I am given such a question:
Let $(X,Y)$ be a point randomly chosen on region $R = \{(x,y): |x| + |y| \leq 1 \}$. Find the marginal densities of X and Y.
The graph looks like a diagonal square with vertices at $x=1,x=-1,y=1,y=-1$. In order to find the marginal density of $x$, I need to find the following integral:
$$f_x(x)=\int^{x+1}_{-x-1}f_{x,y}(x,y)dy$$
I had no idea how to proceed given the absolute values, until I referred to the solution which gave the next step as $$f_x(x)=\int^{x+1}_{-x-1}f_{x,y}(x,y)dy=\int^{x+1}_{-x-1}\frac{1}{2}dy$$
Can anyone point out to me where did the $\frac{1}{2}$ come from?
Thank you!
Since we are dealing with absolute values, we are going to split this problem into two cases: $x > 0$ and $x < 0$.
If $x \geq 0$, then $|y| \leq 1-x$ and thus $x-1 \leq y \leq 1-x$.
If $x < 0$, then $|y| \leq 1 + x$ and thus $-x-1 \leq y \leq 1+x$.
Now, since we assume every point within $R$ has the same chance of occurring, the joint density function of $(X,Y)$ is
$$f_{X,Y}(x,y) = \frac{1}{\text{area of diagonal square}} = \frac{1}{2}.$$
To obtain the marginal density of $X$ we have to integrate over all the values that $y$ can take on, and thus
\begin{align} f_X(x) &= \int_{-x-1}^{1+x} \frac{1}{2}\ dy \ \mathbb{1}_{[-1,0)}(x) \ + \int_{x-1}^{1-x} \frac{1}{2}\ dy \ \mathbb{1}_{[0,1]}(x) \\[0.5em] &= (x+1)\ \mathbb{1}_{[-1,0)}(x) \ + \ (1-x)\mathbb{1}_{[0,1]}(x). \end{align}
So, we can write the marginal density as follows: $$f_X(x) = \begin{cases}x+1, & -1 \leq x < 0 \\ 1-x, & \ \ \ 0 \leq x \leq 1 \\ 0,& \ \ \ \text{otherwise}. \tag{1}\end{cases}$$
If you graph $(1)$ you will see that the marginal density of $X$ is has the form of a triangle with vertices on $(-1,0),(0,1)$ and $(1,0)$.