Convolution CDF

Question

I have two CDF: $$ F_X = \begin{cases} 0 \ \ \ \ x<0\\0.8+0.1x \ \ \ x \in (0,1) \\ 1 \ \ \ \ x>1\end{cases} \\ F_Y = \begin{cases} 0 \ \ \ \ x<0\\0.7+0.2x \ \ \ x \in (0,1) \\ 1 \ \ \ \ x>1\end{cases} $$ What is the cumulative probability function of S= X+Y? I started but I can't finish it. I know formula: $ F_S(s)=\int\limits_{-\infty}^{\infty} F_X(s-x)dF_Y(x)$. I write $ F_X$ as $F_X(s)=0.8*1_{[0,\infty)}+0.1*1_{[1,\infty)}+\int_{-\infty}^{s} 1_{[0,1)}(x)dx $ $$ F_S(s)=0 \ \ \ s<0 \\F_S(0)=0.8*0.7=0.56 \\ F_S(s) = 0.8*(0.7+0.2s)+ 0.1\int_{0}^{s}0.7+0.2(s-x)dx=0.56+0.23s+0.01s^2 \ \ \ s \in (0,1) $$ How to calculate CDF for $ s \in (1,2)$?

Answer 1

Let me write a solution without using the convolution formula. I suppose that it is known the triangular distribution of sum of two independent uniform r.v.'s.

Decompose the CDF's to a linear combination of CDF's of discrete degenerate and absolutely continuous distributions.

For $F_X$ we can see that it has two jumps at $0$ and at $1$, and linear inside $(0,1)$. So, it is a linear combination with some weights of a CDF's of degenerate at $0$ and at $1$ distributions and of Uniform $(0,1)$ distribution.
$$F_X(x)=0.8F_{0}(x)+0.1F_1(x)+0.1F_{U(0,1)}(x).$$

Finally, if we have two independent r.v.'s $U_1\sim \text{Uniform}(0,1)$ and $V_1$ such that $$\mathbb P(V_1=0)=0.8,\quad \mathbb P(V_1=1)=0.1, \quad \mathbb P(V_1=2)=0.1$$ and take $X$ as $$ X=0 \cdot {\mathbb 1}_{\{V_1=0\}}+1\cdot {\mathbb 1}_{\{V_1=1\}}+U_1\cdot {\mathbb 1}_{\{V_1=2\}}={\mathbb 1}_{\{V_1=1\}}+U_1\cdot {\mathbb 1}_{\{V_1=2\}}, $$ we get a r.v. with predefined CDF $F_X(x)$.

Similar way, a r.v. $Y$ with CDF $F_Y(x)$ can be constructed with the help of independent r.v.'s $U_2\sim \text{Uniform}(0,1)$ and $V_2$ $$ \mathbb P(V_2=0)=0.7,\quad \mathbb P(V_2=1)=0.1, \quad \mathbb P(V_2=2)=0.2 $$ as $$ Y={\mathbb 1}_{\{V_2=1\}}+U_2\cdot {\mathbb 1}_{\{V_2=2\}}. $$

Consider the sum $X+Y$: $$ X+Y={\mathbb 1}_{\{V_1=1\}}+U_1\cdot {\mathbb 1}_{\{V_1=2\}}+{\mathbb 1}_{\{V_2=1\}}+U_2\cdot {\mathbb 1}_{\{V_2=2\}}. $$

Note that for two independent r.v.'s $V_1, V_2$, nine disjoint cases of values of this pair are possible: $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$, $(2,0)$, $(0,2)$, $(2,1)$, $(1,2)$ and $(2,2)$. Consider $X+Y$ on each of this events: \begin{align} X+Y & = 0\cdot {\mathbb 1}_{\{V_1=0,V_2=0\}}+1\cdot\left({\mathbb 1}_{\{V_1=0,V_2=1\}}+{\mathbb 1}_{\{V_1=1,V_2=0\}}\right)+ 2\cdot {\mathbb 1}_{\{V_1=1,V_2=1\}} \cr & + U_1\cdot {\mathbb 1}_{\{V_1=2,V_2=0\}} + U_2\cdot {\mathbb 1}_{\{V_1=0,V_2=2\}} \cr & +(U_1+1)\cdot {\mathbb 1}_{\{V_1=2,V_2=1\}}+(U_2+1)\cdot {\mathbb 1}_{\{V_1=1,V_2=2\}} \cr & +(U_1+U_2)\cdot {\mathbb 1}_{\{V_1=2,V_2=2\}}.\end{align} The decomposition of the CDF of $X+Y$ can be written in a more simple way. First calculate the probabilities $$\mathbb P(V_1=0,V_2=0)=0.56, \ \mathbb P(V_1=0,V_2=1\text{ or } V_1=1,V_2=0)=0.15, \ldots $$ and then $$ F_{X+Y}(x)=0.56F_0(x)+ 0.15 F_1(x)+0.01 F_2(x)+0.23 F_{U(0,1)}(x)+0.03 F_{U(1,2)}(x)+0.02F_{U_1+U_2}(x). $$ Here $F_{U(1,2)}(x)$ appears as the CDF of $U_1+1$ or $U_2+1$, and $$F_{U_1+U_2}(x)=\begin{cases}0, & x<0\cr \frac{x^2}{2}, & 0\leq x < 1\cr 1-\frac{(2-x)^2}{2}, & 1\leq x < 2\cr 1, & x\geq 2\end{cases} $$

Finally calculate the CDF on any interval separately and get $$ F_{X+Y}(x)=\begin{cases}0, & x<0\cr 0.56+0.23x+0.02\frac{x^2}{2}, & 0\leq x < 1\cr 0.56+ 0.15+0.23+0.03(x-1)+0.02\left(1-\frac{(2-x)^2}{2}\right), & 1\leq x < 2\cr 1, & x\geq 2\end{cases} = \begin{cases}0, & x<0\cr 0.56+0.23x+0.01 x^2, & 0\leq x < 1\cr 0.89+0.07x-0.01x^2, & 1\leq x < 2\cr 1, & x\geq 2\end{cases} $$