Using central limit theorem to prove most of an n dimensional cube is contained in a given annulus

Question

Let x = (x1, . . . , xn) ∈ $R^n$

Let ||x|| := $\sqrt{(x_1^2 + . . . + x_n^2)}$

Let ε > 0.

Show that for all sufficiently large n, “most” of the cube [−1, 1]$^n$ is contained in the annulus

A := $(x ∈ R^n : (1-\epsilon)\sqrt{n/3}\le||x||\le(1+\epsilon)\sqrt{n/3})$

That is, if X1, . . . , Xn are each independent and identically distributed in [−1, 1], then for n sufficiently large P((X1, . . . , Xn) ∈ A) ≥ 1 − ε.

(Hint: apply the weak law of large numbers to $X_1^2, X_2^2 ...$)

I have absolutely no idea how to approach this. Any help would be greatly appreciated.

Answer 1

Write down what the law of large numbers implies for $X_1^2,X_2^2,\ldots$.

$$\frac{1}{n} \sum_{i=1}^n X_i^2 \overset{p}{\to} \mathbb{E} X_1^2 = \frac{1}{3}.$$

Then note that the condition defining the annulus $A$ can be rewritten as follows. \begin{align} (1-\epsilon)\sqrt{n/3} \le \|X\| \le (1+\epsilon)\sqrt{n/3} \\ (1-\epsilon)^2/3 \le \frac{1}{n} \sum_{i=1}^n X_i^2 \le (1+\epsilon)^2/3 \end{align}