Let $X\in\mathbb{R}^{n}$, be a random variable with probability density function $\rho_{X}(x)$. If one makes a change of variable $Z=f(X)\in\mathbb{R}^{n}$, where $f$ is invertible everywhere then
\begin{align*}
\rho_{X}(x)\mathrm{d}x&=\rho_{X}\left(f^{-1}(z)\right)\text{det}\left(\frac{\partial x}{\partial z}\right)\mathrm{d}z \\
&=\rho_{X}\left(f^{-1}\left(z\right)\right)\left(\text{det}\left(\frac{\partial f}{\partial x}\left(x=f^{-1}\left(z\right)\right)\right)\right)^{-1}\mathrm{d}z
\end{align*}
from which one obtains the probability density function for the random variable $Z$
\begin{align*}
\rho_{Z}\left(z\right)&=\rho_{X}\left(f^{-1}\left(z\right)\right)\left(\text{det}\left(\frac{\partial f}{\partial x}\left(x=f^{-1}\left(z\right)\right)\right)\right)^{-1}
\end{align*}
the previously used logic cannot be used if $Z$ is not the same dimension as $X$ and so $\textbf{I would like to understand how to find the distribution for}$ $Z$ $\textbf{when it is not the same dimension as}$ $X$. In particular I would like to know how to do it in general.
To show how I've given this some thought I will now show an example $X=\left(X_{1},X_{2}\right)\in\mathbb{R}^{2}$, with pdf $\rho_{X}\left(x_{1},x_{2}\right)$, define $Z=X_{1}+X_{2}$. The idea here is that we need to utilize the density function on $X$ and integrate over all pairs that are equal to a fixed $Z$ to obtain the distribution for $Z$.
\begin{align*}
\rho_{Z}\left(z\right)\mathrm{d}z&=\int_{\mathbb{R}}\rho_{X}\left(x_{1},z-x_{1}\right)\mathrm{d}x_{1}\mathrm{d}z\\
\rho_{Z}\left(z\right)&=\int_{\mathbb{R}}\rho_{X}\left(x_{1},z-x_{1}\right)\mathrm{d}x_{1}
\end{align*}
Best Answer
If we have random variables $X_1,X_2,\dots,X_n$ and new random variables $Z_1=g_1(\textbf X),Z_2=g_2(\textbf X),\dots,Z_m=g_m(\textbf X)$ with $m\le n$, we can set up dummy variables $Z_{m+1}=g_{m+1}(\textbf X),\dots,Z_n=g_n(\textbf X)$. Under the assumption that $g_1,g_2,\dots,g_n$ define a one-to-one transformation from the preimage $S\subseteq \mathbb R^n$ to $T\subseteq \mathbb R^n$ [and hence has an inverse transformation], we can use the standard transformation method to obtain the joint distribution of $Z_1,Z_2,\dots,Z_m,Z_{m+1},\dots,Z_n$, $h(\textbf Z)$, and obtain the desired joint distribution by integrating out $Z_{m+1}$ through $Z_n$: $$h(Z_1,\dots,Z_m)=\int_{-\infty}^\infty\dots\int_{-\infty}^\infty h(\textbf Z)dZ_{m+1}\dots dZ_n$$