# Finding the limiting distribution of $\frac{X_n -n}{\sqrt{2n}}$ [closed]

Let $X_n \sim \chi^2 (n)$ for all $n$. I would like to show when $n\to\infty$

$$\frac{X_n -n}{\sqrt{2n}}\to N(0,1)$$

where $N(0,1)$ is normal distribution. Could someone help me?

## 1 Answer

Let $\left(Z_i\right)_i$ be a sequence of i.i.d. random variable following standard normal distribution, such that $X_n = \sum_{i=1}^nZ_i^2$, then

$\frac{X_n-n}{\sqrt{2n}} = \frac{\left(\sum_{i=1}^nZ_i^2\right)-n}{\sqrt{2n}} = \frac{n^{-1}\left(\sum_{i=1}^nZ_i^2\right)-1}{\sqrt{2/n}} = \sqrt{n}\left[\frac{n^{-1}\left(\sum_{i=1}^nZ_i^2\right)-1}{\sqrt{2}}\right]$.

Since $n^{-1}\left(\sum_{i=1}^nZ_i^2\right)$ can be viewed as the sample mean of $Z_i^2$, which follows a $\chi^2(1)$, it follows that

$\frac{X_n-n}{\sqrt{2n}} =\sqrt{n}\left[\frac{n^{-1}\left(\sum_{i=1}^nZ_i^2\right)-1}{\sqrt{2}}\right] \overset{d}\to N(0,1)$.

• Please avoid just giving complete solutions to obvious homework-style questions. See our help center in relation to homework, which calls for guidance and hints. More direct answers for problems with serious attempts are fine but to all appearances this was just a flat out request for you to do their work for them. Actually giving complete answers deprives people of the opportunity to learn to do these for themselves -- which is the entire point of setting such exercises in the first place! (Any substantive benefit only comes by actually working at it, not by posting it here) Commented Aug 12, 2017 at 21:44