I have a c.d.f of variable X with a mass point at the end point,
$$F(x) = \begin{cases} 0 & x<a,\\ 1-\frac{m}{x+m-a} & a\le x < r-a,\\ 1 & x\ge r-a. \end{cases} $$ where m>0. Is it possible to get the analytic solution for the c.d.f of Y=X+X?
Thank you.
For simplicity, let's translate and scale so $a=0$ and $r=1$. One way to realize a random variable $X$ with your cdf is to have $X = B + (1-B) Y$ for independent random variables $B$ and $Y$, where $B$ is Bernoulli with $P(B = 1) = m/(m+1)$ and $Y$ is continuous with cdf $$F_Y(x) = \cases{0 & if $x \le 0$\cr (m+1) F_X(x) = \dfrac{(m+1) x}{x + m} & if $0 \le x \le 1$\cr 1 & if $x \ge 1$\cr}$$ and density $$f_Y(y) = \dfrac{m^2+m}{(x+m)^2} \ \mbox{for } 0 < x < 1, \ 0\ \mbox{otherwise}$$ Let $X_1 = B_1 + (1-B_1) Y_1$ and $X_2 = B_2 + (1-B_2) Y_2$ where $B_1, Y_1, B_2, Y_2$ are independent, so $X_1$ and $X_2$ have these distributions and are independent. Then $X_1 + X_2 = B_1 + B_2 + (1-B_1) Y_1 + (1-B_2) Y_2$. You can compute probabilities for $X_1 + X_2$ by conditioning on $B_1$ and $B_2$. For example, given $B_1 = B_2 = 0$, $X_1 + X_2 = Y_1 + Y_2$. The latter has a distribution which can be computed using some rather tedious integrals.