4
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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 $(X,Y,Z).$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.