# 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 $$f_{x,y,z}(x,y,z)$$?

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:

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

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

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

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 $$(X,Y,Z).$$