Sum of random variables uniformly distributed (0,1) and (0,2) I'm trying to get $P(0.9<Y<=1.8)$ for the sum of 2 random and uniform values x1,x2 (so that y=x1+x2) where $x1$~$u(0,1)$ and $x2$~$(0,2)$ and I'm trying to do the convolution for it.  Seems like $\int\limits_a^b\int_0^2 xf(x)\,\mathrm{d}x$ where a=0.9, b=1.8 and which seems like a logical way to start.  I'm not comfortable with convolution but I'm trying to understand the step-by-step reason that this is the proper equation, and how the problem can be solved.  I'd like to understand as much about this as possible so I'd like to also compare that problem to finding the $P(0.9<Y<=1.8)$ for just $x1$~$u(0,1)$ where the values summed are independent but still over the same (0,1) area, and also compare it to $P(0.9<Y<=1.8)$ for $exp(2)$ where lambda is 2, which I'm also not really understanding the summed distribution. 
 A: The answer to your question is $\frac{179}{400}$. First you need to make the convolution of two $p.d.f.$s. However note that this operation is valid only for the independent random variables $X_1$ and $X_2$. I assume they are independent and continue with the solution.
The result of the convolution will be on the positive $y$ axis, having non zero values between $0$ and $3$. The density of $Y$ is a linear increasing function from $0$ to $1/2$ for $y\in [0,1]$, a constant function $p_Y(y)=1/2$ when $y\in [1,2]$ and a linear decreasing function from $1/2$ to $0$ when $y\in [2,3]$. 
What remains to do is to find the area under this $p.d.f.$. If you draw and calculate the area either with integrals or using some simple geometric relations, you will find that the area under the curve w.r.t. the square is $0.4$ and the area with respect to the left triange is $19/400$, which adds up to $\frac{179}{400}$.
A: Here is a blog post describing something similar to the problem you are attempting to solve. Convolution is the way to go.
