1

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!

1 Answers1

4

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 $\Phi\left(\sqrt{15}x\right)$.

Samuel Barkes
  • 70
Reinier
  • 361
  • 1
    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 $[-1,1]^n$. – Will Orrick Dec 28 '20 at 04:47