the degree of freedom of a chi-squared random variable goes to infinity

Question

Hi guys, I've used an calculator to show that this is correct. I am just wondering what is the other way to prove that this statement is true. Thanks!

Answer 1

The chi-squared distribution with $k$ degrees of freedom can be defined by the following sum:

$$\chi^2_k = \sum_1^{k}Z_i^2\;\;\mathrm{where}\; Z \sim N(0,1)$$

We know that $E[Z_i^2]=\mu < \infty$ and $\mathrm{Var}\left[Z_i^2\right] = \sigma^2 < \infty$

Therefore, the RHS is a sum of iid random variables with finite mean and standard deviations. By the standard CLT, we can conclude that

$$\frac{\chi^2_k-k\mu}{\sqrt{k}\sigma} \xrightarrow{d} N(0,1)$$