# convergence to standard normal distribution

Let random variables $(X_n)$ be i.i.d, $\mathbb{E}X_n = a > 0$, $\operatorname{Var}X_n = \sigma^2 > 0$. Let $S_n = X_1+X_2+\cdots+X_n$. Show that the sequence of random variables

$$\eta_n = \frac{2\sqrt{a}}{\sigma}\left(\sqrt{|S_n|} - \sqrt{na}\right)$$

converges by law to standard normal distribution.

I was trying to use CLT, but the absolute value in $S_n$ is quite problematic. I was also trying to look at cumulative distribution function convergence, but also not result. Some hints are greatly appreciated.

• Delta method – A.S. Dec 9 '15 at 18:24
• Can't it be done in another way? I didn't have it yet. – Prold Dec 9 '15 at 18:26
• $\sqrt {x}-\sqrt{y}=\frac {x-y}{\sqrt x+\sqrt y}$ and note that $P(S_n<0)\to 0$ so you can drop the absolute value. – A.S. Dec 9 '15 at 19:14

As suggested by A.S., use $\sqrt {x}-\sqrt{y}=\frac {x-y}{\sqrt x+\sqrt y}$ with $x=\left\lvert S_n\right\rvert$ and $y=na$ to get $$\sqrt{|S_n|} - \sqrt{na} =\frac{|S_n|-na}{\sqrt{|S_n|}+ \sqrt{na}}=\frac{\sqrt n}{\sqrt{|S_n|}+ \sqrt{na}}\frac{|S_n|-na}{\sqrt n}.$$ By the law of large numbers, $S_n/n\to a$ almost surely hence $\frac{\sqrt n}{\sqrt{|S_n|}+ \sqrt{na}}\to (2\sqrt a)^{-1}$. By Slutsky's theorem, we are reduced to prove that $\left( \frac{|S_n|-na}{\sigma\sqrt n}\right)_{n\geqslant 1}$ converges in distribution to a standard normal distribution. To this aim, write $$\frac{|S_n|-na}{\sigma\sqrt n}=\frac{|S_n|-S_n}{\sigma\sqrt n}+\frac{ S_n-na}{\sigma\sqrt n}.$$ It suffices to prove that $\left(\frac{|S_n|-S_n}{\sigma\sqrt n}\right)_{n\geqslant 1}$ converges to zero in probability. Since $S_n/n\to a>0$ almost surely, $|S_n|-S_n\to 0$ almost surely hence in probability.