Show that $P\left ( \left | \sum_{i=1}^{n}X_i \right |\leq 2\sqrt{n} \right )\to 1-2\Phi(-2)$ - solution explanation

Question

Show that $$P\left ( \left | \sum_{i=1}^{n}X_i \right |\leq 2\sqrt{n} \right )\to 1-2\Phi(-2)$$

I have been given that $EX_i=0$ and $Var(X_i)=1$. Now use CLT and thus

$$P\left ( \left| \sum_{i=1}^{n}X_i \right| \leq x\sqrt{n} \right )=P\left ( \left|\frac{1}{n} \sum_{i=1}^{n}X_i-0 \right| \leq x\frac{1}{\sqrt{n}} \right )\to 1-2\Phi(-x)$$

Insert $x=2$ and we are done

Questions:

1. Why does the convergence follow as that? From my understanding we can rewrite the expression inside $P(*)$ and notice that it is the cdf of a normal distribution $\Phi(\frac{x}{\sigma} )$

2. Should the r.v $Z_n=\frac{\bar{X}-\mu}{\sigma /\sqrt{n}}$ not be the one we want to obtain in $P(*)$ to then use CLT (so that $\lim_{n\to\infty}P(Z_n\leq x)=\Phi(x)$?.

I am just unsure on how to approach the problem as I only seem to know how to get it of the form $\Phi(x)$ and not $1-$ some multiple of my CDF

Answer 1

I'll assume you forgot to mention that the $X_i$ are also independent.

The central limit theorem implies that $Z_n := \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$ converges in distribution to a standard normal random variable $Z \sim N(0,1)$. This implies that for any subset $A$, we have $$P(Z_n \in A) \to P(Z \in A).$$

In your example, $A$ is the interval $[-x, x]$, since $P(|Z_n| \le x) = P(-x \le Z_n \le x)$. Thus the limiting probability is $$P(-x \le Z \le x) = \Phi(x) - \Phi(-x) = (1 - \Phi(-x)) - \Phi(-x) = 1 - 2 \Phi(-x),$$ where the equality $\Phi(x) = 1-\Phi(-x)$ follows from symmetry of the normal distribution.