I have to determine the first order probability density function for $w(t, \epsilon)$, with the hypothesis that I have two statistically independent functions $x(t, \epsilon)$ and $y(t, \epsilon)$ uniformly distributed in $[-\frac12,\frac12]$.
$$ w(t,\epsilon) = z^2(t,\epsilon) \quad\text{and}\quad z^2(t,\epsilon) = (x(t,\epsilon)+y(t,\epsilon))^2 $$
I set a $t$ for the evalutation, and $\epsilon$ is given, so my functions become random variable, defined this way: $$ W = Z^2 = (X+Y)^2 $$
For any variable I have a probability density like: $p_x(x)$ for $x(t,\epsilon)$
So I can define that $p_z(z) = p_x(x) \cdot p_y(y)$.
Having $p_x(x) = \operatorname{rect}(x)$ and $p_y(y) = \operatorname{rect}(y)$, then $p_z(z) = \operatorname{tri}(z)$.
By having triangle as a result for $z$, I have to consider the $p_w(w)$ in two parts $(-\infty, 0)$ and $(0, +\infty)$, so in first part I got \begin{gather*} p_z(z) = (1+z) \cdot \operatorname{rect}\left( z + \frac{1}{2} \right), \\ z_1 = -\sqrt{w}, \\ p_z(z_1) = (1 - \sqrt{w}) \cdot \operatorname{rect}\left( w - \frac{1}{2} \right). \end{gather*} With this assumptions, how can I say that: $$ \operatorname{rect}\left( \frac{1}{2} - \sqrt{w} \right) = \operatorname{rect}\left( w - \frac{1}{2} \right) $$
Please, help me, I'm really going crazy.
Edit: $\operatorname{rect}$ is the rectangular function.