Find the limit joint distribution of the random vector uniformly distributed in the ball

Question

Given an integer $n \geq 1$, define $\left(X_{1}^{(n)}, X_{2}^{(n)}, \ldots, X_{n}^{(n)}\right)$ as a random vector uniformly distributed in the ball $$ \left(X_{1}^{(n)}\right)^{2}+\left(X_{2}^{(n)}\right)^{2}+\cdots+\left(X_{n}^{(n)}\right)^{2} \leq n $$ Find the limit joint distribution of the random vector $\left(X_{1}^{(n)}, X_{2}^{(n)}, X_{3}^{(n)}\right)$ as $n \rightarrow \infty$.
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Answer 1

If $(X_1^{(n)},\ldots,X_n^{(n)})$ is uniformly distributed in $\Big\{\vec{x}\in\mathbb{R}^n:||\vec{x}||^2< n\Big\}$ then the pdf of $(X_1^{(n)},\ldots,X_n^{(n)})$ is the function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ defined by $$f(x_1,\dots,x_n)=\frac{\Gamma\Big(\frac{n}{2}+1\Big)}{(n\pi)^{n/2}}$$ whenever $x_1^2+\dots+x_n^2 < n$ and $f(x_1,\dots, x_n)=0$ elsewhere. We find the joint pdf of $(X_1^{(n)},X_2^{(n)},X_3^{(n)})$ to be $$f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z)=\int_{\mathbb{R}^{n-3}}f(x,y,z,x_4,\dots, x_n)dx_4\ldots dx_n$$ If $x^2+y^2+z^2\geq n$ then $f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z)=0$ so assume $x^2+y^2+z^2 < n$ for the remainder of this post. We get $$f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z)=\frac{\Gamma\Big(\frac{n}{2}+1\Big)}{(n\pi)^{n/2}} \int_{\{x_4^2+\dots +x_n^2 < n-x^2-y^2-z^2\}}dx_4 \dots dx_n$$ Using the fact that a ball of radius $R$ in $\mathbb{R}^{k}$ is $\frac{\pi^{k/2}}{\Gamma\Big(\frac{k}{2}+1\Big)}R^k$ we have $f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z)$ equals the following expression $$\frac{\Gamma\Big(\frac{n}{2}+1\Big)}{\Gamma\Big(\frac{n-3}{2}+1\Big)}\Bigg(1-\frac{x^2+y^2+z^2}{n}\Bigg)^{n/2} \frac{1}{\Big(\pi(n-x^2-y^2-z^2)\Big)^{3/2}}$$ Using Stirling's approximation you will quickly see $$ f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z) \rightarrow \frac{1}{(2\pi)^{3/2}}e^{-\frac{1}{2}(x^2+y^2+z^2)}$$ as $n \rightarrow \infty$ i.e. the limit joint distribution of $(X_1^{(n)},X_2^{(n)},X_3^{(n)})$ is $N\Big(\vec{0},I_{3\times 3}\Big)$.