Finding the Marginal PDF from a Joint PDF with strange variable ranges

I am trying to find the $$f_Y(y)$$ marginal pdf of a joint pdf that has some particular ranges on the variables:

$$f(x,y) = \begin{cases} \frac{4x^2}{15}, & \text{1 \leq x \leq 2 and 0 \leq y \leq x} \\ 0, & \text{otherwise} \end{cases}$$

Specifically, I need to find the marginal $$f_Y(y)$$ for two different scenarios:

1. 0 $$\leq$$ $$y$$ $$\leq$$ $$1$$
2. 0 $$<$$ $$y$$ $$\leq$$ $$2$$

I normally would just integrate over $$x$$ to get to $$f_Y(y)$$, but I don´t understand what do I need to do different in splitting the marginal pdf in these two scenarios or how to find the marginal for each scenario.

The way you want to approach the problem is drawing the region of integration (it should be a trapezoid with the top being the line $$y=x$$). If you fix $$0\leq y \leq 1$$, you will notice that the range of integration in the trapezoid is the entire base (i.e. $$x=1$$ to $$x=2$$).

On the other hand, if $$1, the $$x$$ integration range is from the line $$x=y$$ to the vertical line $$x=2$$.

Outside of these ranges, there are no $$x$$ in the range to integrate.

• Thank you so much! I could not get the second part before but now I get it :) Mar 10, 2020 at 21:42

To evaluate the marginal for $$Y$$, you need to express the support in terms of $$y$$ relative to contants, and $$x$$ relative to $$y$$.   The support is provided as an expression the other way around, so we need to do a little work.

We know that the maximum for $$x$$ is $$2$$ so since $$y\leq x$$ is a constraint, the maximum for $$y$$ must also be $$2$$.   So now look at values for $$y$$ from $$0$$ to $$2$$ and find how $$x$$ is constrained.   Well when $$y<1$$, the minumum value for $$x$$ is $$1$$, but otherwise it is $$y$$.

The support may be partitioned by whether $$y\lt 1$$ or not. $${\quad\{\langle x,y\rangle: 0\leq y\leq x ~\land~ 1\leq x\leq 2\}\\=\{\langle x,y\rangle: (0\leq y\lt 1~\land~ 1\leq x\leq 2)~\lor~(1\leq y\leq 2~\land~ y\leq x\leq 2)\}}$$

So therefore

$$f_Y(y)=\mathbf 1_{0\leq y\lt 1} \int_1^2 f_{X,Y}(x,y)\,\mathrm d x+\mathbf 1_{1\leq y\leq 2}\int_y^2f_{X,Y}(x,y)\,\mathrm d x$$

Or if you prefer:

$$f_Y(y)=\begin{cases}\displaystyle\int_1^2 f_{X,Y}(x,y)\,\mathrm d x&:&0\leq y\lt 1\\[1ex]\displaystyle\int_y^2f_{X,Y}(x,y)\,\mathrm d x&:&1\leq y\leq 2\\[1ex]0&:&\textsf{otherwise}\end{cases}$$