Let $X_1,X_2, \cdots$ be i.i.d. random variables with mean $\mu$ and variance $1$. Show that $\lim\limits_{n\to\infty} \Bbb P\ (L_n \le \mu \le R_n) = 1 - \alpha\ ,$
where
$$R_n = \frac {1} {n} \sum\limits_{i=1}^{n}X_i - \frac {1} {\sqrt n} \Phi^{-1} \left (\frac {\alpha} {2}\right )\ ,$$
$$ L_n = \frac {1} {n} \sum\limits_{i=1}^{n}X_i + \frac {1} {\sqrt n} \Phi^{-1} \left (\frac {\alpha} {2}\right )\ ,$$
and $\Phi$ is the standard normal C.D.F. It should be noted that since $\alpha<1,$
$$\Phi^{-1} \left (\frac {\alpha} {2} \right ) < 0.$$
I have applied Chebychev's inequality and found that
$$\Bbb P\ (L_n \le \mu \le R_n) \ge 1 - \frac {n} {\left [\Phi^{-1} \left (\frac {\alpha} {2} \right )\right ]^{2}}.$$
Now how do I proceed? Please help me in this regard.
Thank you very much.
You need to rearrange the inequalities and apply CLT (central limit theorem). Namely, $L_n \leq \mu$ is equivalent to $$ \frac{1}{n}\sum\limits_{i=1}^n (X_i - \mu) \leq -\frac{1}{n^{1/2}}\Phi^{-1}(\alpha/2) \Longleftrightarrow \frac{1}{n^{1/2}}\sum\limits_{i=1}^n (X_i - \mu) \leq -\Phi^{-1}(\alpha/2) . $$ Doing a similarly rearrangement for the inequality with $R_n $, and applying the central limit theorem to pass to the limit we get $$ \tag{1} \mathbb{P}(L_n \leq \mu \leq R_n) = \mathbb{P}\left(\Phi^{-1}(\alpha/2) \leq \frac{1}{n^{1/2}} \sum\limits_{i=1}^n (X_i - \mu) \leq - \Phi^{-1}(\alpha/2) \right) \\ \to \Phi( -\Phi^{-1}(\alpha/2) ) -\Phi( \Phi^{-1}(\alpha/2) ) . $$ In view of the definition of $\Phi^{-1}$ we have $\Phi(\Phi^{-1}(\alpha/2)) = \alpha/2$. Using the fact that $\Phi(-a) + \Phi(a) = 1$ when $a\in \mathbb{R}$ it follows that $\Phi(-\Phi^{-1}(\alpha/2)) = 1 - \alpha/2$. Thus the limit in $(1)$ equals $1-\alpha$.