What is the limiting marginal distribution of a fixed number of coordinates of a random point drawn uniformly on large-dimensional sphere?

Question

Let $X=(X_1,\ldots,X_d)$ be uniformly-distributed on the sphere of radius $\sqrt{d}$ in $\mathbb R^d$. It is well-known that in the limit $d \to \infty$, the marginal distribution of $X_1$ converges weakly to the standard normal distribution $N(0,1)$.

Question. Let $k \ge 2$ be a fixed integer. Is it true that marginal distribution of $(X_1,\ldots,X_k)$ converges (say weakly) to $N(0,I_k)$ in the limit $d \to \infty $?

Note. From this post https://mathoverflow.net/a/359747/78539, we know that the marginal distribution of $(X_1/\sqrt{d},\ldots,X_k/\sqrt{d})$ has density $$ p_k(x_1,\ldots,x_k) \propto \begin{cases}(1-\sum_{j=1}^k x_j^2)^{(d-k)/2},&\mbox{ if }\sum_{j=1}^k x_j^2 < 1,\\ 0,&\mbox{ else.} \end{cases} $$

Answer 1

A simple way to prove this is to note that the distribution of $(X_1,\dots,X_d)$ is the distribution of $$\frac{(G_1,\dots,G_d)}{\sqrt{G_1^2+\dots+G_d^2}}\sqrt d,$$ where the $G_i$'s are independent standard normal random variables. So, for a fixed $k$ and $d\ge k$, the distribution of $(X_1,\dots,X_k)$ is the distribution of $$(G_1,\dots,G_k)\frac{\sqrt d}{\sqrt{G_1^2+\dots+G_d^2}}\to(G_1,\dots,G_k)\sim N(0,I_k)$$ in probability as $d\to\infty$, by the law of large numbers and Slutsky's theorem.

Answer 2

You can deduce the result if it is already known that $X_1\to^d N(0,1)$.

The convergence in law of $(X_1,...,X_k)$ boils down, for fixed reals $t_1,...,t_k$, to the convergence of the characteristic function $E[\exp(i\sum_j t_j X_j)]$ to $\exp(-\|t\|^2/2)$. For each $d$, pick a rotation $P\in O(d)$ such that the first row is $(t_1,...,t_k,0,0,...,0)/\|t\|$. Then $Y=PX$ is also uniformly distributed on the sphere, hence $Y_1=(PX)_1 \to^d N(0,1)$ by what is already known, and consequently $E[\exp(i\sum_j t_j X_j)]=E[\exp(i \|t\| Y_1)]$ converges to $\exp(-\|t\|^2/2)$.