Y is a uniform distribution on [0,1] Let $U=Y_1+Y_2$ Find the Cdf of U.
For $1\leq u\leq2$ we can describe cdf of U as the proportional volume of the shaded area below to 1, for every u.
The example goes on to find the shaded area by 1-A where $A=\int_{u-1}^{1} {\int_{u-y_1}^{1} 1 dy_2} dy_1$.
This makes sense, but when I tried calculating the shaded region using 2 other methods, I got different answers. My issues may not just be with the random variable part but with the double integration part too.
Method 1:
Instead of calculating the area of A, I calculated the area of the red shaded region, which by the (-1) gradient of $y_2=-y_1 +u$ should by the same.
Then the integral becomes: $\int_{u-1}^{1} \int_{u-1}^{u-y_1} 1 dy_2 dy_1$ but this evaluates to something different than $A=\int_{u-1}^{1} {\int_{u-y_1}^{1} 1 dy_2} dy_1$. Why is this?
Method 2:
Just integrating the $u=y_1+y_2$ line from $0\leq y_1\leq<1$ and then subtracting from it the red triangle of area $(u-1)^2$
I know the fact that the function we're integrating =1 gives us a lot of leniency in terms of evaluating the problem as simple prisms, but if we were technical here, does the fact that the pdf of Y is discontinuous at 1 mean that we can't integrate the interval containing 1? ie [0,u] where u>1
All three of your methods should give the same answer $\dfrac{(2-u)^2}{2}$
Method 0: $$\int_{u-1}^{1} {\int_{u-y_1}^{1} 1 \,dy_2} \,dy_1= \int_{u-1}^{1} {(1-u+y_1)} \,dy_1 \\= \frac{(2-u)^2}{2}$$
Method 1: $$\int_{u-1}^{1} \int_{u-1}^{u-y_1} 1 \,dy_2 \,dy_1= \int_{u-1}^{1} {(1-y_1)} \,dy_1 \\= \frac{(2-u)^2}{2}$$
Method 2: $$1 - \left(\int_{0}^{1} {(u-y_1)} \,dy_1 - \int_{0}^{u-1} {(u-y_1-1)} \,dy_1 \right) =1-\left(\left(u-\frac12\right) -\frac{(u-1)^2}{2}\right) \\= \frac{(2-u)^2}{2}$$