# Math Help - sums of independent

1. ## sums of independent

If X and Y are both uniformly distributed on (0,1), calculate the probabily of X+Y
Here's what I have:
fx(a)=fy(a)=1 because both uniformaly distributed.

Then fx+y(a)=integral (0,1) of fx(a-y)dy
just found by rearranging x+y=a and fy(a)=1

Now, it says that fx+y(a)=integral (0,a) dy=a
My question is how they changed the integration limits

Why isn't it fx+y(a)=integral (0,1) dy=1?

Intuitively the first answer makes more sense, but I'm just confused with the math.

2. the sum of two indep U(0,1) is a triangular density
that's 0 when S=X+Y is less than 0 or greater than 2.
It's s on 0<s<1 and it's 2-s on 1<s<2.
The way to figure that out is via the CDF.
$P(S\le s)=P(X+Y\le s)$.
Then draw the unit square of x and y between (0,1) for both
and find this area WITHOUT integrating, use geometry.