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Note: I have already seen this Wikipedia article, and similar questions on this website: 1.

Given two dependent multivariate Gaussian random variables, is the sum also a multivariate Gaussian?

$X \sim N(b_X,\Sigma_X), Y \sim N(b_Y,\Sigma_Y)$
$X = A_X w_1 + b_X$
$Y = A_Y w_2 + b_Y$

where $A_X$ and $A_Y$ have dimensions $d\times n_1$ and $d \times n_2$ respectively, and $A_XA_X^T=\Sigma_X$, $A_YA_Y^T=\Sigma_Y$. Vectors $w_1$ and $w_2$ have dimensions $n_1$ and $n_2$ respectively, and their entries are i.i.d. standard Gaussian RVs.

Now,

$Z = X+Y $
$Z = A_Xw_1 + A_Yw_2 + b_X + b_Y $
$Z = [A_X \ A_Y][w_1\ w_2]^T + (b_X+b_Y) $
$Z = A_Zw_3+b_Z$

So, $Z$ is a multivariate Gaussian with mean $b_Z$ and covariance matrix $A_ZA_Z^T$.

I don't think this uses independence of $X$ and $Y$ anywhere. Is it correct?

Edit: Based on the comments by whuber and Dilip Sarwate:

I have understood that $X+Y$ won't always be a multivariate Gaussian. In particular, this might happen when all entries of $[w_1\ w_2]^T$ are not independent.

If it is ensured that all entries of $[w_1\ w_2]^T$ are pairwise independent, does the claim hold? And no restrictions on $b_X$ and $b_Y$ are needed then?

EngrStudent
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muser
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    The general claim you state at the outset is false. Your demonstration fails because to justify the "So," you implicitly make additional independence assumptions about $(w_1,b_X,w_2,b_Y).$ For the simplest case (the dimension is $1$) see https://stats.stackexchange.com/questions/30159 for counterexamples and further explanation. Now if the $b_X$ and $b_Y$ are constants, there is no problem--but there's nothing new here, because you already knew linear combinations of multivariate Normals are multivariate Normal and $(X,Y)$ is a linear combination of $(w_1,w_2).$ – whuber Aug 09 '22 at 16:36
  • Vectors $w_1$ and $w_2$ might well individually be vectors of i.i.d. standard Gaussian RVs, and yet not be independent as vectors. For example, $w_1 = [n_1, n_2, n_3]$ and $w_2 = [n_3, n_4]$ are individually 3-vector and 2-vector of i.i.d. standard Gaussians but nonetheless are not independent vectors. – Dilip Sarwate Aug 09 '22 at 21:37
  • Thanks for the comments. If it is ensured that all entries of $[w_1\ w_2]^T$ are pairwise independent, does the claim hold? (and no restrictions on $b_X$ and $b_Y$ are needed then?) – muser Aug 10 '22 at 02:30
  • in the case where $n_1 \neq n_2$ the addition of biases $w_1$ and $w_2$ is problematic. You could zero append, but you could zero prepend, and both require assumptions about dimension. The product of the covariances is also problematic. – EngrStudent Aug 11 '22 at 11:47
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    @EngrStudent Addition of $w_1$ and $w_2$? I don't see where this happens... $w_1$ and $w_2$ are row concatenated to get $[w_1\ w_2]^T$. And $b_X$, $b_Y$ have the same dimensions ($d\times 1$) – muser Aug 11 '22 at 13:57
  • @abs - I am not used to the "$+$" being a concatenate operation. Above it indicates that $b_z = b_x + b_y$. The conventional interpretation of it is addition, but that only works if they are the same size. – EngrStudent Aug 11 '22 at 14:23
  • @EngrStudent I also am referring to $+$ as addition. $b_X + b_Y$ is well-defined, since both vectors are of size $d\times 1$. Concatenation was for $w$, not $b$. – muser Aug 11 '22 at 15:06

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