I am stuck on a problem and wonder if anyone can give me some suggestions.

$X_1, X_2, X_3$ all follow a $\text{Uniform}[0,1]$ distribution and are subject to the constraint $X_1+X_2+X_3\leq 1$.

What's the joint distribution for $(X_1, X_2, X_3)$, that is, what's $p(X_1, X_2, X_3)$, and what's the variance-covariance matrix for it?

I get the joint distribution by geometrical way, that the pdf should be $1/6$.

However, I can't calculate the variance-covariance matrix for it. I wonder how to get it?

  • $\begingroup$ Is this for some subject? $\endgroup$ – Glen_b Oct 13 '13 at 3:58
  • $\begingroup$ As it stands the question seems to be underspecified, but maybe I missed something. $\endgroup$ – Glen_b Oct 13 '13 at 4:18
  • $\begingroup$ Is this homework or self-study? If so, please add the appropriate tag... Try deriving the PDF using $p(x_1, x_2, x_3) = p(x_1)p(x_2|x_1)p(x_3|x_1,x_2)$. $\endgroup$ – jbowman Oct 13 '13 at 15:17
  • $\begingroup$ The joint distribution for which the pdf is a constant $1/6$ where $0\le X_i$ and $X_1+X_2+X_3\le 1$ does not have uniform marginals. @jbowman How are these conditional probabilities to be obtained given that only the marginal distributions are specified? $\endgroup$ – whuber Oct 13 '13 at 15:52

Such a distribution does not exist.

To see why not, let $0 \lt t \lt 1/2$ and notice that $X_2\gt 1-t$ entails $X_1\le t$ and $X_3\gt 1-t$ also implies $X_1\le t$, for otherwise in either situation the sum of all the $X_i$ would exceed $1.$ The latter two events are disjoint, because we cannot simultaneously have $X_2\gt 1-t \gt 1/2$ and $X_3\gt 1-t\gt 1/2.$ Consequently the chance that $X_1\le t$ is no less than the sum of the chances that $X_2\ge 1-t$ and $X_3\ge 1-t$, each of which equals $t$ by the uniform distribution assumptions. This shows that $t \ge t+t,$ which for $t\gt 0$ obviously is false.

This contradiction forces us to give up at least one of the assumptions: if indeed $X_1+X_2+X_3\le 1$, then the only other assumptions used in this argument are that each $X_i$ has a Uniform$[0,1]$ distribution. Therefore at least one of the $X_i$ cannot have a Uniform$[0,1]$ distribution, QED.


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.