Why is this true? (sum of 2 uniform distributions) If $X\sim U[0,1]$ and $Y\sim U[-1,0]$ and they are independent, then the distribution of $X+Y$ is not simply $U\sim [-1, 1]$, but it is the sum of 2 independent $U\sim [-0.5 ,0.5]$ distributions.
Why are these not equal?
 A: Intuitively, $X+Y$ has much more chance of being around 0 than around 1 or -1, so it is definitely not uniform.
EDIT
To derive they are actually the same distributions, find the cdf by conditioning. Notice if $Z = X+Y$ then
$$\begin{split}
F_Z(z) &= \mathbb{P}[Z \leq z] = \mathbb{P}[X+Y \leq z] \\
       &= \int_y \lim_{a \to 0} \mathbb{P}[X \leq z-y | Y \in (y-a,y+a)]
                       * \mathbb{P}[Y \in (y-a,y+a)] \\
       &= \int_y \lim_{a \to 0} \mathbb{P}[X \leq z-y]
                       * \mathbb{P}[Y \in (y-a,y+a)] \\
       &= \int_y F_X(z-y) f_Y(y) dy
\end{split}
$$
both of which are known. Same argument applies for sum of 2 independent uniforms, and you should get identical cdf's
The next to last step
$$
\mathbb{P}[X \leq z-y | Y \in (y-a,y+a)] = \mathbb{P}[X \leq z-y]
$$
can be made because $X,Y$ are assumed independent.
A: Another way to see this is as follows: if $X \sim U[0,1]$ and $Y \sim U[-1,0]$, then we can write them as $X = 0.5 + W$ and $Y = -0.5 + Z$, where $W \sim U[-0.5, 0.5]$ and $Z \sim U[-0.5, 0.5]$.  Then $X + Y = (0.5 + W) + (-0.5 + Z) = W + Z$.
