# Checking if jointly CRVs are independent via the marginal density

I'm trying to determine whether the following continuous random variables are independent or not given their joint probability density.

$$f_{X,Y}(x,y) = 2(x+y)$$ for $$0\le y\le x\le 1$$

I calculated the marginal densities as follows:

$$f_x(x) = \int_0^1{2(x+y)} dy$$ = $$2x+1$$

Using the same method for $$f_y(y)$$ I got $$f_y(y) = 2y+1$$

Since $$f_{X,Y}(x,y)\neq f_X(x)f_Y(y)$$ they are not independent.

My concern is that my calculation of the marginal densities is incorrect, could someone verify whether my method is correct? Or if this is the best way to check if the two are independent.

• The two marginal densities should not be the same, though they should both integrate to $1$ over $[0,1]$. Note $y \le x$ is a constraint, which affects the limits of integration. (The same constraint means that you obviously do not have independence, even without calculation) Dec 14, 2020 at 18:07
• @Henry What should my densities be, should I recalculate y with the limits of integration being x and 0 and do the same with x, changing the limits of integration to y and 1? Dec 14, 2020 at 18:10
• Yes $f_x(x) = \int\limits_0^x{2(x+y)} \, dy$ and $f_y(y) = \int\limits_y^1 {2(x+y)}\, dx$ Dec 14, 2020 at 18:13

It is not correct. $$f_{\small Y}(y)=\int_y^1 2(x+y)\,\mathbf 1_{0\leqslant y\leqslant 1}\,\mathrm d x\\f_{\small X}(x)=\int_0^x 2(x+y)\,\mathbf 1_{0\leqslant x\leqslant 1}\,\mathrm d y$$
However, there is a easier way.   The support for the joint p.d.f. is $$0\leqslant y\leqslant x\leqslant 1$$.   That is all you need to know.
Without even integrating, you can assert that: \begin{align}\mathsf P(Y>1/2,X<1/2)&=0\\\mathsf P(Y>1/2)&>0\\\mathsf P(X<1/2)&>0\end{align}