This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 5 based on central limit theorem . Let’s give it a try !!

**Problem**– ISI MStat PSB 2012 Problem 5

Let \( X_{1}, X_{2}, \ldots, X_{j} \ldots \) be i.i.d. \(N(0,1)\) random variables. Show that for any \(a>0\)

\( \lim {n \rightarrow \infty} P(\sum_{i=1}^{n} {X_i}^2 \leq a) = 0 \)

**Prerequisites**

Limit

Central Limit Theorem

Normal Distribution

Chi- Square Distribution

## Solution :

\( X_{1}, X_{2}, \ldots, X_{j} \ldots \) are i.i.d. \(N(0,1)\) random variables .

Let \( S_n = \sum_{i=1}^{n} {X_i}^2 \) , then \( S_n \sim \chi^{2}(n) \) , where \( \chi^{2}(n) \) is Chi-Square distribution with n degrees of freedom .

Therefore , \( E(S_n)= n \) and \( Var(S_n)=2n \) .

In this type of problems obvious thing that would come to our mind is to apply Central Limit Theorem right ! Let’s try to apply it .

Now by Lindeberg Levy Central Limit Theorem we can say \( \frac{S_n-E(S_n)}{\sqrt{Var(S_n)}} \) = \( \frac{S_n-n}{\sqrt{2n}} {\to }^{d} N(0,1) \) , as n approaches infinity.

So, \( \lim {n \rightarrow \infty} P(\sum_{i=1}^{n} {X_i}^2 \leq a) \)

= \( \lim {n \rightarrow \infty} P( \frac{S_n-n}{\sqrt{2n}} \le \frac{a-n}{\sqrt{2n}} ) \)

= \( \lim {n \rightarrow \infty} \Phi(\frac{a-n}{\sqrt{2n}}) \)

= \( \lim {n \rightarrow \infty} \Phi(\frac{a}{\sqrt{2n}}- \sqrt{\frac{n}{2}}) = \Phi(0- \infty) \) (Since \( \Phi(x) \) is right continuous ) \( = 0 \) .

Hence Proved .

## Food For Thought

Let \( \{X_{1}: i \geq 1 \}\) be a sequence of independent random variables each having a normal distribution with mean 2 and variance 5.Then \( (\frac{1}{n} \sum_{i=1}^{n} x_{i})^{2} \) converges in probability to ?

We can solve this problem using only the Strong Law (SLLN). We do not need the CLT here.

Using Reverse Fatou’s lemma, we have

\[ \limsup \mathbb{P}(\frac{1}{n}\sum_{i=1}^n X_i^2 \leq \frac{a}{n}) \leq \mathbb{P}(\limsup \frac{1}{n} \sum_{i=1}^n X_i^2 \leq 0) = \mathbb{P}(1\leq 0)=0.\]

It’s okay if you want to solve in that way .