Can I find $f_{x,y,z}(x,y,z)$ from $f_{x,y+z}(x,y+z)$? Suppose I know densities $f_{x,y+z}(x,y+z)$, $f_y(y)$, $f_z(z)$, and $y$ and $z$ are independent. Given this information, can I derive $f_{x,y,z}(x,y,z)$?
 A: It is tempting to think so, but a simple counterexample with a discrete probability distribution shows why this is not generally possible.
Let $(X,Y,Z)$ take on the eight possible values $(\pm1,\pm1,\pm1).$  Let $0\le p\le 1$ be a number and use it to define a probability distribution $\mathbb{P}_p$ as follows:

*

*$\mathbb{P}_p=1/8$ whenever $Y+Z\ne 0.$


*$\mathbb{P}_p(1,-1,1) = \mathbb{P}_p(-1,-1,1)=p/4.$


*$\mathbb{P}_p(1,1,-1) = \mathbb{P}_p(-1,1,-1)=(1-p)/4.$
These probabilities evidently are positive and sum to $1:$
$$\begin{array}{crrr|rc}
\mathbb{P}_p & X & Y & Z & Y+Z & \Pr(Y,Z)\\
\hline
\frac{1}{8} & \color{gray}{-1} & \color{gray}{-1} & \color{gray}{-1} & -2 & \frac{1}{4}\\
\frac{1}{8} & 1 & \color{gray}{-1} & \color{gray}{-1} & -2 & \cdot\\
\frac{p}{4} & \color{gray}{-1} & \color{gray}{-1} & 1 & 0&\frac{1}{4}\\
\frac{1-p}{4} & 1 & \color{gray}{-1} & 1 & 0 & \cdot\\
\frac{1-p}{4} & \color{gray}{-1} & 1 & \color{gray}{-1} & 0 & \frac{1}{4}\\
\frac{p}{4} & 1 & 1 & \color{gray}{-1} & 0 & \cdot\\
\frac{1}{8} & \color{gray}{-1} & 1 & 1 & 2 & \frac{1}{4}\\
\frac{1}{8} & 1 & 1 & 1 & 2 & \cdot\\
\end{array}$$
Note two things:

*

*$Y$ and $Z$ are independent Rademacher variables.  That is, $\mathbb{P}_p(Y=y,Z=z)=1/4$ for all $y,z\in\{-1,1\}.$


*The joint distribution of $(X,Y+Z)$ does not depend on $p,$ as you may check by adding (2) and (3) to deduce that $\mathbb{P}_p(X=x\mid Y+Z=0)=1/2$ for $x\in\{-1,1\}.$  (Thus, $X$ is independent of $Y+Z.$)
The value of $p$ does not appear in the marginal distributions of $Y$ and $Z$ nor in the joint distribution of $(X,Y+Z).$  Thus, these distributions do not determine $p.$  Nevertheless, different values of $p$ produce different distributions of $(X,Y,Z)$: that's the counterexample.
If you must have an example involving continuous distributions (with densities), then add an independent standard trivariate Normal variable to $(X,Y,Z).$
