Find the common PDF of $|X|,|Y|$ 
Consider $X,Y$ R.Vs which their common PDF is uniform on the triangle $(-1,-1),(-1,1),(1,-1)$
Find the common PDF of $|X|,|Y|$

I've found that $f_{X,Y}=1/2$ but didn't really know how to work with $|X|,|Y|$ any hint will be welcome :)
 A: Well, I'll do it for $X$.
Clearly $\mathbb{P}(|X|\leq1)=1$. Now, with $0\leq k \leq 1$,  $\mathbb{P}(|X|\leq k) = \frac12\int\int_T \chi_{|x|\leq k} dx dy = \frac12\int_{-k}^k\int_{-1}^{-x}dydx = \frac12\int_{-k}^k(1-x)dx = \frac12\big(2k\big) = k$.
So $|X|$ is uniformly distributed on $[0,1]$.
A: We know that for $0 \le a < b$ and $0 \le c < d$
$$
\begin{align}
\mathbb P((|X|, |Y|) \in (a, b) \times (c, d))  =& \mathbb P((X, Y) \in (a, b) \times (c,d)) \\
&+ \mathbb P((X, Y) \in -(a, b) \times (c,d)) \\
&+ \mathbb P((X, Y) \in (a, b) \times -(c,d)) \\
&+ \mathbb P((X, Y) \in -(a, b) \times -(c,d)) .
\end{align}
$$
But we know what the summands on the right hand side are: Suppose that $f_{X, Y}$ equals the distribution function of $(X, Y)$. Then
$$
\begin{align}
\mathbb P((X, Y) \in (a, b) \times (c,d))& = \int_a^b \int_c^d f_{X, Y}(x, y) dx dy, \\
\mathbb P((X, Y) \in -(a, b) \times (c,d))& = \int_{-b}^{-a} \int_c^d f_{X, Y}(x, y) dx dy \\
& = \int_a^b \int_c^d f_{X, Y}(-x, y) dx dy, \\
\mathbb P((X, Y) \in (a, b) \times -(c,d)) &= \int_a^b \int_c^d f_{X, Y}(x, -y) dx dy, \text{ and} \\
\mathbb P((X, Y) \in -(a, b) \times -(c,d)) &= \int_a^b \int_c^d f_{X, Y}(-x, -y) dx dy.
\end{align}
$$
From this, we get that the probability distribution function on the upper right quadrant is
$$
f_{X, Y}(x, y) + f_{X, Y}(-x, y) + f_{X, Y}(x, -y) + f_{X, Y}(-x, -y).
$$
Since both components of $(|X|, |Y|)$ are non-negative, the probability distribution function is zero on the other quadrants.
Now you just note that
$$
f_{X, Y}(x, y) = \mathbf 1_{x + y \le 0 \wedge x \ge -1 \wedge y \ge -1}
$$
and substitute that into the above expression.
