# Linearity of expected value of random vectors

Let $$Z_1, Z_2, \ldots, Z_k$$ of $$Z$$ i.i.d. in $$\mathbb{R}^n$$. A sample mean

$$$$\bar{Z}_k = \frac{1}{k} \sum_{j=1}^k Z_j$$$$

by the strong law of large numbers is

$$$$\bar{Z}_k \to x$$$$

almost surely as $$k \to \infty$$.

The variance of $$\bar{Z}_k$$ is given by

$$$$\frac{1}{k^2} \mathbb{E} \left( \left\| \sum_{j=1}^k (Z_j -x ) \right\|_2^2 \right)$$$$

in which cases can obtain the equality?

$$$$\frac{1}{k^2} \mathbb{E} \left( \left\| \sum_{j=1}^k (Z_j - x ) \right\|_2^2 \right) = \frac{1}{k^2} \sum_{j=1}^k \mathbb{E} \left\| Z_j - x \right\|_2^2$$$$

It is true by independence of $$Z_j$$'s. Note that $$Z_i-x, 1\leq i \leq n$$ are orthogonal since they have mean $$0$$ and they are independent. For orthogonal vectors $$(v_i)$$ we have $$\|\sum v_i\|^{2}=\sum \|v_i\|^{2}$$.
• How can we relate orthogonality with zero mean? Do we assume mean 0, since $\bar{Z_k} \to x$ almost surely for large enough $k$ ? Can you please elaborate a little more about it?
• If $X$ and $Y$ are independent random vectors with mean $0$ then $E \langle X, Y \rangle=E\sum X_iY_i=\sum EX_iY_i=\sum EX_i EY_i=0$. @Lin Jan 22 '20 at 7:59