I am studying Central Limit Theorem and have seen the equation, $\sqrt{n}(\bar{X}-\mu) \sim N(0, \sigma^2)$, but I am wondering why the $\sqrt{n}$ is multiplied in the equation. At the end of the calculation, the equation becomes $(\bar{X}-\mu) \sim N(0, 1)$. The problem is the lecturer does say why he has put $\sqrt{n}$. Can somebody tell me why I can put the $\sqrt{n}$?


Think about what happens to $\bar X-\mu$ as the sample size $n$ increases. We expect the sample mean to go toward the true mean, right? So if we looked at this quantity, it would converge to zero.

However, if we multiply by $\sqrt{n}$ it turns out that compensates for the above convergence to zero, and the random quantity $\sqrt{n}(\bar X-\mu)$ has a nice standard normal distribution.

The reason the correct quantity is $\sqrt{n}$ is because the variance of $\bar X$ is $$ Var(\bar X) = Var\left(\frac{\sum_i X_i}{n}\right) = \frac{1}{n^2}\sum_iVar(X) = \frac{Var(X)}{n} $$ So that $$ Var(\sqrt{n}(\bar X-\mu)) = nVar(\bar X) = 1. $$

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