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?
Best Answer
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.