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

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

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 $0<x<y$

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

  • 1
    $\begingroup$ 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 $F_{X,Y}(y,y)$ $\endgroup$ – Henry May 23 at 17:09
  • $\begingroup$ Sorry, woulld that mean my bounds are incorrect? I thought x<y is omitted as it's 0<x<y $\endgroup$ – Jerry May 23 at 17:14
  • $\begingroup$ $$ \mathbb P(X\leqslant x, Y\leqslant y) = \int_0^{x\wedge y}\int_0^y f_{X,Y}(s,t)\ \mathsf dt\ \mathsf ds $$ $\endgroup$ – Math1000 May 23 at 17:24
  • $\begingroup$ @math1000 can you explain that please ? $\endgroup$ – Jerry May 23 at 17:33
  • $\begingroup$ 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 $0 \le y \lt x$ $\endgroup$ – Henry May 23 at 17:57

As we have $0<x<y<\infty$, 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 - $$\iint_R f_{XY}dx dy= \int_0^\infty\int_x^\infty f_{XY}dydx$$

  • $\begingroup$ For the first case, I got 1, and for the second case, I got -1. $\endgroup$ – Jerry May 24 at 0:30
  • $\begingroup$ Sorry, both are 1 $\endgroup$ – Jerry May 24 at 1:26

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