Suppose that $\mathbf{y=Ax}$ and that a probability density function over $\mathbf{x}$ is defined as $p(\mathbf{x})$.

If $\mathbf{A}$ has an inverse then the PDF over $\mathbf{y}$ is given by $q(\mathbf{y}) = \frac{1}{|\det \mathbf{A}|}p(\mathbf{A}^{-1}\mathbf{y})$

However if $\mathbf{x}$ has higher dimensionality than $\mathbf{y}$, so $\mathbf{A}$ is singular, then this doesn't work. So what then is the transformed PDF, $q(\mathbf{y})$? It seems to me that this PDF is still well defined, although some dimensions have been marginalized out.

I'm also interested in the independent case where $p(\mathbf{x}) = \prod_i \phi(x_i)$, which might simplify things.

  • 1
    $\begingroup$ How about $p(y \in \delta R) = \sum_{x \in (\delta R)^{-1}} p(x)$? This is the definition which in this case since the transformation is not bijective is defined as a sum over the cases which should be takn into account $\endgroup$
    – Nikos M.
    Jul 5, 2014 at 4:06


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