Joy of Mathematics

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 $g$ ,

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

Let $r_n$ be a random variable that returns one of the digits $2, 0, 1, 6$ each with equal probability for all positive integers $n$ . Find the value of

$\displaystyle \mathcal{V}=\mathbb{E} \left[ \sum_{n=1}^{\infty} \frac{r_n}{10^n} \right]$

where $\mathbb{E}$ denotes the expected value of $x$ .

Solution

Since each $r_n$ are independent and each occurs in only one term, we may interchange the expectation and the summation, thus:

$\begin{align*} \mathbb{E} \left [ \sum_{n=1}^{\infty} \frac{r_n}{10^n} \right ] &=\sum_{n=1}^{\infty} \mathbb{E} \left [ \frac{r_n}{10^n} \right ] \\ &= \sum_{n=1}^{\infty} \frac{\mathbb{E}\left [ r_n \right ]}{10^n}\\ &= \sum_{n=1}^{\infty} \frac{\frac{2+0+6+1}{4}}{10^n}\\ &= \frac{1}{4} \end{align*}$

Tag: Probabilities

Limit of a multiple integral

Expected value of a summation