Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In my study of probability I've come across exercise questions talking about random variables uniformly distributed in disks and triangles.

I'd like to know if there are "standard" ways of approaching questions like "Find the density of the random variable $X$", "Find the joint density of the random vector $(X,Y)$" and "Find the joint distribution function of the random vector $(X,Y)$" when $X$ and $Y$ are random variables uniformly distributed over a miscellaneous two dimensional shape.

Note: My definitions of disk and triangle are $D := \{ (x,y) : x^2 + y^2 \leq r^2 \}$ and $T := \{ (x,y) : x \leq 1, 0 \leq y \leq x \}$ but I have seen others.

So far the only concrete thought I have come across is that the joint density of a random vector $(X,Y)$ is $\frac{1}{A}$ where $A$ is the area of the shape in question. This holds true since the integral of the joint density over it's domain is defined to be $1$, and integrating $1$ over that same domain is equivalent to finding it's area.

share|cite|improve this question
up vote 2 down vote accepted

The vector $(X,Y)$ is uniformly distributed on the domain $S$ of finite positive area $A$ if and only if the density $f$ of $(X,Y)$ is $f(x,y)=A^{-1}\mathbf 1_S(x,y)$ (and not just $A^{-1}$). In particular, the density $f_X$ of $X$ is such that $$ f_X(x)=A^{-1}\int_{\mathbb R}\mathbf 1_S(x,y)\mathrm dy, $$ and the joint distribution function $F$ of $(X,Y)$ is such that $$ F(x,y)=P[X\leqslant x,Y\leqslant y]=\iint_{u\leqslant x,v\leqslant y}f(u,v)\mathrm du\mathrm dv=A^{-1}\iint_{u\leqslant x,v\leqslant y}\mathbf 1_S(u,v)\mathrm du\mathrm dv. $$

share|cite|improve this answer
This was a perfect answer. Thanks so much! – Gil Apr 5 '13 at 6:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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