Suppose $X_1, X_2, X_3$ are iid random variables with uniform$[0,1]$. I'm trying to find $\mathbb{P}(X_1 + X_2 > X_3)$. I know that there are answers elsewhere (it's $\frac{5}{6}$). To improve my familiarity with the convolution formula, I decided to do it this way. First, $Y = X_1 + X_2$ has a pdf that looks like a triangle, i.e. $f_Y(y) = \chi_{[0,1]} y + \chi_{[1,2]} (2-y)$. Then I define $Z = Y - X_3$, where $-X_3$ follows uniform$[-1,0]$ and apply the convolution formula
$$f_Z(z) = \int f_Y(y) f_{-X_3}(z - y) dy$$
To figure out the limits of integration, note that $z - y \in [-1,0] \implies y \in [z, z+1]$ so
$$f_Z(z) = \int_{z}^{z+1} f_Y(y)$$
When $z$ is between $-1$ and $0$,
$$f_Z(z) = \int_{z}^{1} z dz + \int_{1}^{1-z} 2 - z dz = 1/2 - z^2 - z$$
And integrating $\int_{-1}^0 f_Z(z) = \int_{-1}^0 1/2 - z^2 - z dz = \frac{2}{3}$.
Then $\mathbb{P}(Z > 0) = \frac{1}{3}$ (but it's supposed to be $\frac{5}{6}$).
What went wrong? I'm completely stuck.
When $-1 <z<0$ we have $\int_z^{z+1} f_Y(y)dy=\int_0^{z+1}ydy$ because $z+1 $ is between $0$ a and $1$ and $f_Y(y)$ vanishes for $y <0$. The value of this integral
is $\frac 1 6$ so the required probability is $1-\frac 16=\frac 5 6$.