# what's the pdf and covariance for this distribution?

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?

• Is this for some subject? – Glen_b Oct 13 '13 at 3:58
• As it stands the question seems to be underspecified, but maybe I missed something. – Glen_b Oct 13 '13 at 4:18
• Is this homework or self-study? If so, please add the appropriate tag... Try deriving the PDF using $p(x_1, x_2, x_3) = p(x_1)p(x_2|x_1)p(x_3|x_1,x_2)$. – jbowman Oct 13 '13 at 15:17
• 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? – whuber Oct 13 '13 at 15:52

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.