# Two independent uniformly distributed random variables on $A$ in $\Bbb R^2$

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?

Thanks!

• Strictly speaking you cannot prove such a thing. The question is badly formulated. Commented Dec 31, 2020 at 9:22
• There has to be something missing in the question? Commented Dec 31, 2020 at 10:45
• 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 Commented Dec 31, 2020 at 14:55

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.