Marginal probability density function for a Joint Variable with two upper bounds

Question

My joint variable function is $f(x,y) = \frac{2x+y}{36}$, $ 0 \leq y \leq x$, $x + 2y \leq 6$, zero otherwise.

My question is when trying to find the marginal p.d.f. of X, how can I do that when y has two upperbounds. Do I have to find the higher upperbound or the lower bound?

Otherwise the answer can be written as

$$f_X(x) = \int_{0}^{6-2y} f(x,y) \quad \textrm{ or } \quad f_X(x) = \int_{0}^{y} f(x,y)?$$

Answer 1

You wish to integrate with respect to $y$, so the bounds need to be functions in $x$, which the parameter of the marginal pdf.

Now, $x+2y\leqslant 6$ is equivalent to $y\leqslant 3-x/2$, so the support for this distribution is thus: $$\{\langle x,y\rangle: 0\leqslant x\leqslant 6\land 0\leqslant y\leqslant\min(x,3-x/2)\}$$ which is the triangle $\triangle(0,0)(2,2)(6,0)$, aka: $$\{\langle x,y\rangle: (0\leqslant x\lt 2~\land~ 0\leqslant y\leqslant x)\lor(2\leqslant x\leqslant 6~\land~ 0\leqslant y\leqslant 3-\tfrac x2\} $$

So $$f_X(x)=\mathbf 1_{x\in[0..2)}\int_0^x f(x,y)\mathsf d y+\mathbf 1_{x\in[2..6]}\int_0^{3-x/2} f(x,y)\mathrm d y$$

Answer 2

The first equation implies $y \le x$, the second implies $y \le 3-x/2$. Draw the region over which these bounds apply, and then your integral will separate in 2, one with one constraint and the other with the other...