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Given independent Gaussian $d$ dimensional vectors $G_i$,

Let $ \sigma^2_n=\mathbb{E}(\sum_{i \le n} G_i) \cdot (\sum_{i \le n} G_i)^T$. $||\sigma_n^2||$ is norm of $\sigma_n^2$.

Is there a $d$-dimension Brownian motion $B_t$ with covariance matrix $\sigma^2$ s.t. $\sum_{i \le n} G_i=B_{||\sigma_n^2||}+o({||\sigma_n^2||}^{\frac{1}{2}-\epsilon})$ and

$\mathbb{E}(\sum_{i \le n} G_i)(\sum_{i \le n} G_i)^T=||\sigma_n^2||\cdot \sigma^2+o({||\sigma_n^2||}^{1-\epsilon})$ ?

the crucial part of this question is $\frac{\mathbb{E}(\sum_{i \le n} G_i)(\sum_{i \le n} G_i)^T}{||\sigma_n^2||}\to \sigma^2$? is it possible to converge?

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The answer is no. E.g., let $d=2$ and let $Z_1,Z_2,\dots$ be iid standard normal random variables. Let then $G_i=(a_iZ_i,0)$ if $i$ is odd and $G_i=(0,a_iZ_i)$ if $i$ is even, where the $a_i$'s are positive real numbers increasing fast enough in $i$ so that $\sum_1^{n-1}a_i^2=o(a_n^2)$; the convergence everywhere here is as $n\to\infty$. (For instance, one may take $a_i=i!$ or $a_i=2^{2^i}$.) For any random vector $X$, let $\|X\|_2:=(E\|X\|^2)^{1/2}$. For any sequence $(X_n)$ of random vectors and any sequence $(b_n)$ of positive real numbers, write $X_n=o_2(b_n)$ to mean $\|X_n\|_2=o(b_n)$. Then $\sum_1^{n-1} a_{2j-1}Z_{2j-1}=o_2(a_{2n-2})$ and $\sum_1^{n-1} a_{2j}Z_{2j}=o_2(a_{2n-1})$. Let \begin{equation} S_n:=\sum_1^n G_i. \end{equation} Then for natural $n$ we have \begin{equation} S_{2n}=\Big(\sum_1^n a_{2j-1}Z_{2j-1},\sum_1^n a_{2j}Z_{2j}\Big)=a_{2n}(0,Z_{2n})+o_2(a_{2n}), \end{equation} \begin{equation} S_{2n-1}=\Big(\sum_1^n a_{2j-1}Z_{2j-1},\sum_1^{n-1} a_{2j}Z_{2j}\Big)=a_{2n-1}(Z_{2n-1},0)+o_2(a_{2-1}), \end{equation} \begin{equation} \sigma_{2n}^2=ES_{2n}S_{2n}^T=a_{2n}^2 \begin{pmatrix} o(1)&o(1)\\o(1)&1+o(1) \end{pmatrix}, \end{equation} \begin{equation} \sigma_{2n-1}^2=ES_{2n-1}S_{2n-1}^T=a_{2n-1}^2 \begin{pmatrix} 1+o(1)&o(1)\\o(1)&o(1) \end{pmatrix}, \end{equation} \begin{equation} \|\sigma_n\|^2\sim a_n^2. \end{equation} So, the matrix \begin{equation} M_n:=\frac1{\|\sigma_n\|^2}E\Big(\sum_1^n G_i\Big)\Big(\sum_1^n G_i\Big)^T=\frac{\sigma_n^2}{\|\sigma_n\|^2} \end{equation} does not converge to any matrix; rather, $M_n$ asymptotically oscillates between $\begin{pmatrix} 0&0\\0&1 \end{pmatrix}$ and $\begin{pmatrix} 1&0\\0&0 \end{pmatrix}$.

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  • $\begingroup$ Thanks for your answer. but I do not figure out why your $\sigma_{2n}^2$ has zero diagonal entry? I think its minimum eigenvalue should go to $\infty$? $\endgroup$
    – jason
    Commented Oct 28, 2018 at 0:46
  • $\begingroup$ @jason : Thank you for your comment. I have corrected the expressions for $\sigma_{2n}^2$ and $\sigma_{2n-1}^2$. $\endgroup$ Commented Oct 28, 2018 at 17:03

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