How to derive the joint distribution of Y=AX and Z=BX given a random vector X with known pdf?

Question

Given a random vector $X \in \mathbb{R}^k$, with a known pdf given by $f_X$. If $Y, Z \in \mathbb{R}^k$ are defined by $Y = AX$, $Z = BX$, where $A,B \in \mathbb{R}^{k\times k}$ are different, given, real-valued matrices.

I know how to calculate pdfs of $Y$ and $Z$ on their own. But how do I derive the joint pdf of $Y$ and $Z$?

If it helps to be more specific, $f_X$ is a mixture of $0$-mean multivariate gaussians, each component in the mixture with a different, diagonal covariance matrix (but not of the form $\Sigma = \sigma^2 I$).

Any help would be much appreciated.

For some context:

My goal is to check for the $f_X$ mentioned above, and a specific $A$ and $B$, whether the vectors $Y$ and $Z$ are independent. This means I need to check whether the joint distribution of $Y$ and $Z$ factorises into the product of the marginals. There are at least some cases when this is true: if, for example, $X \sim \mathcal{N}(0,\sigma^2 I)$ and $A$ and $B$ are projections onto orthogonal subspaces. But proving it is not true in my case would also be helpful. Hence my need to derive the joint distribution of $Y$ and $Z$.

Answer 1

Both $A$ and $B$ are square matrices, so there are a couple of basic points to note:

• If $A$ is non-singular then $Z = BX = B (A^{-1} A X) = B A^{-1} Y$; and similarly
• If $B$ is non-singular then $Y = AX = A (B^{-1} B X) = A B^{-1} Z$.

We can see from these results that if $A$ is non-singular then there is a function $Y \mapsto Z$ and if $B$ is non-singular then there is a function $Z \mapsto Y$. In either case the joint density of the two random vectors follows trivially from the marginal density of the first, and the random vectors will not be independent. (Indeed, one is a deterministic function of the other.)

This should cover almost all cases that you encounter. Indeed, it means that the only non-trivial case of interest is the case where both $A$ and $B$ are singular matrices. In this latter case things get trickier, and you will need to figure out the sets of values of one random vector that correspond to an individual value of the other random vector. Even in this case it is highly unlikely that you can obtain independence of the random vectors, since that would require special construction.