# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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.