Cfrgtkky

$X$ and $Y$ are random variables with joint density function given below

$f{_X}{_Y}(x,y)$ = $8xy$, where $0 < y < x < 1$

I know I am supposed to draw a region and then find new limits of integration

Could any one find the limits of integration for this one?

This one seems rather tricky since y is dependent on x.

So the entire region looks like a triangle contained within the unit square, and the integral over the whole region looks like the following:
$$
int_0^1int_0^x 8xy,dy,dx = 1
$$
We wish to calculate $P[X<2Y]$, which is the portion of the triangle above the line $y=x/2$. If I am not mistaken, this gives us the following integral:
$$
int_0^1int_{x/2}^x 8xy,dy,dx
$$

For the general problem of this kind you set up a double integral as follows. $$int_0^1int_0^x xy.dy.dx$$

Imagine a vertical strip of infinitesimal thickness $dx$ at some distance $x$ along the $x$-axis: you first integrate from $y=0$ to $y=r$, because the limit is the line $y=x$. The variable $x$ behaves as a constant in this integration, as you are integrating along a vertical strip along which $x$ is contant - at the value $x$. Because of this, the integral can be recast as $$int_0^1xint_0^x y.dy.dx .$$ Doing the inner integral first, you get $$int_0^1x.{x^2over2}.dx .$$ Now the integrand is a finction of $x$ alone, so you have $$int_0^1{x^3over2}.dx$$$$=$$$$left[{x^4over8}right]_0^1 = {1over8} .$$

I've just realised that your function had a factor of 8 in to begin with ... so the result here is 1.

This is a very pleasant double integral, in which you can use that trick of 'integrating-out' one of the variables first to get a function purely of the other, and then integrating with respect to that one. They aren't anywhere near so pleasant generally - mainly by reason of the region of integration not being a simple shape such as a triangle. Complexities of the function itself tend to be less of a problem ... as long as you do the inner integration over a region - in this case a vertical strip - over which the other variable is constant.

Actually, you could simplify this even more by noting that the integral is symmetrical in $x$ & $y$, whence you could have just put the limits [0,1] into both the inner and the outer integral, and halved it at the end. Tricks like that - and much more cunning ones - are often extremely efficacious in the solution of complicated double (or triple ... or higher-order, even) integrals.