# an exercise related central limit theorem

I'm working on the following problem in Durrett:

Let $X_1, X_2, ...$ be i.i.d, nonnegative, $EX_i=1$ and $Var(X_i)=\sigma ^2$. Then we have $2(\sqrt{S_n}-\sqrt{n})$ converge to $\sigma \chi$ in distribution, where $S_n=\Sigma_{i=1}^{n}X_i$.

I was thinking about using central limit theorem for this but it seems like some kind of transform is needed. And it has to connect the original $S_n$ with $\sqrt{S_n}$, while I don't know how. Could I ask for a hint? Thanks a lot.

Hint: $$2(\sqrt{S_n}-\sqrt{n})=\frac{S_n-n}{\sqrt{n}}+o_p(1)$$

First, $\frac{S_n}{n}\rightarrow^p 1$. Using the mean value theorem (expanding $\sqrt{S_n/n}$ around 1)

$$2\sqrt{n}\left(\sqrt{S_n/n}-1\right)=2\sqrt{n}\left(1+\frac{1}{2\sqrt{V_n}}(\frac{S_n}{n}-1)-1\right)$$ $$=\frac{1}{1+o_p(1)}\frac{S_n-n}{\sqrt{n}}$$

where $V_n$ lies between $S_n/n$ and $1$.

• Excuse me, I've never seen this before. Could you tell me more about why $S_n/n$ converges to 1 in probability implies the first equality you wrote? And what is the relation between $S_n'$ and $S_n$? – guest625 Jul 1 '15 at 0:06
• I expanded $\sqrt{S_n}$ around $\sqrt{n}$. However, you can make more rigorous by considering the difference $\left| 2(\sqrt{S_n}-\sqrt{n})-\frac{S_n-n}{\sqrt{n}} \right|$ and showing that it converges to 0 in probability. – d.k.o. Jul 1 '15 at 0:12
• Yes, you're right. Thank you for your help. I got it now. – guest625 Jul 1 '15 at 0:22
• Yes, this would work too. Thanks for this other method. (I think you missed two square root symbols within the integral.) – guest625 Jul 1 '15 at 0:37
• Oh thank's. There should be square roots: $$\left| 2(\sqrt{S_n}-\sqrt{n})-\frac{S_n-n}{\sqrt{n}} \right|=\left|\int_n^{S_n}\frac{1}{\sqrt{z}}-\frac{1}{\sqrt{n}}dz\right|$$ – d.k.o. Jul 1 '15 at 2:37