# Can we use chi-square distribution and central limit theorem to find the approximate normal distribution?

If $$X_1,\ldots,X_i,\ldots,X_n$$ are same normal distribution, $$X_i \sim \operatorname{Normal}(0,σ^2)$$, and they are independent. $$Z = \frac{\sum_{i=1}^n X_i^2} n.$$

What is the distribution of the square of the normal distribution?like $$X_i^2$$,and,what is it mean and variance?

I am trying to turn this Z into a normal distribution

can we use chi-square distribution and central limit theorem to find the approximate normal distribution ?

How to do it？

I do not quite understand the chi-square distribution and central limit theorem, could you answer this question in detail?

Any help would be much appreciated!

## re-edit：

I do this works: $$Z = \frac{\sum_{i=1}^n X_i^2} n= σ^2\sum_{i=1}^n \left(\frac{X_i}{σ}\right)^2.$$ this is a chi-square distribution,and mean $$= nσ^2$$, var$${}=2nσ^2$$.

is this right?

and how to use CLT to find the approximate normal distribution?

• Could you show us what you've tried? Also are those $X_i$ independent? – Yujie Zha Jul 27 '17 at 4:07
• thanks,and sorry,I wrote it wrongly,Z should be$Z = \frac{\sum_{i=1}^n |X_i|^2} n$.,I have re-edited the question. – ysbbs Jul 27 '17 at 5:08
• "$Z\sim N\left(σ^2,\frac{2σ^2} n \right)\text{ ?}$(Someone gave me this answer, but did not explain)" Well, then "someone" was very wrong. – Did Jul 27 '17 at 9:08
• @Did,Can you help me explain the correct answer?any help would be much appreciated! – ysbbs Jul 27 '17 at 9:26
• This post would be helpful: math.stackexchange.com/questions/260391/… – Jung-Hyub Lee Jul 27 '17 at 9:35

The binomial distribution can not reach negative value but can still be approximated by normal approximation, so I don't think Murthy's answer is persuasive enough. And actually, $$(\frac{X_{i}}{\sigma })^{2}\sim \chi ^{2}(1), with\:finite\: mean\: 1\: and\: finite\:variance\: 2; since\:Xi\sim i.i.d,\\ (\frac{X_{i}}{\sigma })^{2}\sim i.i.d,\\by\:Central\:Limit\:Theorem,\\ \sqrt{n}(\frac{\sum_{i=1}^{n}((\frac{X_{i}}{\sigma })^{2}-1)}{n})=\sqrt{n}(\frac{\sum_{i=1}^{n}(\frac{X_{i}}{\sigma })^{2}}{n}-1) \overset{d}{\rightarrow} N(0,2)\:as\: n \rightarrow +\infty \\$$