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

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?

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 $$\forall(t,s)\in \mathbb{R}^2, F_{X,Y}(t,s)=F_X(t)F_Y(s)$$.

• Your thoughts/work? – StubbornAtom Dec 26 '18 at 17:40
• By sketching a picture of the region $S=\{(x,y):|x|+|y|<1\}$ you will see that Support$(X_1)\times$ Support$(X_2)\ne S$, violating a necessary condition of independence of $X_1$ and $X_2$. – StubbornAtom Dec 26 '18 at 18:28
• Just to complement @StubbornAtom: also see math.stackexchange.com/questions/663175/… – Just_to_Answer Dec 26 '18 at 18:34
• and one more note: from the picture of the support (the diamond centered at the origin), think about the locations inside the unit square outside the diamond. For those the marginals will be non-zero, but the joint will be zero. – Just_to_Answer Dec 26 '18 at 18:36
• An illustration of @StubbornAtom's comment can be found my answer where I called it the eyeball test. If the support of the joint pdf is not a rectangle with sides parallel to the axes, then the random variables are dependent: no need for finding the marginal densities and checking whether or not $f_{X,Y}(x,y)$ equals the product of $f_X(x)$ and $f_Y(y)$ everywhere in the plane. – Dilip Sarwate Dec 26 '18 at 23:17

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.

• I only know that $f_{X,Y}$ is constant in the diamond, but it could be equal to $c=0$. Doesn't it contradict the support of $f_{X,Y}$? @StubbornAtom – J. Doe Dec 27 '18 at 12:22
• @J.Doe How could $f_{X,Y}$ equal zero ? The region $S$ has a finite area, $f_{X,Y}$ equals the reciprocal of that area whenever $|x|+|y|<1$. – StubbornAtom Dec 27 '18 at 12:34
• @J.Doe "contradict the support of $f_{X,Y}$" ?? – StubbornAtom Dec 27 '18 at 12:40