Multivariate probability function and interval of integral Cheers, I am given a density function of $f_{x,y} = \begin{cases} 2 , 0 < y < x < 1 \\ 0 , \text{elsewhere} \end{cases}$
I am asked to find $P(X > 2Y)$, but I don't know what methodology I should follow to solve it. My mind first tried solving for $\int \int_{2y}^1 2dxdy$ but I don't know what interval to place on the second one.
Also, the solution I am given tells me to graph the intervals, but I don't know how, and also it states that the right interval is $\int_0^1\int^\frac{x}{2}_0 2dydx$. My question is why is dx's integral on [0,1] if $x$ only goes from $y<x<1$? Any help would be greatly apperciated =)
 A: You want to integrate over the intersection of $x>2y$ and $0<y<x<1$.
  So if you place $y$ as the outer integral ...
$$\int_0^{1/2}\int_{2y}^1 2\,\mathrm d x\,\mathrm dy$$
Because when $y>1/2$ then the bound for $x$ would be contradictory (ie $x<1$ and $x>2y$ when $2y>1$).

  Or if you place $x$ as the outer integral ...
$$\int_0^{1}\int_{0}^{x/2}\,\mathrm d y\,\mathrm dx$$

My question is why is dx's integral on [0,1] if x only goes from y<x<1? Any help would be greatly aoperciated =)

The bounds for the outer integral cannot directly reference the inner integral's bound variable.  Bound variables do not reach outside the scope of their integrand.
Rather, as above, you just need to ensure the inner interal's domain is supported for all of the outer integrals' values. .


Also, the solution I am given tells me to graph the intervals,

Just graph the triangle $\{\langle x,y\rangle:0<y<x<1\}$ and area under $2y<x$.  Where they intersect is where you integrate, namely the triangle $\langle 0,0\rangle\langle 1,0\rangle\langle1,1/2\rangle$ .
