# How can I use the Central Limit Theorem to calculate the distribution of $\bar{X}$?

The central limit theorem says that $$\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}} \stackrel{\mathcal{D}}{\rightarrow} N(0,1)$$

What is the distribution of $$\bar{X}$$? I've seen it given as $$\sum X \sim N\left((n)\left(\mu_x\right),(\sqrt{n})\left(\sigma_x\right)\right)$$, but don't know how to prove this.

## 2 Answers

If $$X_i$$ are iid with mean $$\mu_X$$ and variance $$\sigma^2_X$$ then

• $$\sum X_i$$ has mean $$n \mu_X$$ and variance $$n\sigma^2_X$$
• $$\bar X =\frac1n\sum X_i$$ has mean $$\mu_X$$ and variance $$\frac1n\sigma^2_X$$

The central limit theorem gives you a standard normal as the limiting distribution for $$\dfrac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}$$ or equivalently for $$\dfrac{\sum X_i-n\mu}{\sqrt{n}\sigma}$$. This does not mean they have that normal distribution for any particular finite $$n$$ but instead that their distribution is "close" to normal for large enough $$n$$, with "enough" depending on the original distribution of $$X_i$$.

If you look at the distribution for $$\sum X_i$$, it diverges as $$n$$ increases, while the distribution for $$\bar X$$ concentrates on the point $$\mu_X$$ (the law of large numbers). Hence the need to rescale and relocate.

There are times when you might want to claim that the approximation to a standard normal is good enough for a given distribution and $$n$$; you might then undo the rescaling and relocation in the sort of way you suggest. Remember that there will be other times when the approximation is not good enough, and you may not be able to spot which without further investigation.

There's probably an easier way to prove this, simply based on linearity of expectation and variance, and the fact that Gaussians are defined by these two moments, however:

CLT implies convergence by distribution (also called weak convergences), which means that the Cumulative Density Function converges for $$n \rightarrow \inf$$

This means that for $$Y = \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}},$$ the CDF will converge to the one of $$N(0,1)$$, which is: $$F_Y \rightarrow \frac{1}{2}\left[1+\text{erf}\left(\frac{y}{\sqrt{2}}\right)\right]$$ Now if we call $$Z = \sum X = n\bar{X}$$, we have that: \begin{align}F_Z &\rightarrow \frac{1}{2}\left[1+\text{erf}\left(\frac{\frac{Z}{n}-\mu}{\sqrt{2}\sigma}\sqrt{n}\right)\right] \\&= \frac{1}{2}\left[1+\text{erf}\left(\frac{Z-n\mu}{\sqrt{2}\sigma\sqrt{n}}\right)\right] \end{align}

which is the CDF of $$N(n\mu, \sigma\sqrt{n}).$$