# The Distribution of the Sample Mean and Sum

Posted by Beetle B. on Wed 07 June 2017

Let $$X_{1}\dots X_{n}$$ be a random sample from a distribution with mean $$\mu$$ and standard deviation $$\sigma$$. Then:

1. $$E(\bar{X})=\mu_{\bar{X}}=\mu$$
2. $$V(\bar{X})=\frac{\sigma^{2}}{n}$$

If $$T=\sum_{i}X_{i}$$, then $$E(T)=n\mu$$ and $$V(T)=n\sigma^{2}$$

So averaging squeezes the distribution, and summing it widens it.

## The Case of A Normal Distribution

If the distribution is normal, then so is the distribution of the mean and total.

## The Central Limit Theorem

Let $$X_{i}$$ be a random sample from any distribution with mean $$\mu$$ and variance $$\sigma^{2}$$. As $$n$$ gets larger, $$\bar{X}$$ converges to a normal distribution, as does $$T$$.

How large should $$n$$ be? Unfortunately, it depends on the underlying distribution. As a rule of thumb, we use $$n>30$$ (although this may not work well for all distributions).

If $$X_{1}\dots X_{n}$$ is a random sample, and only positive values are allowed, then $$Y=X_{1}\dots X_{n}$$ converges to lognormal. This is because $$\ln Y=\ln X_{1}+\dots+\ln X_{n}$$