# Gaussian sum VS Brownian motion

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?

• I'm afraid the term "Gaussian sum" is occupied Commented Oct 26, 2018 at 18:25

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 $$$$S_n:=\sum_1^n G_i.$$$$ Then for natural $$n$$ we have $$$$S_{2n}=\Big(\sum_1^n a_{2j-1}Z_{2j-1},\sum_1^n a_{2j}Z_{2j}\Big)=a_{2n}(0,Z_{2n})+o_2(a_{2n}),$$$$ $$$$S_{2n-1}=\Big(\sum_1^n a_{2j-1}Z_{2j-1},\sum_1^{n-1} a_{2j}Z_{2j}\Big)=a_{2n-1}(Z_{2n-1},0)+o_2(a_{2-1}),$$$$ $$$$\sigma_{2n}^2=ES_{2n}S_{2n}^T=a_{2n}^2 \begin{pmatrix} o(1)&o(1)\\o(1)&1+o(1) \end{pmatrix},$$$$ $$$$\sigma_{2n-1}^2=ES_{2n-1}S_{2n-1}^T=a_{2n-1}^2 \begin{pmatrix} 1+o(1)&o(1)\\o(1)&o(1) \end{pmatrix},$$$$ $$$$\|\sigma_n\|^2\sim a_n^2.$$$$ So, the matrix $$$$M_n:=\frac1{\|\sigma_n\|^2}E\Big(\sum_1^n G_i\Big)\Big(\sum_1^n G_i\Big)^T=\frac{\sigma_n^2}{\|\sigma_n\|^2}$$$$ does not converge to any matrix; rather, $$M_n$$ asymptotically oscillates between $$\begin{pmatrix} 0&0\\0&1 \end{pmatrix}$$ and $$\begin{pmatrix} 1&0\\0&0 \end{pmatrix}$$.
• 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 $\infty$? Commented Oct 28, 2018 at 0:46
• @jason : Thank you for your comment. I have corrected the expressions for $\sigma_{2n}^2$ and $\sigma_{2n-1}^2$. Commented Oct 28, 2018 at 17:03