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

\begin{equation} \bar{Z}_k = \frac{1}{k} \sum_{j=1}^k Z_j \end{equation}

by the strong law of large numbers is

\begin{equation} \bar{Z}_k \to x \end{equation}

almost surely as $k \to \infty$.

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

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

in which cases can obtain the equality?

\begin{equation} \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 \end{equation}

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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 $\|\sum v_i\|^{2}=\sum \|v_i\|^{2}$.

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  • $\begingroup$ 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? $\endgroup$
    – Lin
    Jan 22 '20 at 7:57
  • $\begingroup$ 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 $\endgroup$ Jan 22 '20 at 7:59

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