0, 1]$.

Sangchul Lee
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  • Does this approach also give us an idea about the limiting value? @Sangchul Lee – Sam May 19 '24 at 23:01
  • Is it $\frac{3}{4}$, if use steps similar to this? – Sam May 19 '24 at 23:05
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    @Sam, The resulting integral representation indeed helps compute the limiting value. If $I$ denotes the integral, then by symmetry, $$I=6\iiint\limits_{0<x<y<z<1} z,\mathrm{d}x\mathrm{d}y\mathrm{d}z=6\iint\limits_{0<y<z<1} yz,\mathrm{d}x\mathrm{d}y\mathrm{d}z=3\int\limits_{0<z<1} z^3,\mathrm{d}x\mathrm{d}y\mathrm{d}z=\frac{3}{4}.$$ Of course, the method illustrated in the link is much simpler and more general. – Sangchul Lee May 20 '24 at 01:07