how to get chi square distribution to N(0,1) [closed]

Let $$X_1,X_2,...,X_n$$ be a random sample from chi square distribution. Let $$x̄$$ be the sample mean. How would you use the central limit theorem to get this approximation:

$$\frac{\left(x̄-1\right)}{\sqrt{\frac{2}{n}}}$$ ~ $$Normal(0,1)$$

also how would one derive a normal approximation for $$X=\sum _{i=1}^n X_i$$

closed as off-topic by StubbornAtom, Leucippus, mrtaurho, NCh, ShaileshFeb 18 at 9:05

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• Your first question is fine, but your second question doesn't make any sense. What is $X$? – parsiad Feb 17 at 21:14
• sorry ill edit it – user520403 Feb 17 at 21:18
• By chi squared you must have more specifically meant $\chi_1^2$. Do you know the mean & variance of the $\chi_1^2$ distribution? – J.G. Feb 17 at 22:00

CLT

One version of the central limit theorem says that if $$X_{1},X_{2},\ldots$$ are i.i.d. with the same mean $$\mu$$ and finite variance $$\sigma^{2}$$, then $$\frac{\sqrt{n}}{\sigma}\left(\overline{X}_{n}-\mu\right)\xrightarrow{\mathcal{D}}\mathcal{N}(0,1) \qquad\text{as }n\rightarrow\infty$$ where $$\xrightarrow{\mathcal{D}}$$ denotes convergence in distribution and $$\overline{X}_{n}=(X_{1}+\cdots+X_{n})/n$$. In this case, convergence in distribution means that the left-hand side's cumulative distribution function converges to the cumulative distribution function of $$\mathcal{N}(0,1)$$.

The above suggests that when $$n$$ is large, $$\frac{\sqrt{n}}{\sigma}\left(\overline{X}_{n}-\mu\right)$$ can be approximated by a standard normal random variable. Stated equivalently, when $$n$$ is large, we can approximate $$\overline{X}_n$$ by $$\mathcal{N}\left(\mu,\frac{\sigma^{2}}{n}\right).$$

In your first question, $$X_i \sim \chi_1^2$$ where $$\chi_1^2$$ is a chi-squared distribution with parameter $$k=1$$ (this just means that $$X_i$$ is a squared standard normal random variable). In this case, $$\mu \equiv \mathbb{E}X_{i}=1$$ and $$\sigma^2 \equiv \operatorname{Var}X_{i}=2$$.
Fixing $$n$$ large and letting $$\overline{X} \equiv \overline{X}_n$$ to match your notation, the previous section tells us that we can approximate $$\sqrt{\frac{n}{2}}\left(\overline{X}-1\right)$$ by a standard normal random variable. Equivalently, we can approximate $$\overline{X}$$ by $$\mathcal{N}\left(1,\frac{2}{n}\right).$$
Lastly, you ask how we can approximate $$X \equiv X_1 + \cdots + X_n$$. Note that this is just the sample mean multiplied by $$n$$ (i.e., $$X = n\overline{X}$$). It stands to reason that you can approximate this by $$\mathcal{N}(n, 2n)$$.