0
$\begingroup$

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.

$\endgroup$
  • 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
2
$\begingroup$

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 - $$\iint_R f_{XY}dx dy= \int_0^\infty\int_x^\infty f_{XY}dydx$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.