# probability – On the independence of multivariate gaussians.

Hello I have some perplexities about independence of multivariate gaussian distributions:
before getting there, I have studied the following results:

1)Let $$m$$ r.v $$X_1, dots , X_m$$ which range in $$mathbb{R}^{d_1} , dots , mathbb{R}^{d_m}$$ respectively. They are called independent if for every $$A_1 subset mathbb{R}^{d_1}, dots , A_m subset mathbb{R}^{d_m}$$ we have

$$P(X_1 in A_1 , dots , X_m in A_m) = P(X_1 in A_1) dots P(X_m in A_m)$$

Also, if $$phi_1 : mathbb{R}^{d_1} rightarrow mathbb{R} , dots, phi_m:mathbb{R}^{d_m} rightarrow mathbb{R}$$ respect some regularity conditions, then $$phi_1(X_1), dots , phi_m(X_m)$$ are independent.

Now, let $$X_1, X_2, X_3$$ independent r.v $$mathcal{N}(0,1)$$ and consider

$$U = 2X_1 – X_2 – X_3 qquad V = X_1 + X_2 + X_3, qquad W = X_1 -3X_2 +2X_3$$

What can I say about the independence of r.v $$U,V,W$$?

The just stated result would bring me to think that these random variables $$(U,V,W)$$ are all independent (since $$phi$$ are here simple linear transformations)..

Of course $$X = (X_1, X_2, X_3) sim mathcal{N}(0,I)$$ where $$I$$ is the identity $$3 times 3$$ matrix. Then $$U, V, W$$ are jointly gaussian because linear transformations of jointly gaussian, so to check the independence is it sufficient to prove they are not correlated. But it results $$Cov(U,W) neq 0$$ so $$U,W$$ seem not to be independent. Why does the initial statement fail here? Thanks