Get inspired by the success stories of our students in IIT JAM 2021. Learn More

# ISI MSTAT PSB 2011 Problem 4 | Digging deep into Multivariate Normal

This is an interesting problem from ISI MSTAT PSB 2011 Problem 4 that tests the student's knowledge of how he visualizes the normal distribution in higher dimensions.

## The Problem: ISI MSTAT PSB 2011 Problem 4

Suppose that $X_1,X_2,...$ are independent and identically distributed $d$ dimensional normal random vectors. Consider a fixed $x_0 \in \mathbb{R}^d$ and for $i=1,2,...,$ define $D_i = \| X_i - x_0 \|$, the Euclidean distance between $X_i$ and $x_0$. Show that for every $\epsilon > 0$, $P[\min_{1 \le i \le n} D_i > \epsilon] \rightarrow 0$ as $n \rightarrow \infty$

## Prerequisites:

1. Finding the distribution of the minimum order statistic
2. Multivariate Gaussian properties

## Solution:

First of all, see that $P(\min_{1 \le i \le n} D_i > \epsilon)=P(D_i > \epsilon)^n$ (Verify yourself!)

But, apparently we are more interested in the event $\{D_i < \epsilon \}$.

Let me elaborate why this makes sense!

Let $\phi$ denote the $d$ dimensional Gaussian density, and let $B(x_0, \epsilon)$ be the Euclidean ball around $x_0$ of radius $\epsilon$ . Note that $\{D_i < \epsilon\}$ is the event that the gaussian $X_i$ will land in this Euclidean 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 as Gaussians decay as we move away from the centre, this infimum exists and is given by $\phi(x_0 + \epsilon \frac{x_0}{||x_0||})$ . (To see that this is indeed a lower bound, note that $B(x_0, \epsilon) \subset B(0, \epsilon + ||x_0||)$.

So, basically what we have shown here is that exists a $\delta > 0$ such that $P(D_i < \epsilon )>\delta$.

As, $\delta$ is a lower bound of a probability , hence it a fraction strictly below 1.

Thus, we have $\lim_{n \rightarrow \infty} P(D_i > \epsilon)^n \leq \lim_{n \rightarrow \infty} (1-\delta)^n = 0$.

Hence we are done.

## Food for thought:

There is a fantastic amount of statistical literature on the equi-density contours of a multivariate Gaussian distribution .

Try to visualize them for non singular and a singular Gaussian distribution separately. They are covered extensively in the books of Kotz and Anderson. Do give it a read!

# Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy