Htykuut

From SOA sample 138:

A machine consists of two components, whose lifetimes have the joint density function

$$f(x,y)=
begin{cases}
{1over50}, & text{for }x>0,y>0,x+y<10 \
0, & text{otherwise}
end{cases}$$

The machine operates until both components fail.
Calculate the expected operational time of the machine.

I was able to get the solution by deciding on case $Y>X$, drawing a triangle with vertices at $(0,0)$, $(0,10)$, and $(5,5)$ and then integrating $int_0^5 int_{x}^{10-x}{yover50} dy dx$ to get $2.5$, and then doubling by symmetry for the case of $X>Y$ to get $5$.

However the SOA solution does it a different way that I am trying to understand:

Suppose the component represented by the random variable $X$ fails last. This is represented by
the triangle with vertices at $(0, 0)$, $(10, 0)$ and $(5, 5)$. Because the density is uniform over this
region, the mean value of $X$ and thus the expected operational time of the machine is 5. By
symmetry, if the component represented by the random variable $Y$ fails last, the expected
operational time of the machine is also $5$. Thus, the unconditional expected operational time of
the machine must be $5$ as well.

I bolded the part that I do not understand. Where do they get $5$ just by looking at the triangle, which is $25$ in area?

It has naught to do with the area.

The isosceles triangle, $triangle(0,0)(10,0)(5,5)$, is symmetrical about the vertical line through $overline{~(5,0)(5,5)~}$, and the density inside that shape is uniform.   Therefore the (conditional) mean value of $X$ given it is inside that shape is $5$.

$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$