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I'm having trouble finding the a different order of integration.

Here is the problem:

The joint probability distribution of $X$ and $Y$, f($x,y$),is given by:

$\frac{6}{7}(x^2 + \frac{xy}{2})$, where $0 < x < 1, 0 < y < 2$.

I'm supposed to calculate $P(X>Y)$.

I $can$ calculate $P(X>Y)$, by doing this:

$$\int_0^1 \int_0^x \! f(x,y) \, \mathrm{d} y \mathrm{d} x,$$ where $f(x,y)$ is the joint probability distribution of X and Y mentioned above. (FWIW, the solution I get is 15/56.)

$Now$, $this$ $is$ $where$ $my$ $problem$ $lies$. If I would have happened to integrate dx first, instead of dy, I need to change my limits of integration. I would have changed them to:

$$\int_0^2 \int_y^1 \! f(x,y) \, \mathrm{d} x \mathrm{d} y$$

This does not give the right answer. Thus, my question is: how do I set up this integral?

ps. Yes, this is homework (which isn't graded, btw). But I have already solved it. I'm just trying to understand a different way of doing the problem.

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    ZulfiqarIII has given a good answer. Put simply, in your second integral you need $y \le 1$ for the integral over $x$ and this needs to feed through into the limits of the integral over $y$.2012-07-20

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