# Convergence of uniformly distributed random variables on a sphere

I am reading "Asymptotic Statistics" by A.W van der Vaart and I am stuck with an exercise of chapter 2. Here is the question : for each $$n \in \mathbb{N}$$, let $$U_n$$ be uniformly distributed on the unit sphere $$S^{n-1} \subseteq \mathbb{R}^n$$. Show that the random vectors $$\sqrt{n}(U_{n,1},U_{n,2})$$ converge in distribution to a pair of independent standard normal variables.

Maybe the solution is extremely stupid but I don't know where to start. Could you provide me some hint ?

Also, I am sorry if the solution is already available on the internet, I couldn't find it.

• Are $U_{n,1}$ and $U_{n,2}$ the first and second coordinates of $U_n$? Commented May 8, 2021 at 20:55
• @fblundun yes exactly. Sorry for the imprecision Commented May 9, 2021 at 6:14
• Although I wouldn't recommend it -- I like solutions that require the least work possible because they tend to be the most insightful -- it is possible to obtain an explicit formula for the distribution of $(U_{n,1},U_{n,2}):$ see stats.stackexchange.com/a/520811/919. You can then obtain the limit easily.
– whuber
Commented May 11, 2021 at 17:18
• @whuber oh thanks for sharing, appreciate it. I was thinking about polar coordinates but couldn't write the argument. Thanks a lot for your awnser. Commented May 11, 2021 at 18:24

In outline: one approach is to think of generating $$U_n$$ by generating $$n$$ iid standard Normals $$Z_{n,1},\ldots,Z_{n,n}$$ and defining $$U_{n,i}=\frac{Z_{n,i}}{\sqrt{\sum_j Z_{n,j}^2}}$$

As $$n\to\infty$$, the denominator converges to its expected value (eg, by Chebyshev's inequality) and can be treated as a constant. The expected value is a multiple of $$\sqrt{n}$$, so rescaling any finite set of $$U_{n,i}$$ by $$\sqrt{n}$$ will asymptotically give independent Gaussians that are just multiples of the corresponding $$Z_{n,i}$$.

Update: the result is fairly straightforward but the implications are non-intuitive. $$U_{n,1}= O_p(n^{-1/2})$$, for $$U_n$$ uniformly distributed on $$S^n$$, so nearly all of the area of $$S^n$$ is within $$O(n^{-1/2})$$ of the equator for large $$n$$(!!).

• I see ! Thanks a lot for your awnser. Commented May 9, 2021 at 6:20

This answer is essentially similar to @Thomas Lumley's, but hopefully to add more clarity by explicitly justifying some key steps.

Let $$X_n = (X_{n, 1}, \ldots, X_{n, n}) \sim N_n(0, I_{(n)})$$ (i.e., $$X_{n, 1}, \ldots, X_{n, n} \text{ i.i.d.} \sim N(0, 1)$$), then it follows by a property of spherical distribution (see, e.g., Theorem 1.5.6 in Aspects of Multivariate Statistical Theory by Robb J. Muirhead) that $$X_n/\|X_n\| \sim \text{Uniform}(S_{n - 1})$$, hence \begin{align} \sqrt{n}(U_{n, 1}, U_{n, 2}) \overset{d}{=} \frac{\sqrt{n}}{\|X_n\|}(X_{n, 1}, X_{n, 2}) = \frac{1}{\sqrt{\frac{X_{n, 1}^2 + \cdots + X_{n, n}^2}{n}}}(X_{n, 1}, X_{n, 2}). \tag{1} \end{align} By the weak law of large numbers, $$\frac{X_{n, 1}^2 + \cdots + X_{n, n}^2}{n}$$ converges to $$E[Z^2] = 1$$ in probability, where $$Z \sim N(0, 1)$$, whence $$\frac{1}{\sqrt{\frac{X_{n, 1}^2 + \cdots + X_{n, n}^2}{n}}}$$ converges to $$1$$ in probability by the continuous mapping theorem. On the other hand, $$(X_{n, 1}, X_{n, 2}) \sim N_2(0, I_{(2)})$$ for all $$n$$. It thus follows by Slutsky's theorem that
\begin{align} \frac{1}{\sqrt{\frac{X_{n, 1}^2 + \cdots + X_{n, n}^2}{n}}}(X_{n, 1}, X_{n, 2}) \to_d N_2(0, I_{(2)}). \tag{2} \end{align}

Combining $$(1)$$ and $$(2)$$ gives $$\sqrt{n}(U_{n, 1}, U_{n, 2}) \to_d N_2(0, I_{(2)})$$.

• This is indeed very complementary. Thanks also for the reference ! Commented Feb 1, 2023 at 10:58