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

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

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 \(\ln Y=\ln X_{1}+\dots+\ln X_{n}\)