This is the formula to help you get the marginal pdf from a joint distribution:
$$f(x)= \int_{}^{} f(x,y) \, dy\\$$ $$f(y)= \int_{}^{} f(x,y) \, dx\\$$
where the first integral is over all points in the range of (X,Y) for which X=x and the second inegral is over all points in the range of (X,Y) for which Y=y.
I have no problem to get the marginal in sample case, like f(x,y)=1/x, for 0<x<10, 0<y<5.
However, I don't think that those formula will work on the case below: $$f(x,y)= \frac{8}{3}xy $$ $$for\;0\le x\le1,x\le y\le 2x$$
Y is dependent on X for Y's region.
Can anyone give me a key word or hint? And then I can search more for this.
You can still find the marginal PDF - it's just going to be a piecewise function. Firstly - finding ${f(x) = \int_{x}^{2x}f(x,y)dy}$ is nice and straight forward. For ${f(y)}$ - we need to rewrite the bounds. If you do so, you'll notice it splits in two: $$ 0\leq y\leq 1,\frac{y}{2}\leq x\leq y $$ $$ 1 < y \leq 2,\frac{y}{2}\leq x < 1 $$ So $$ f(y) = \left\{ \begin{array}{cc} \int_{\frac{y}{2}}^{y}f(x,y)dx&\text{ if ${0\leq y\leq 1}$}\\ \int_{\frac{y}{2}}^{1}f(x,y)dx&\text{ if ${1 < y \leq 2}$} \end{array} \right. $$ Notice now that if you integrate ${\int_{0}^{2}f(y)dy}$, we should get $1$ - and you see that $$ \int_{0}^{2}f(y)dy = \int_{0}^{1}\int_{\frac{y}{2}}^{y}f(x,y)dxdy + \int_{1}^{2}\int_{\frac{y}{2}}^{1}f(x,y)dxdy = 1 $$