This is a homework question from my probability class.

Let $X$, $Y$ be two random variables. Suppose $X$ is uniformly distributed on $A$, $Y$ is uniformly distributed on $A$, $A$ $\subseteq$ $\Bbb R^2$. Further assume $X, Y$ are independent.
Prove/Disprove: $A$ is a rectangle with lines parallel to the axes.

Because $X,Y$ are uniformly distributed on $A$ I know that their PDF is equal to:
$f_X(x)=f_Y(y)=1/\lambda(A)$ where $\lambda$ is the Lebesgue measure on the plane. because $X,Y$ are independent we get: $f_{(X,Y)}(x,y)= 1/\lambda(A)^2$

Now I got confused... How can I do this?


  • 1
    $\begingroup$ Strictly speaking you cannot prove such a thing. The question is badly formulated. $\endgroup$ Commented Dec 31, 2020 at 9:22
  • $\begingroup$ There has to be something missing in the question? $\endgroup$
    – Clement C.
    Commented Dec 31, 2020 at 10:45
  • $\begingroup$ This is the question. I ask my lecturer, she says it's perfectly fine... If you say it's false, can you give me a counterexample? every X, Y I try to take doesn't work $\endgroup$ Commented Dec 31, 2020 at 14:55

1 Answer 1


Since $X$ and $Y$ are independent, $ f_{XY}(x,y)= \begin{cases} f_X(x)f_Y(y), & (x,y)\in A \\ 0, & \text{otherwise} \end{cases}$.

Suppose that $f_{XY}(x,y)>0, \forall (x,y)\in A$. So, for any $x_0$ that $f_X(x_0)>0$, $f_{XY}(x_0,y)>0$ for all $y$ that $f_Y(y)>0$, and $f_{XY}(x_0,y)=0$ for all $y$ that $f_Y(y)=0$, i.e., the support set of non-zero probability for $Y$ remains the same for any arbitrary $x_0$ that $f_X(x_0)>0$. A similar argument applies to $Y$. Hence, $A$ is a rectangle.

The uniformity of distribution is not necessary for $A$ to be a rectangle.


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