How do I find the cdf of $X_1 + X_2$?

Question

$X_1$ uniform $(0,1)$ and $X_2$ uniform $(0,2)$

$$ \begin{cases} f(x_1,x_2) = \frac{1}{2}, &\quad \mbox{for} \ 0<x_1<1, 0<x_2<2 \\ 0, & \quad \mbox{otherwise} \end{cases} $$

The density of $X_1+X_2$: $$ f(x_1,x_2) = \begin{cases} \displaystyle \frac{z}{2} & \quad 0<z<1\\ \displaystyle \frac{1}{2} & \quad 1<z<2\\ \displaystyle \frac{3-z}{2} & \quad 2<z<3\\ \end{cases} $$

The question is how do I find the cdf of $X_1+X_2$?

I would do: $$ f(x_1,x_2) = \begin{cases} \displaystyle \int_0^z \frac{z}{2} dx =\frac{z^2}{4}, & \quad 0<z<1\\ \displaystyle \int_0^1 \frac{1}{2} dx =\frac{1}{2}, & \quad 1<z<2\\ \end{cases} $$

That just seems wrong.. Help

Answer 1

The PDF of $Y = X_1 + X_2$ is the convolution of the PDFs of $X_1$ and $X_2$

$$f_Y (y) = \begin{cases} \dfrac{y}{2} & \text{if } y \in [0,1]\\\\ \dfrac{1}{2} & \text{if } y \in [1,2]\\\\ \dfrac{3-y}{2} & \text{if } y \in [2,3]\\\\ 0 & \text{otherwise}\end{cases}$$

Integrating, we obtain the CDF

$$F_Y (y) = \begin{cases} 0 & \text{if } y < 0\\\\ \dfrac{y^2}{4} & \text{if } y \in [0,1]\\\\ \dfrac{1}{4} + \dfrac{y-1}{2} & \text{if } y \in [1,2]\\\\ \dfrac{3}{4} + \dfrac{3}{2} (y-2) - \dfrac{1}{4} (y^2 - 2^2) & \text{if } y \in [2,3]\\\\ 1 & \text{if } y \geq 3\end{cases}$$

Answer 2

We have been given that: $$~f_{X_1,X_2}(s,t)= \tfrac 12 \mathbf 1_{s\in [0;1],t\in [0;2]}$$

  Where $~\mathbf 1_{E}=\begin{cases}1 & : E\\ 0 & :\textsf{otherwise}\end{cases}~~$ is an indicator function of event $E$.

Are you aware that the density of the sum is the convolution:?

$$\begin{align}f_{X_1+X_2}(z)~=~&\int_\Bbb R f_{X_1,X_2}(s, z-s)\operatorname d s \\[1ex] =~& \tfrac 12 \int_\Bbb R\mathbf 1_{s\in[0;1], s\in[z-2;z], z\in[0;3]}\operatorname d s \\[1ex] =~& \tfrac 12 \int_\Bbb R \mathbf 1_{s\in[0,z],z\in[0;1)}+\mathbf 1_{s\in[0,1],z\in[1;2)}+\mathbf 1_{s\in[z-2,1],z\in[2;3]}\operatorname d s \\[1ex] =~& \tfrac 12\big(z\,\mathbf 1_{z\in[0;1)}+\mathbf 1_{z\in[1;2)}+(3-z)\,\mathbf 1_{z\in[2;3]} \big) \\[2ex] f_{X_1+X_2}(z)~=~& \begin{cases}z/2 & : 0\leq z< 1\\ 1/2 & : 1\leq z< 2 \\ (3-z)/2 & : 2\leq z\leq 3 \\ 0 & : \textsf{otherwise} \end{cases} \end{align}$$

Then the CDF is $$\begin{align}F_{X_1+X_2}(z) ~=~& \int_{-\infty}^z f_{X_1+X_2}(u)\operatorname d u \\[1ex] =~& \tfrac 12 \begin{cases} 0 & : z<0 \\ \int_0^z u\operatorname d u & : 0\leq z< 1 \\ \int_0^1 u\operatorname d u + \int_1^z 1\operatorname d u & : 1\leq z< 2 \\ \int_0^1 u\operatorname d u + \int_1^2 1\operatorname d u + \int_2^z (3-u)\operatorname d u & : 2\leq z\leq 3 \\ 1 & : 3 < z \end{cases} \\[1ex] ~=~& \tfrac 12 \begin{cases} 0 & : z<0 \\ z^2/2 & : 0\leq z< 1 \\ 1/2 + (z-1) & : 1\leq z< 2 \\ 1/2 + 1 + (3z-z^2/2)-4 & : 2\leq z\leq 3 \\ 1 & : 3 < z \end{cases} \\[2ex] F_{X_1+X_2}(z) ~=~& \tfrac 14 \begin{cases} 0 & : z<0 \\ z^2 & : 0\leq z< 1 \\ 2z-1 & : 1\leq z< 2 \\ 4-(3-z)^2 & : 2\leq z\leq 3 \\ 1 & : 3 < z \end{cases} \end{align}$$