Finding marginal pdf from a joint pdf I am trying to do part 1c of this question:

I am not sure how to get P(Y>1/2). I tried getting the marginal density of y by using $$\int_{-1}^1 cy \,dx= 2cy$$
Is this right? Then I tried getting P(Y>1/2) by using $$\int_{1/2}^{|x|} 2cy \,dx= 3x^2-3/4$$
This doesn't look right; it is supposed to be a real number right?
 A: You are wrong in computing the marginal of $Y$. Note that $f(x,y)=cy$ only when $|x|\geq y$ and $-1\leq x\leq 1$. So, the marginal of $Y$, which I shall denote as $f_{Y}(y)$, should be computed as follows:
\begin{align}
f_{Y}(y)&=\int\limits_{\{x:-1\leq x\leq 1, |x|\geq y\}}f(x,y)\,dx\\
        &=\int\limits_{-1}^{-y}cy\,dx + \int\limits_{y}^{1}cy\,dx\\
        &=2cy(1-y),
\end{align}
valid for $0\leq y\leq 1$. Further, since the joint PDF has to integrate to $1$, we have
\begin{align}
1=\int\limits_{x=-1}^{1}\quad\int\limits_{y=0}^{|x|}cy\,dy\,dx=c\int\limits_{x=-1}^{1}\frac{x^{2}}{2}\,dx=\frac{c}{3},
\end{align}
from which we get $c=3$. Thus, $f_{Y}(y)=6y(1-y)$ for $0\leq y\leq 1$. You may now go ahead and compute $P(Y\geq 0.5)$ or whatever quantity is of interest to you.

EDIT: The conditional density $f_{X|Y}(x|y)$ should be computed as follows:
\begin{equation}
f_{X|Y}(x|y)=\frac{f(x,y)}{f_{Y}(y)}.
\end{equation}
Now, notice that $f_{Y}(y)=6y(1-y)$ whenever $0\leq y\leq 1$, and the numerator term is $cy$ only when $|x|\geq y$. So, we get
\begin{equation}
f_{X|Y}(x|y)=\frac{3y}{6y(1-y)}=\frac{1}{2(1-y)},~0\leq y\leq 1,~|x|\geq y.
\end{equation}
It is easy to check that $\int\limits_{|x|\geq y}f_{X|Y}(x|y)\,dx=1$ as should be.
