Uniform Distribution and Distribution function technique Let $X_1$ and $X_2$ be independent random variables having the uniform density with $\alpha = 0$ and $\beta = 1$. Find expressions for the function $Y =X_1 + X_2$.
(a)$y \le 0$
(b)$0<y<1$
(c)$1<y<2$
(d)$y\ge2$
I'm thinking $f(x_1)=f(x_2) = 1$ for $0 \le x_1\le 1$ and $0 \le x_2\le 1$. I suppose this implies that $f(x_1,x_2) = 1$ so $F(x_1,x_2) = \int_0^y\int_0^{y-x_2} f(x_1,x_2) dx_1dx_2 $ Then I would need $\frac{d}{dy} F(x_1,x_2)$ to find an appropriate $f(y)$
I already have the answers to these in my book but I don't know how to come upon them as they don't correspond at all to what I have here. Where is my fault?

You have $P(X_1 \leq u) = P(X_2 \leq u) = u$ for $0 \leq u \leq 1$ and $X_1,X_2$ independent. Thus $$
  P(X_1 + X_2 \leq u) = \int_{(x_1,x_2) \in A_u}  \,d(P_{X_1} \times P_{X_2}) \text{ where } A_u = \{(x_1,x_2) \in [0,1]^2 \mid x_1 + x_2 \leq u\} \text{.}
$$
which yields the CDF (note that the density of $X_1$ and $X_2$ on $[0,1]$ is $1$)  for $u \in [0,2]$ $$ \begin{eqnarray}
  F_{X_1+X_2}(u) &=& \int_0^u \int_0^{u-x_1} dP_{X_2}(x_2)\,dP_{X_1}(x_1)
  = \int_0^{\min\{u,1\}} \int_0^{\min\{u-x_1,1\}}  \,d x_1 \,d x_2 \\
  &=& \int_0^{\min\{u,1\}} \min\{u-x_1,1\} \,dx_1 \\
  &=& \begin{cases}
     \int_0^{u-1} 1 \,dx_1 +  \int_{u-1}^1 u - x_1 \,dx_1 & \text{$u > 1$} \\
     \int_0^u u - x_1 \,dx_1 &\text{$u \leq 1$}
  \end{cases} \\
  &=& \begin{cases}
     (u-1) + (1 - (u-1))u - \frac{1}{2}(1^2 - (u-1)^2)  & \text{$u > 1$} \\
     u^2 - \frac{1}{2}(u^2 - 0^2) &\text{$u \leq 1$}
  \end{cases} \\
  &=& \begin{cases}
      1 &\text{$u > 2$} \\
      -\frac{1}{2}u^2 + 2u - 1 & \text{$u > 1$} \\
      \frac{1}{2}u^2 &\text{$u \leq 1$} \\
      0 &\text{$u  < 0$.}
  \end{cases}
\end{eqnarray}$$
Note that the CDF is continuous as expected, so for any particular $u$, $P(X_1 + X_2 = u) = 0$. Using the CDF and this fact, you get


*

*$P(Y \in (0,1)) = P(Y \leq 1) - P(Y \leq 0) = \frac{1}{2} - 0$

*$P(Y \in (1,2)) = P(Y \leq 2) - P(Y \leq 1) = 1 - \frac{1}{2} = \frac{1}{2}$


For the other two it's immediately obvious even without the CDF that


*

*$P(Y \leq 0) = 0$

*$P(Y \geq 2) = 0$

