I need to calcuate a convolution of two uniform distributed random variables, both are defined as $f_z\left(z\right)=\frac{1}{h},z\in\left(0,h\right)$, to get the pdf for Z.
As a convolution of two continious random variables is defined $fz\left(z\right)=\int fx\left(z-y\right)fy\left(y\right)dy$ hence z = x + y, so x = z-y.
Based on this explanation my current idea for the integrand and the borders are:
$\int_{0}^{z}{\frac{1}{2h}dz}\ for\ 0\le\ z\ \le\ h $
and
$\int_{z-h}^{h}{\frac{1}{2h}dz}\ for\ z-h\le\ z\ \le\ 2h$
Which leads to a triangular function like
$ fz(z) = \begin{cases} \frac{1}{2h} \text{for 0 < z < h} \text{ and } 1 - \frac{1}{2h} \text{for z-h < z < 2h} \end{cases}$
Is that correct ?
