So in the problem I was given the joint pdf $f(x,y) = x + y$, $0< x <1$, $0< y <1$, $0$ elsewhere. I am tasked to find $P( x+y \le 1 )$.
My intuition was to $\int_0^1\int_{1-x}^1 (x+y) \, dy \, dx $ working that out, i got the answer of 50%. Is my intuition correct?
It should be $$ \int_0^1\int_0^{1-x}(x+y) \, dy \, dx $$ which evaluates to $\frac{1}{3}$.
Because the density is $x+y$, regions with high $x,y$ are more likely. So intuition should tell you that $\{x+y\leq 1\}$, the lower triangle of the unit rectangle, should have probability less than that of the upper triangle of the same rectangle. That is, you should expect $\Pr(x+y\leq 1)\leq\frac{1}{2}$.
