Let $X_1,X_2,...X_n$ be independently distributed $d$ variate normal random variables.Let $D_i$ be the Euclidean distance between $X_i$ and a fixed $x_0 \in \mathbb{R}^d$.(i.e.$||{X_i-x_0}||$) Prove that for any $\epsilon >0$, $ P(\min_{1 \le i \le n} D_i > \epsilon)$ goes to $0$ as $n \rightarrow \infty$.
See that $P(\min_{1 \le i \le n} D_i > \epsilon)=P(D_i > \epsilon)^n$. Now, by Markov's inequality,$P(D_i>\epsilon) \le \frac{E(D_i)}{\epsilon}$ But I cannot give a bound to $E(D_i)$. Can anyone help?
You don't really have to. Let $\phi$ denote the $d$-dimensional Gaussian density, and let $B(x_0, \epsilon)$ be the Euclidian ball around $x_0$ of radius $\epsilon$. Note that $\{D_i < \epsilon\}$ is the event that the gaussian $X_i$ will land in this Euclidian ball, so if we can show that this event has positive probability for any given $x_0, \epsilon$ pair, we will be done, since then in the limit, we will be exponentiating a number strictly less than 1 by a quantity that is growing larger and larger.
In particular, we have that:
$P(D_i < \epsilon)= \int_{B(x_0, \epsilon)} \phi(x) dx \geq |B(x_0, \epsilon)| \inf_{x \in B(x_0, \epsilon)} \phi(x)$ , and we know that by rotational symmetry and monotonicity of Gaussian, this infimum is $\phi(x_0 + \epsilon \frac{x_0}{||x_0||})$. (To see that this is indeed a lower bound, note that $B(x_0, \epsilon) \subset B(0, \epsilon + ||x_0||) $.
You have just shown that $P(D_i < \epsilon) > \delta$ for some $\delta > 0$, and thus you have that $\lim_{n \rightarrow \infty} P(D_i < \epsilon)^n \leq \lim_{n \rightarrow \infty} (1-\delta)^n = 0$.