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? 



Gaussian sum VS Brownian motion
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 $\begin{pmatrix}
 1&0\\0&0
\end{pmatrix}$. 


