How to find a joint PDF?

Question

I have this question, $$ f_{X,Y}(x,y) = ax^2 \qquad \text{if } 1 \leq x \leq 2 \quad \text{and} \quad 0 \leq y \leq x $$ I found out $~a= \frac{4}{15}$.

Now I am supposed to find marginal PDF $~f_Y(y)~$ for the cases of

$1.\quad$ If $~0 ≤ y ≤ 1 :$

$~f_Y(y)~=$

$2.\quad$ If $~1< y ≤ 2 :$

$~f_Y(y)~=$

I don't know how to proceed.

Answer 1

In case 1) $f_Y(y)=\int_1^{2} ax^{2}\, dx$ and in case 2) $f_Y(y)=\int_y^{2} ax^{2}\, dx$.

Answer 2

The support is $\{\langle x,y\rangle: 1\leqslant x\leqslant 2, 0\leqslant y\leqslant x\}$, which may be partitioned on where $y$ is relative to $1$.

That is, the support is the union:$$\{\langle x,y\rangle: 0\leqslant y < 1\leqslant x\leqslant 2\}\cup\{\langle x,y\rangle: 1\leqslant y\leqslant x\leqslant 2\}$$

So, the integral is piecewise over those parts: $$f_Y(y)=\begin{cases}\displaystyle\int_1^2 a x^2~\mathrm d x&:& 0\leqslant y<1\\[1ex]\displaystyle\int_y^2 ax^2~\mathrm d x&:& 1\leqslant y\leqslant 2\\[1ex]0&:& \textrm{elsewhere}\end{cases}$$