Can't determine limits for double integral

[nadroj]

This is from statistics homework, but the problem I'm having I think is just basic integration.

Given:
, where x > 0, y > 0, x+y < 1

To find P( X <=a, Y <=b), that is, the probability that X is less than or equal to some constant a, and Y is less than or equal to some constant b, you integrate f from 0 to a, with respect to x, and then 0 to b, with respect to y. (Or the other order).

Now, the actual question is that I need to find P(X+Y > 2/3), that is, the probability that (X+Y) is greater than (2/3). I never know how to setup the limits for questions like this. I think I have to basically find the intersection of all of the ranges, and those are the limits. I also think that the "outer" integral always has constants in the limits (opposed to variables say x and y). I think this is true because the result has to be a number, not a function of x or y. And if the outer limits do not contain constants, the result isn't a number (right?).

The answer is 5/9, but there is no solution of how to derive to that. Of course the actual integration is trivial, it's just setting up the question I don't know how to do.

I have tried things like and but I end up with 2/9 or -2/9. My reason that I think one of the limits must go from 2/3 to 1 is that the Y (or X) can be 0, in which case the other number must go from 2/3 to 1.

Also, the next one asks for P(X > 2Y) and the answer is 1/3. I don't have a clue for that one. The only one I got correct was P(X < 1/2, Y < 1/2), so the limits are straightforward.

Thanks for any help

[skipjack]

[nadroj]

You're good

The first one I seem to understand and can derive now.

Not too long ago, our GA posted handwritten solutions. For the second one, he did 1 - P(X < 2Y). The limits he chose were 0 < x < 2y, and 0 < y < 1/3. The result was 7/9. I had emailed someone asking which answer was correct (book or GAs). But yours seems to confirm that 1/3 is the correct answer.

I see that if we are given and , can we do this: let then so that the limits are and ? Is this how you go about determining the limits?

Alternatively, for this same question, can we let then so that the limits are and ?

Thanks again

EDIT: Apparently not! I just tried that and got . Is there only the one unique solution above?

[skipjack]

Consider the region of the x-y coordinate plane defined by the given inequalities. Are there integration limits that correspond to the same region? As the integrand is the constant 2, the integral gives twice the area of the region, and the answer can alternatively be obtained by geometry.

[nadroj]

I've tried to work it out geometrically, or at least trying to find the region that the limits define and was able to get it for the limits you proposed. I tried to do it geometrically but couldnt draw the picture. I imagine, whatever it may be, is just the same region, rotated about the line y = x. Oh well, Ill try to stick with the algebraic interpretation as to not confuse myself. Thanks again