I have a question regarding the definition of a uniform distribution for a bivariate random vector. For example, I am doing a few exercises and the premise of the questions are as follows:

Let $(X, Y)$ be a random vector and suppose that it has a joint uniform distribution over the square $(-1, 1) \times (-1, 1)$.

Let $(X, Y)$ be a random vector and suppose that it has a joint uniform distribution over the ball $\{(x, y) \in \mathbb{R}^2: x^2+y^2 \le 1\}$.

My question is simply what is the of a uniform distribution for a bivariate random vector? For example, I know that the pdf (of a single random variable ) of the uniform distribution is:

$$f(x)=\begin{cases}{\frac {1}{b-a}}&\mathrm {for} \ a\leq x\leq b,\\[8pt]0&\mathrm {for} \ x<a\ \mathrm {or} \ x>b\end{cases}$$

So what is $f_{X, Y}(x, y)$ in each of the premises above?

  • $\begingroup$ By definition, the probability of any event for a uniform distribution on any plane region is proportional to the area of that region. That criterion, together with the axiomatic requirement that the total probability equal $1,$ determines the distribution. $\endgroup$
    – whuber
    Sep 12, 2023 at 12:00

2 Answers 2


The definition of a "uniform distribution" is that the density function is constant for all $x,y$ within the support region. So one must have $$f_{X,Y}(x,y) = \frac{1}{A}$$ where $A$ is the area of either the square or the circle.

The same formula will hold for the density function of a "uniform distribution" on any geometric region.

Note though that this idea of uniform-ness applies only to the Cartesian coordinates ($x,y$). If you re-parametrized the circle in terms of radius and angle ($r,\theta$), for example, then the radial distance would not be uniformly distributed. The angle $\theta$ would be uniformly distributed on $(0,2\pi)$ but the radial distance $r$ would have to follow a triangular distribution for the bivariate distribution to be uniform in terms of $x$ and $y$.

  • $\begingroup$ But why $1$? Why can't it be $0.00001/A$? Since there can be an infinite amount of points in $A$, if you say $1/A$ and $A=100$, then you're saying that point $(x,y)$ account for 1% of the total area, which is not true because points have no area. $\endgroup$
    – NoName
    Jul 20, 2021 at 22:15
  • 1
    $\begingroup$ @NoName The definition I gave is correct. I suggest that you do some reading on the concepts of continuous probability distributions and density functions. You are confusing the density function with a probability mass function, which is what one would define for a discrete distribution. $\endgroup$ Jun 12, 2022 at 0:42

The volume under fxy(x, y) must be equal to 1, over the x, y support, since fxy(x, y) is a probability density function. The x, y support defines an area A. This area can be any area. Since the volume equals V = A x fxy(x, y) = 1, then fxy(x, y) = 1/A. Which is constant over all the x, y support.


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