Let $ X \sim (0,1) $ and $ Y \sim (-1,2)$ be independent. Compute the distribution function of $Z=X+Y$ - how to break into cases? 
Let $ X \sim (0,1) $ and $ Y \sim (-1,2)$ be independent. Compute the
  distribution function of $Z=X+Y$ - how to break into cases?

I first found the density functions:
$$
f_x(t) =\begin{cases}
1 && t\in[0,1] \\
0 && else 
\end{cases}
$$
$$
f_y(t) =\begin{cases}
\frac{1}{3} && t\in[-1,2] \\
0 && else 
\end{cases}
$$
Now:
$ F_z(t)=P(Z\leq t)=P(X+Y\leq t)=\int_{-1}^{2}\int_{0}^{t-y}\frac{1}{3}dxdy$
But now I am stuck with breaking the result into cases.
How could it be done? I simply can't understand how to break that into cases when we have two differently distributed variables? 
With one variable I would usually draw the function and then break to cases according to it's behavior, but how could it be done here?
Thanks
 A: By convolution theorem we get 
$$f_Z(z) = \int_{0}^{1} f_X(x) f_Y(z-x)\ \text{dx}.$$
In order to make non vanishing integrand we should have $-1 \leq z-x \leq 2$ i.e. $z-2 \leq x \leq z+1$. Observe that $-1 \leq z \leq 3 .$ Now there are three cases to consider
$(1)$ $z-2 < 0 \leq z+1 < 1.$ 
$(2)$ $z-2 < 0 < 1 \leq z+1.$
$(3)$ $0 \leq z-2 \leq 1 < z+1.$
If you analyze each of these cases you will find that the probability density function $f_Z$ of $Z = X + Y$ is defined as follows $:$
$$f_Z(z) = \begin{cases}
\frac 1 3(z+1) & \text{for $-1 \leq z < 0$} \\
\frac 1 3 & \text{for $0 \leq z < 2$} \\
\frac 1 3 (3-z) & \text{for}\ 2 \leq z \leq 3 \\
0 & \text{elsewhere}
\end{cases}$$
A: First notice that for the inside integral to contribute something you must have:
$0 \leq t-y\leq 1$
Which means that:
$\int_{-1}^{2}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy= \int_{t-1}^{t}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy$
But you should notice that:
$\int_{t-1}^{t}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy\neq \int_{-1}^{2}\int_{0}^{t-y}\frac{1}{3}dxdy$
But instead: $t\leq -1$ then $\int_{t}^{t-1}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy=0$, and if $t>2$ $\int_{t}^{t-1}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy=0$.  else:
$\int_{t-1}^{t}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy=\begin{cases}0 & ,t<-1 \\
 1 & ,t>2 \\
 \int_{(t-1)\vee -1}^{t}\int_{0}^{t-y} \frac{1}{3} dxdy & ,\text{else}
 \end{cases}$
A: Assuming that you know that the sum of two independent random variables is their convolution, we have 
$$F_z(t)= \int_{-\infty}^{\infty}f_X(t-\tau)f_Y(\tau) \ d \tau = \frac{1}{3}  \int_{-1}^{2}f_X(t-\tau)  \ d \tau = \frac{1}{3}\int_{t-2}^{t+1}f_X(x) \ dx$$
Now, we have cases according to the integral boundaries and the PDF of $X$, namely:
1) If $t+1<0$, i.e. $t<-1$, the integral evaluates to zero.
2)If $t-2>1$, i.e. $t>3$, the integral evaluates to zero.
3) If $t-2<0$ ($t<2$) given that $t+1<1$, i.e. $t<0$ So for $-1 < t < 0$
$$F_z(t)=\frac{1}{3}\int_{0}^{t+1}f_X(x) \ dx = \frac{1}{3}(t+1)$$
4) If $t-2 > 0$ ($t > 2$). So for $2 <t < 3$
$$F_z(t)=\frac{1}{3}\int_{t-2}^{1}f_X(x) \ dx = \frac{1}{3}(3-t)$$
5) Now $0 < t < 2$
$$F_z(t)=\frac{1}{3}\int_{t-2}^{t+1}f_X(x) \ dx =\frac{1}{3}\int_{0}^{1} 1 \ dx =\frac{1}{3}$$
