The mutual density of $X,Y$ in $\{|t|+|s|<1\}$ is constant, are $X,Y$ independent?

Question

Let $X,Y$ absolutely continuous random variables with density finctions $f_X,f_Y$. Assume that the mutual density $f_{X,Y}$ equals to a constant $c$ in $\{(t,s)\in\mathbb{R}^2:|t|+|s|<1\}$. Are $X,Y$ independent?

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I guess I need to use $\int_{-1}^0 \int _{1+s}^{−1−s}f_{X,Y}(x,y)dxdy+\int_0^1\int_{1−s}^{s−1}f_{X,Y}(x,y)dxdy=c$, But I'm not sure how can it help me.

The defenition of independent random variables is random variables $X,Y$ such that $\forall(t,s)\in \mathbb{R}^2, F_{X,Y}(t,s)=F_X(t)F_Y(s)$.

Answer 1

The joint support of $(X,Y)$ is the set

$$S=\left\{(x,y)\in\mathbb R^2:|x|+|y|\le 1\right\}$$

Sketch the region $S$. It should look like this picture:

Let $S_1$ and $S_2$ be the supports of $X$ and $Y$ respectively. Clearly,

$$S_1=\{x\in\mathbb R:-1\le x\le 1\}=\{y\in\mathbb R:-1\le y\le 1\}=S_2$$

A necessary condition of independence of two jointly distributed random variables is that their joint support must be the Cartesian product of their marginal supports.

That is, $X$ and $Y$ are independent only if $\text{supp}(X,Y)=\text{supp}(X)\times \text{supp}(Y)$.

[Simplest example: Consider $(X,Y)$ uniform on the unit square.]

Here of course $S\ne S_1\times S_2$, as should be evident from the picture above. Hence $X$ are $Y$ are dependent.

Equivalently, observe that $$P\{0.5\le X\le 1,0.5\le Y\le 1\}=0\ne P\{0.5\le X\le 1\}P\{0.5\le Y\le 1\}$$

So no need really to check whether $F_{X,Y}=F_{X}F_Y$ or $f_{X,Y}=f_Xf_Y$ for independence of $X$ and $Y$.

Also see this relevant answer from Dilip Sarwate.