Linear transformation of a random vector with pseudo-inverse I have already asked this question in cross-validated, but nobody answered, so I thought I may have more luck here.
If $$ \mathbf{X} = (X_1,\ldots,X_n)^t$$ is a random variable drawn according to a probability density function (pdf) $$ f_{X_1,\ldots,X_n}(x_1,\ldots,x_n) $$ then $$ \mathbf{Y} = A\mathbf{X} = (Y_1,\ldots,Y_n)^t$$ with $A$ a square non-singular matrix, has a pdf given by:
$$ f_{Y_1,\ldots,Y_n}(y_1,\ldots,y_n)=\frac{f_\mathbf{X}(A^{-1} \mathbf{y})}{|A|} $$
I have heard that this can be generalised to the case of a non-square matrix with the Moore–Penrose pseudo-inverse concept, where now $$ \mathbf{Y} = A\mathbf{X} = (Y_1,\ldots,Y_m)^t \qquad m<n $$
and 
$$ f_{Y_1,\ldots,Y_m}(y_1,\ldots,y_m)=\frac{f_\mathbf{X}(A^{+} \mathbf{y})}{|A|_{+}} $$ 
with $A^+$ the pseudo-inverse of $A$ and $|A|_{+}$ the pseudo-determinant.
If this is right, how can it be proved? and more important, what is the intuition behind this generalisation?
I've only found this related question, but I can't understand how the OP finds the general expression for $f_{Y_1,\ldots,Y_m}(y_1,\ldots,y_m)$.
 A: I can only help you for to understand $Y=Ax$ and it is a concept I am new and tried to learn searching on the web when reading your question. Remember the Moore–Penrose pseudo-inverse concept solves the problem in the least squared error sense. In general there is no exact solution.
Let be the pseudo invers-matrix :
\begin{align}
A^§&=(A^TA^{-1})A\\
\end{align}
Which is used when your equation as more rows than colums, i.e more rows than variable:it can't be solved 
Properties :
\begin{align}
A^§A&=I\\
AA^§&\neq I \mbox{ unless you have the usual invert}
\end{align}
Why does it helps you
\begin{align}
A^§\vec y&=A^§A \vec x\\
A^§\vec y&\approx \vec x
\end{align}
Example : 
\begin{cases}
x_1&+3x_2&=17\\
5x_1&+7x_2 &= 19\\
11x_1&+13x_2&=23
\end{cases}
There is more rows than variables, it is undecidable !
$$A=\begin{pmatrix}
1 & 3 \\
5 & 7\\
11 & 13
\end{pmatrix},
A^§=\begin{pmatrix}
-0.51 & -0.21 & 02\\
0.42 & 0.20 & -0.13
\end{pmatrix}$$
$$\begin{pmatrix}
x_1\\x_2
\end{pmatrix}\approx\begin{pmatrix}
-0.51 & -0.21 & 02\\
0.42 & 0.20 & -0.13
\end{pmatrix}
\begin{pmatrix}
17\\19\\23
\end{pmatrix}$$
And there you are. I hope it will help you, don't hesitate to take a tour on youtube. It is full of (yet to be-checked and corrected) resources.
