# Disprove independence of vector of Gaussians by independence of marginals

If we have three random variables $X,Y,Z$, then if $X$ and $Z$ are independent, and $Y$ and $Z$ are independent, it doesn't follow that $Z$ is independent of the vector $(X,Y)$.

There is a simple counter example for this. But, I can't find a counter example in the case where all three are Normal, ie $X,Y,Z$ are Gaussian variables, and not multivariate Gaussian.

Appreciate any help.

Example: Let $X$ and $Z$ be iid standard Gaussian (zero mean) variables. And let $$Y = \left\{ \begin{array}{ll} X & \mbox{if } |Z| \leq 1 \\ -X & \mbox{elsewhere} \end{array} \right.$$

Then, $Y$ is marginally gaussian, and independent of $Z$: $P(Z|Y) = P(Z)$, as $P(Z|X)=P(Z)$. However, $P(Z|X Y) \ne P(Z)$, because knowing $X$ and $Y$ we know whether $|Z|<1$ or not.

• That's great! Thanks. BTW, you mean $P(Z|X)=P(Z)$, right?
– yoki
Jan 7, 2013 at 20:53
• Oops, yes, of course, fixed. Jan 7, 2013 at 21:16

If I understand your question right, you are assuming that the joint distribution of $X,Y,Z$ is Gaussian and has the covariance matrix $$C_{XYZ} = \begin{bmatrix} c_{XX} & c_{XY} & 0\\ c_{XY} & c_{YY} & 0\\ 0 & 0 & c_{ZZ} \end{bmatrix}\ .$$ It can be easily shown that $$N([X\ Y\ Z]^T, C_{XYZ}) = N([X\ Y]^T,C_{XY})\, N(z, c_{zz})$$ hold, where $N(xx,C)$ is the Normal Distribution of the vector $xx$ with covariance matrix $C$. So for scalar random Gaussian variables the independence marginals results in the independence of the variables. This holds not for non-Gaussians of course.

• The OP ask for X Y Z marginally gaussian but "not multivariate gaussian" Jan 7, 2013 at 13:43
• @leonbloy is right, I don't let them be multivariate, so they don't have this distribution and covariance matrix.
– yoki
Jan 7, 2013 at 14:51