# Showing proportion of volume of n-dimensional cube within distance ${n/3}^{1/2} + x$ converges as $n\to\infty$

For a fixed $x\in\Re$, show that the proportion of the volume of the cube within distance $(n/3)^{(1/2)} + x$ of the origin converges as as $n \to \infty$ and find the limit.

I am having trouble interpreting this question, specifically the geometry of the question. I know that the Central Limit Theorem applies in this question but do understand how to show the convergence. Any help would be very appreciated!

We can view points in the $$n$$ dimensional cube $$[-1,1]^n$$ as $$n$$ dimensional coordinate vectors $$(x_1,x_2,\ldots,x_n)$$ with $$-1 \leq x_i \leq 1$$. Now the distance from a point to the origin is $$\sqrt{x_1^2+x_2^2+\cdots+x_n^2}$$.

We can now take random points inside the cube, the probability that such a point is within distance $$(n/3)^{1/2}+x$$ of the origin equals the desired propotion of the volume.

So let $$X_i$$ be the $$i$$'th coordinate of our random point, now $$X_i$$ is uniformly distributed between $$-1$$ and $$1$$. If we define $$Y_i = X_i^2$$, then we have that the distance from the origin to our point equals $$\sqrt{Y_1+Y_2+\cdots+Y_n}$$. This means that the probability we want to compute equals

$$\mathbb{P}(\sqrt{Y_1+Y_2+\cdots+Y_n} \leq (n/3)^{1/2}+x) = \mathbb{P}(Y_1+Y_2+\cdots+Y_n \leq n/3+2x(n/3)^{1/2}+x^2).$$

We can compute that $$\mathbb{E}(Y_i) = \frac{1}{3}$$ and $$Var(Y_i)=\frac{4}{45}$$, so now we can apply the central limit theorem to the sum of $$Y_i$$ to see that $$\frac{Y_1+Y_2+\cdots+Y_n-n/3}{(4n/45)^{1/2}}$$ converges to a random variable $$Z$$ with a standard normal distribution. This means that we have for the probability that a random point is in the given area that

$$\mathbb{P}(Y_1+Y_2+\cdots+Y_n \leq n/3+2x(n/3)^{1/2}+x^2) = \mathbb{P}\left(\frac{Y_1+Y_2+\cdots+Y_n-n/3}{(4n/45)^{1/2}} \leq \frac{n/3+2x(n/3)^{1/2}+x^2-n/3}{(4n/45)^{1/2}}\right) = \mathbb{P}\left(\frac{Y_1+Y_2+\cdots+Y_n-n/3}{(4n/45)^{1/2}} \leq \sqrt{15}x+\frac{3\sqrt{5}}{2\sqrt{n}}x^2\right) \to \mathbb{P}\left(Z \leq \sqrt{15}x\right) = \Phi\left(\sqrt{15}x\right)$$

So the asked proportion equals $$\Phi\left(\sqrt{15}x\right)$$.

• Worth mentioning some implications: Setting $x=\pm\frac{1}{2}$ and subtracting, one computes $\Phi\left(\frac{1}{2}\sqrt{15}\right)-\Phi\left(-\frac{1}{2}\sqrt{15}\right)\approx0.9472$. So, for large $n$, $94.72\%$ of the volume of the hypercube $[-1,1]^n$ lies within a hyperspherical shell of radius $\sqrt{n/3}$ and thickness $1$ centered at the origin. (This shell is bounded by two $(n-1)$-dimensional surfaces.) The same holds for the hypercube $[0,1]^n$ as the shell centered at the origin has equal-volume intersections with each of the $2^n$ hypercubes of side $1$ that make up $[-1,1]^n$. Commented Dec 28, 2020 at 4:47