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, Shailesh Feb 18 at 9:05

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


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 $$\mathcal{N}\left(\mu,\frac{\sigma^{2}}{n}\right).$$

Your questions

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 $\mathcal{N}(n, 2n)$.


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