Given two independent random variables X and Y, where CDF and PDF for X are: \begin{equation} F_X(x) = \Biggl\{ \begin{array}{ll} 0 & \mbox{if $x \le 0$}\\ 1-(x-a)^2/a^2 & \mbox{if $0\le x \le a$}\\ 1 & \mbox{if $x\geq a$} \end{array} \ \end{equation} \begin{equation} f_X(x) = \Biggl\{ \begin{array}{ll} \frac{2}{a}\Big(1-\frac{x}{a}\Big) & \mbox{if $0\le x \le a$}\\ 0 & \mbox{otherwise} \end{array} \ \end{equation}
$Y$ has the very same distributions.
Consider now a new random variable $D$ given by $D = X+Y$.
Obtain the CDF of $D$ ( $F_D(d)$ ).
What I've done so far is to use the convolution formula over two intervals: when $d \le a$ and $a\le d$.
When $d \le a$: \begin{equation} F_D(d)=\int^d_0 f_X(x) F_Y(d-x)dx\\ F_D(d)=\int^d_0 \frac{2}{a}\Big(1-\frac{x}{a}\Big)(1-(d-x-a)^2/a^2)dx\\ F_D(d)=\frac{d^2(d-6a)(d-2a)}{6a^4} \end{equation}
For $a \le d$:
\begin{equation} F_D(d)=\int^a_{d-a} f_X(x) F_Y(d-x)dx\\ F_D(d)=\int^a_{d-a}\frac{2}{a}\Big(1-\frac{x}{a}\Big)(1-(d-x-a)^2/a^2)dx\\ F_D(d)=-\frac{(d-2a)^2(d^2-4ad-2a^2)}{6a^4} \end{equation}
However, when I checked if the CDFs converges to 1 as $d$ increases, I obtained nonesense. Can someone tell me what I'm doing wrong?? Thank you
When $a\leq d$, there may be values of $X\leq d-a$ such that $X+Y\leq d$. You are missing the corresponding term in your $a\leq d$ case, i.e. the first line of that case should read $F_D(d) = \int_{d-a}^ a f_X(x) F_Y(d-x) dx + F_X(d-a)$.