Double Integral Calculation

Question

I am confused as to how the red arrow step was preformed.

If I type the same integral into Maple I get $1-e^{-x}-e^{-y}+e^{-x-y}$ which is the same that I manually calculated, clearly not the same answer?

Thanks

Answer 1

It seems that the pdf's in question are $\equiv0$ on the negative real axis, so that we can begin right away with $$F_{X.Y}(x,y):=\int_0^y\int_0^x e^{-u-v}\ du\>dv\qquad(x\geq 0, \ y\geq0)$$ and $\equiv0$ otherwise.

Now, since $e^{-u-v}=e^{-u}\>e^{-v}$, we have a "cartesian product situation" in every respect, and it is immediately obvious that $$\int_0^y\int_0^x e^{-u-v}\ du\>dv=\int_0^x e^{-u}\ du\cdot\int_0^y e^{-v}\ dv=\bigl(-e^{-u}\bigr)\biggr|_{u=0}^{u=x}\cdot \bigl(-e^{-v}\bigr)\biggr|_{v=0}^{v=y}\ .$$ But it is also possible to use Fubini's theorem: $$\int_0^y\int_0^x e^{-u-v}\ du\>dv=\int_0^y\left(\int_0^x e^{-u-v}\ du\right) dv=\int_0^y e^{-v}\bigl(1-e^{-x}\bigr)\> dv=(1-e^{-x})(1-e^{-y})\ .$$