Compute marginal PDF from joint PDF The joint PDF of $X$ and $Y$ are given by $$f(x,y)=3y, for 0<y<1, 0<x<y$$
As usual, I tried to find the marginal PDF of X by $$\int_{0}^{1}f(x,y) dy $$ However, this gives me 1.5 which is greater than 1.
How can I approach the problem?
 A: To fully formulate this problem, it is helpful to introduce the indicator function defined in the following way:
$$
I_A(x) = \begin{cases}
               1 \quad \text{if $x \in A$}\\
               0 \quad \text{if $x \notin A$}
          \end{cases}
$$
Thus, the joint PDF of $X$ and $Y$ can be written as:
$$
f(x, y) = 3 y I_{(0, y)}(x) I_{(0, 1)}(y) = 3 y I_{(x, 1)}(y) I_{(0, 1)}(x)
$$
The marginal PDF of $X$ is given by:
$$
f(x) = \int_{-\infty}^{+\infty} f(x, y) \; d  y = \int_{-\infty}^{+\infty} 3 y I_{(x, 1)}(y) I_{(0, 1)}(x) \; d  y = I_{(0, 1)}(x) \int_{-\infty}^{+\infty} 3 y I_{(x, 1)}(y)  \; d  y = I_{(0, 1)}(x) \int_{x}^{1} 3 y \; d  y = I_{(0, 1)}(x) \bigg ( \frac{3}{2}y^2 \bigg ) \Bigg |_x^1 = \frac{3}{2} (1 - x^2) I_{(0, 1)}(x)
$$
Write this in the usual way, we have:
$$
f(x) = \begin{cases}
               \frac{3}{2} (1 - x^2) \quad \text{if $0 < x < 1$}\\
               0 \quad \text{otherwise}
          \end{cases}
$$
It is not hard to verify that $f(x)$ is a PDF.  

Remark: Always remember to analyze the range of random variables first. A general way to do this is using the indicator function to extend the range of random variables to the entire real line(for a certain real random variable) and compute the integral with the infinite upper and lower bound. Then, consider the "true" bound(range) of the integral indicated by those indicator functions.
A: We have to integrate the joint distribution $f_{X,Y}(x,y)$ with respect to $Y$ to get the marginal $f_X(x)$. The limits of integration are:

So it amounts to getting rid of the $y$'s by integrating:
$$f_X(x)=\int_{y=x}^1 3y\,dy=\frac{3}{2}y^2\Big|_x^1=\frac{3}{2}(1 - x^2).$$
This is a valid pdf, because if we integrate it over the domain of $X: 0<x<1:$
$$\int_0^1 \frac{3}{2}(1-x^2)\,dx=1$$
