probability from joint density function [closed]

Could anyone help with this problem? Thanks

A joint density function is given as follows:

$$f(x,y) =\begin{cases}{} 0.125\cdot (x+y+1) \ \ \text{for} -1<x<1, 0<y<2 \\ 0, \text{otherwise} \end{cases}$$

Calculate $P(X>Y)$

closed as off-topic by Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly TaxelJul 26 '18 at 4:18

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• There is something missing here. Where do we define the function according to the formula? Surely not on the whole plane, it would not be a probability density function. – A. Pongrácz Jul 25 '18 at 9:39
• @A.Pongrácz I´ve fixed it. – callculus Jul 25 '18 at 9:40

Just recall what the density function represents: the probability of an event $A$ is the integral of the density function on $A$. So you have to inegrate the function on the set of points $A= \{(x,y) \mid x>y\}$. So $x$ can be any number in $[-1, 1]$, and $y$ has to be smaller than $x$.
Hence, compute $\int\limits_{-1}^{1} \int\limits_{0}^{x} f(x,y) \, dy \, dx$.
As you integrate 0 in the inner integral whenever $x$ is negative, it is the same as $\int\limits_{0}^{1} \int\limits_{0}^{x} f(x,y) \, dy \, dx$. You can easily compute this.