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, \quad \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 $X_1$ .