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A joint distribution in $X$ and $Y$ is given as;

$$f(x,y) = 2e^{-x}e^{-2y} \ \ \ 0< x< \infty, \ \ 0< y< \infty $$

a) Compute $P\{X>1, Y<1\}$

b) Compute $P\{X < Y\}$

What is a general startegy for approaching these types of questions? Along which variable should I integrate first?

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I take it you mean $f$ is the density function.

Since this factors as a function depending only on $x$ times another depending only on $y$, it follows that the two components are independent. Use that in item (a). Those functions of $x$ and $y$ are the densities, up to scalar multiples (e.g. depending on how you factor it, you may need to multiply one of those by $2$ and the other by $1/2$, etc.).

For part (b), you're integrating over a subset of the plane. You could let $x$ go from $0$ to $\infty$ and then for each value of $x$, let $y$ go from $x$ to $\infty$. Or you could let $y$ go from $0$ to $\infty$ and then for each value of $y$, let $x$ go from $0$ to $y$. In some cases you might get an intractable integral one way and an easy one the other way, and you might want to just try both to see which one works. In this case, I suspect both will work without complications.

Later note in response to comments: I don't quite understand Imray's question below. But just to be more explicit: One way is this: $$ \int_0^\infty \left( \int_x^\infty f(x,y) \, dy \right) \, dx $$ This is the case where $x$ goes from $0$ to $\infty$, and then for each value of $x$, $y$ goes from $x$ to $\infty$. That way you integrate over the whole set $\{(x,y) : 0<x<y\}$. The other way is this: $$ \int_0^\infty \left( \int_0^y f(x,y) \, dx \right) \, dy $$ That is the case where $y$ goes from $0$ to $\infty$, and then for each value of $y$, $x$ goes from $0$ to $y$. Again, you're integrating over the wholse set $\{(x,y) : 0<x<y\}$.

In either case, you would evaluate the inner integral first.

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Quoting your answer: "Or you could let y go from 0 to inf and then for each value of y let x go from 0 to y". Do you mean "let go from to ? (because you want to get the area ABOVE . Do I have that right? –  Imray Oct 18 '12 at 21:56
    
@Imray : Are there some illegible characters in your comment? –  Michael Hardy Oct 18 '12 at 22:12
    
REDO - Quoting your answer: "Or you could let y go from 0 to inf and then for each value of y let x go from 0 to y". Do you mean "let x go from y to inf? (because you want to get the area ABOVE y=x. Do I have that right? –  Imray Oct 19 '12 at 0:06
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First sketch the region you want to integrate over in the $xy$ plane. This lets you see what the bounds of integration should be, in either order. Do it in whichever order seems most convenient.

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Will you always get the same answer regardless of order of integration? –  Imray Oct 18 '12 at 18:23
    
@Imray: Yes you will. –  Stefan Hansen Oct 18 '12 at 18:30
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