Joint Probability Density Function where Y depends on X

Question

Consider the following joint PDF:

$$f_{X,Y}(x,y)=\begin{cases} 1 & \text{if } 0<x<1, -x<y<x \\ 0 & \text{otherwise}\end{cases}$$

I am able to visualize the figure in 3 dimensions as a triangle at height 1 spanning $y=-1$ to $1$ and $x=0$ to $1$. I am trying to study the properties of this PDF, such as conditional expectation, marginal PDF, etc., and I am having a really tough time properly setting up the integrals. Specifically, I want to know $E[Y|X]$ and the marginal PDFs $f_X(x)$, $f_y(y)$, so that I can study the unconditional expectations and covariance. I am frustrated because I cannot visualize what is going on when I compute these objects. Any hints would be very helpful.

Answer 1

Examine the support.

We have $0<x< 1$ and $-x<y<x$, exactly when, $\lvert y\rvert < x < 1$ and, $-1<y<1$.

$$\large\begin{align}f_{X,Y}(x,y) &=\mathbf 1_{x\in(0,1), y\in(-x,x)}\\[1ex] &=\mathbf 1_{y\in(-1,1), x\in (\lvert y\rvert,1)}\end{align}$$

So $$\begin{align}f_X(x) &=\int_{(-x,x)} \mathbf 1_{x\in(0,1)}~\mathrm d y\\[2ex]f_Y(y)&=\int_{(\lvert y\rvert,1)}\mathbf 1_{y\in(-1,1)}~\mathrm d x\end{align}$$

am frustrated because I cannot visualize what is going on when I compute these objects.

The support is a triangle, $\triangle\langle 0,0\rangle\langle 1,-1\rangle\langle 1,1\rangle$.

The marginal density for $X$ at some $x\in(0,1)$ integrates over the vertical lines, where $y$ ranges from $-x$ to $x$. Since the joint density is uniform over the triangle, the marginal is proportional to the length of these lines: $2x$. -- smallest near the origin, widest near the opposite base

The marginal density for $Y$ at some $y\in(-1,1)$ integrates over the horizontal lines, where $x$ ranges from $\lvert y\rvert$ to $1$. Since the joint density is uniform over the triangle, the marginal is proportional to the length of these lines: $1-\lvert y\rvert$. -- smallest near the tips, largest near the central axis.

Oh, and it should be clear what $\mathsf E(Y\mid X)$ equals.