Linearity of the <Bochner integral> gives
$$
\mathbb E\widehat\mu
=\frac1n\sum_{i=1}^n\mathbb EX_i=\mu.
$$
Independence and centering of $X_i-\mu$ imply
$$
\begin{aligned}
\mathbb E\lVert\widehat\mu-\mu\rVert^2
&=\frac1{n^2}\sum_{i,j}
\mathbb E\langle X_i-\mu,X_j-\mu\rangle\\
&=\frac1n\mathbb E\lVert X-\mu\rVert^2
=\frac{\operatorname{tr}C_X}{n}
=O(n^{-1}).
\end{aligned}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 225 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-225/1/d/solution