Arithmetic Mean

$M(t)=\frac{1}{n}{\sum }_{k=1}^n{t}_k$

Properties If we have $\mathbb{E}\left(X_1^2\right)<\infty$ , then for the Arithmetic Mean ${\bar{X}}_{(n)}$ we have: $\mathbb{E}\left({\bar{X}}_{(n)}\right)=\mu ,\mathbb{V}\left({\bar{X}}_{(n)}\right)=\frac{{\sigma }^2}{n}$ Of course the Expectation is $\mu$ as this is exactly what we want to estimate with the arithmetic mean estimator.

Given the Central Limit Theorem the Distribution of the Arithmetic Mean is for large $n$ in good approximation a Normal Distribution with:

$N\left(\mathbb{E}\left({\bar{X}}_{(n)}\right),\mathbb{V}\left({\bar{X}}_{(n)}\right)\right)=N\left(\mu ,{\sigma }^2/n\right)$

independent of the particular distribution of $X_1$.