Convolution of two uniform probability densities (two square waves)

Question

Assuming $X$ and $Y$ are i.i.d. random variables let $Z = X + Y$

$$ f_X(x) = f_Y(y) = \begin{cases} 1/2 & -1 \le x \le 1 \\ 0 & \text{else} \end{cases} $$

Find the density $f_z$ of $Z$.

I am attempting to do this using the convolution of $f_X(x) * f_Y(y)$.

$$ f_X(x) * f_Y(y) = \int_{-\infty}^{\infty} f_X(x)f_Y(z-x) dx \\ = \int_{0}^{2} \frac{1}{2}(z - \frac{1}{2}) dx \\ = \frac{1}{2} \left[ zx - \frac{1}{2}x \right]_{0}^2 \\ = \frac{2z - 1}{2} $$

I understand the intuitive approach to this problem. i.e. the signals & systems geometric approach. And (I think) I also understand what the integral in the definition of convolution is actually doing. That being said I know the answer should be a piecewise function separated at intervals $[-2,0]$ and $[0,2]$. And I know my answer cannot be correct as it has a positive slope for the interval $[0,2]$ which should most certainly have negative slope.

Does anyone know what I am doing wrong here? Additionally, does anyone have any tips for understanding how to change the limits of integration other than imagining the graphical convolution?

Answer 1

It is not correct to replace $f_y(z-x)$ with $z - \frac12$.

There are only two possible values for $f_y(z-x)$: it can be either $\frac12$ or $0$. What you need to do is to determine which values of $x$ make $f_y(z-x) = \frac12$ and which values of $x$ make $f_y(z-x) = 0$. Then split the integral into integrals over disjoint intervals so that you are integrating with $\frac12$ in place of $f_y(z-x)$ over the interval where $f_y(z-x) = \frac12$, and integrating with $0$ in place of $f_y(z-x)$ over the interval where $f_y(z-x) = 0$.

The approach with indicators (see other answer) is a more explicit notation for the same idea.

Answer 2

We have to take into account the indicators: on has $f_X(x)=f_X(x)=2^{-1}\mathbf{1}_{[-1,1]}(x)$ hence we actually have $$ f_Z(z)=\int_{-\infty}^\infty 2^{-1}\mathbf{1}_{-1\leqslant x\leqslant 1}2^{-1}2^{-1}\mathbf{1}_{-1\leqslant z-x\leqslant 1}dx. $$ The product of indicators can be written as $\mathbf{1}_{.\max\{z-1,-1\}\leqslant x\leqslant \min\{1,z+1\}}$. For some values of $z$, this indicator is $0$.