Transform probability density from low to high dimension with Jacobian 
*

*$\vec{X}$ is a vector of random variables in $\mathbb{R}^{n}$

*$\rho_{\vec{X}}(\vec{x})$ is probability density of $\vec{X}$.

*$\vec{y} = f(\vec{x})$ is a vector in $\mathbb{R}^{m}$

*$\vec{Y} = f(\vec{X})$ is a vector of random variables in $\mathbb{R}^{m}$

*$\rho_{\vec{Y}}(\vec{y})$ is probability density of $\vec{Y}$.

*$m \gt n$
The goal is to find $\rho_{\vec{Y}}(\vec{y})$ given $\rho_{\vec{X}}(\vec{x})$ and $f(\vec{x})$.

When does the following trick work?
$\rho_{\vec{Y}}(\vec{y}) d\vec{y} = \rho_{\vec{X}}(\vec{x}) d\vec{x}$
$\rho_{\vec{Y}}(\vec{y}) \det(\left \lvert J \right \rvert) = \rho_{\vec{X}}(\vec{x})$
For $n = 2, m = 4$, 
$$ J = \left[ \begin{matrix} 
a_{11} & a_{12} & 0 & 0 \\
a_{21} & a_{22} & 0 & 0 \\
a_{31} & a_{32} & 1 & 0 \\
a_{41} & a_{42} & 0 & 1 \\
\end{matrix} \right]$$
where:


*

*$ a_{ij} = \frac{\partial{y_i}}{\partial{x_j}} $

*The lower right corner of J is a block of identity matrix

*$\det$ is determinant



The explanation of the trick says it tries to augment the transform from:
$\vec{x} \rightarrow \vec{y}$
to
$(x_{1}, x_{2}, y_{3}, y_{4}) \rightarrow (y_{1}, y_{2}, y_{3}, y_{4})$
and then integrate the extra dimensions out.

What if $\frac{\partial y_{1}(\vec{x})}{\partial{y_3}(\vec{x})} \ne 0$?
If the trick doesn't always work, what is the general method to do this?

Update: add an example
Example (Unit simplex transform or Stick breaking transform)
Let $\vec{x} = (x_1, x_2, x_3, \ldots, x_{K})$ be a point on the probability simplex:


*

*$0 \le x_{n} \le 1$

*$x_1 + x_2 + x_3 + ... x_{K} = 1$
The transformation $\vec{z} = (z_{1}, \ldots, z_{K-1}) = f^{-1}(\vec{x})$ maps from the probability simplex to a hypercube:


*

*$z_{l} = \log(x_{l}) - \log(x_{l+1} + ... + x_{K})$
For ease of derivation, I define the following:


*

*Let ${b_{l} = [1 + \exp(z_{l})]^{-1}}$

*Let $c_{l} = 1 - b_{l} = [1 + \exp(-z_{l})]^{-1}$
Then, 
$$ x_{m} = \begin{cases}
c_{1} & \text{if } m = 1 \\
\prod_{l=1}^{K-1} b_{l} & \text{if } m = K \\
c_{m} \prod_{l=1}^{m-1} b_{l} & \text{otherwise}
\end{cases}$$
For $K = 3$,
$$ J = \left[ \begin{matrix} 
a_{11} & a_{12} & 0  \\
a_{21} & a_{22} & 0  \\
a_{31} & a_{32} & 1  \\
\end{matrix} \right]$$
where $a_{ij} = \frac{\partial x_{i}}{\partial z_j}$
Since $a_{12} = 0$, $ \det(J) = a_{11}a_{22}$
For higher K, only the diagonal terms contribute to the determinant.
$-\log[\det(J)] = \sum_{l=1}^{K-1} (K-l)\log[b_{l}] + \log[c_{l}]$
By testing over many randomly selected values, I verify that the above expression is equivalent to the expression in section "Absolute Jacobian Determinant of the Unit-Simplex Inverse Transform" in Stan, which mentioned the Jacobian is triangular.
But the manual and Betancourt 2010 didn't say the exact form of the Jacobian ...
 A: I also originally believe there is no efficient way to deal with the problem. However, by accident, I read some paper on this topic. If the transformation is bijective, there may be some chance.
Basically, the claim is preservation of volume, the $m \times n$ Jacobian
$$ J_f = \frac{\partial Y}{\partial X} $$
If the matrix is full column rank, its volume is 
$$ vol \:J_f = \sqrt{det J_f^T J_f}$$
So, the density function becomes 
$$ \rho_Y \times vol \: J_f = \rho_X
$$ 
Ben-Israel, Adi, The change-of-variables formula using matrix volume, SIAM J. Matrix Anal. Appl. 21, No. 1, 300-312 (1999). ZBL0955.26007. 
A: If $f$ is differentiable as you assume, then its image $f(\mathbb R^n)$ will be supported on an (at most $n$-dimensional) submanifold of $\mathbb R^m$. Since the ($m$-dimensional) Lebesgue measure of this submanifold will equal zero, $Y$ will never possess a probability density (with respect to the Lebesgue measure).
As far as I know, there is no general method to address this problem, since probability measures (or random variables) supported on submanifolds of $\mathbb R^m$ are complicated for exactly the above reason. In particular cases, solutions can be found which describe the distribution of $\vec Y$ in other ways than through a probability density.
Update:
In your example, I do not see what $f$, $\vec Y$, $\rho_{\vec X}$ and $\rho_{\vec Y}$ are supposed to be.
