Finding a joint CDF F

Given the joint density of two random variables $$X$$ and $$Y$$,

$$f_{XY}(x,y)=2e^{-(x+y)}$$ for $$0

How do I find the joint CDF ?

I know it'll be:

$$F_{XY}(x,y)=\int\int_R f_{XY}(x,y)=\int\int_R2e^{-(x+y)}dxdy$$ for $$0

I am unsure what my regions would be but I am guessing it is from x to infinity and y to infinity.

• Be careful: $\int_{-\infty}^x f(x)\, dx$ confuses free variables and bound variables. You might anslo want to include a calculation of $F_{X,Y}(x,y)$ when $x \gt y$, though you might be able to spot this is equal to $F_{X,Y}(y,y)$ – Henry May 23 at 17:09
• Sorry, woulld that mean my bounds are incorrect? I thought x<y is omitted as it's 0<x<y – Jerry May 23 at 17:14
• $$\mathbb P(X\leqslant x, Y\leqslant y) = \int_0^{x\wedge y}\int_0^y f_{X,Y}(s,t)\ \mathsf dt\ \mathsf ds$$ – Math1000 May 23 at 17:24
• @math1000 can you explain that please ? – Jerry May 23 at 17:33
• The point is that $F_{X,Y}(x,y)$ is a function on the whole of $\mathbb R^2$. It is clearly $0$ if either $x$ or $y$ are negative, but you also need to deal with the case $0 \le x \le y$ as you do, and the case $0 \le y \lt x$ – Henry May 23 at 17:57

As we have $$0, if we take $$x$$ as dependent on $$y$$, we get - $$\iint_R f_{XY}dx dy= \int_0^\infty\int_0^yf_{XY}dxdy$$ If we would take $$y$$ as dependent on $$x$$, we get - $$\iint_R f_{XY}dx dy= \int_0^\infty\int_x^\infty f_{XY}dydx$$