Suppose we have a sequence of independent standard normal r.v. $(X_i)_{i=1}^n$. Also, let $S_r$ be an $n$-dimensional ball of radius $r$ centered at zero. Then why does
\begin{align} \lim_{n \to \infty} P[ (X_i)_{i=1}^n \in S_r]=0 \end{align} for any $r \ge 0$ ?
Here is an argument for the limit. Let's calculate $P[ (X_i)_{i=1}^n \in S_r]$ for any finite $n$. Using result from this question we have that \begin{align} P[ (X_i)_{i=1}^n \in S_r]= \frac{ \Gamma( \frac{n}{2})- \Gamma( \frac{n}{2},r^2) }{\Gamma( \frac{n}{2})} \end{align} where $\Gamma( \frac{n}{2},r^2) $ is an incomple Gamma function. It is not difficult to check that the above expression goes to one as $n \to \infty$.
My questions:
1) From finite dimensional cases, in particular for $n=1$, we know that Gaussian is well centered around the origin. Why is thin not the case for $n \to \infty$?
2) I would like to see more intuition behind this conclusion. Is it saying that the probability that at least one coordinate would be outside of a sphere of any radius is 1.
It may become easier if we enlarge the sphere to a cube $Q_r$of side length $2r$. Then each $X_i$ individually has a positive probability $p$ of not falling into $[-r,r]$, from which $$P[(X_i)_{i=1}^n\in S_r]\le P[(X_i)_{i=1}^n\in Q_r]=p^n $$ and this clearly $\to 0$ as $n\to\infty$.