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Suppose that $x_1, x_2, \dots$ and $y_1, y_2, \dots$ are i.i.d. standard Gaussian random variables. Define the following

$$z_n = \sum_{i=1}^n c_{ni} x_i + y_n$$

We know that $z_n$ is a zero-mean Gaussian random variable. For any $n\geq 1$, we can write

\begin{align} z_{1:n} = \begin{bmatrix} z_1\\ z_2\\ \vdots\\ z_n \end{bmatrix} = \begin{bmatrix} c_{11} & 0 & \dots & 0\\ c_{21} & c_{22} & \dots & 0\\ \vdots\\ c_{n1} & c_{n2} & \dots & c_{nn} \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} + I \begin{bmatrix} y_1\\ y_2\\ \vdots\\ y_n \end{bmatrix} = C_n x_{1:n} + I y_{1:n} \end{align}

$z_{1:n}$ can be written as a linear combination of $n$ i.i.d. standard Gaussian random variables $v_1, \dots, v_n$

$$z_{1:n} = A_n v_{1:n},$$

where $A_nA^\intercal_n = C_nC^\intercal_n + I$. Intuitively, I expect $A_n$ to be lower triangular (similar to $C_n$) because of the causality in the problem. However, my simulations suggest that $A_n$ is not triangular. Could you please explain why my approach/intuition is wrong?

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1 Answer 1

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I figured out that $A_n$ is not unique. In code, I used eigenvalue decomposition to compute $A_n$ and got a non-triangular matrix. To get a lower triangular matrix, use Cholesky decomposition.

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