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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.

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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.

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  • $\begingroup$ Thank you so much for your help. The absolute value was throwing me off. I am able to visualize $f_X(x)$ and $f_Y(y)$ now. I believe we have: $E(Y|X)=0$, $Cov(X,Y)=0$, and $X,Y$ are independent. $\endgroup$ – Austin D. Jul 17 at 0:22
  • $\begingroup$ $X,Y$ are surely not independent, the value of one does constrain the value of the other. $\mathsf{Cov}(X,Y)=0$ merely means the variables are uncorrelated. $\endgroup$ – Graham Kemp Jul 17 at 0:26
  • $\begingroup$ To be independent you need $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ for all $x,y$. $\endgroup$ – Graham Kemp Jul 17 at 0:31
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    $\begingroup$ Yes, of course... Clearly $2x(1-|y|)\neq 1$, so they are not independent. $\endgroup$ – Austin D. Jul 17 at 1:13

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