Suppose that $x_i, y_i$ are i.i.d. standard normal random variables. Consider the following random vector
\begin{align} z = \begin{bmatrix} z_0\\ z_1\\ z_2\\ z_3\\ \vdots\\ z_{n-1} \end{bmatrix} = \begin{bmatrix} x_0 \\ x_1 + y_1\\ x_1 + x_2 + y_2\\ x_1 + x_2 + x_3 + y_3\\ \vdots\\ x_1 + \dots + x_{n-1} + y_{n-1} \end{bmatrix} \end{align}
This random vector is jointly Gaussian because it is written as a linear combination of i.i.d. random variables. However, I think this representation of $z$ is degenerate. Can I represent $z$ as a linear combination of $n$ i.i.d. standard Gaussian variables? If so, how?
Lets see what happen with $n=2$. Define $$ A = \begin{pmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 1 & 1 \end{pmatrix} $$ and $$ B = \begin{pmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} $$ If $X=(X_1,X_2,X_3)^{\top}$ and $Y=(Y_1,Y_2,Y_3)^{\top}$, then $$ Z = AX+BY. $$ Because $X$ and $Y$ are independent, $$ \Sigma:=Cov(Z) = ACov(X)A^{\top} + BCov(Y)B^{\top} = AA^{\top} + BB^{\top} $$ Its easy to see that $\Sigma$ is singular, so Z is "degenerate", i.e., lives in a lower dimensional space.