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 $P(X>Y)$


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  • $\begingroup$ 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. $\endgroup$ – A. Pongrácz Jul 25 '18 at 9:39
  • $\begingroup$ @A.Pongrácz I´ve fixed it. $\endgroup$ – 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.

  • $\begingroup$ Thanks very much. I got stuck with the limits of the integrals- if the limit of the inner integral is from 0 to x, then I thought this would imply that x can be equal to y but we are asked to calculate P(X>Y) (and not P(X>=Y)) $\endgroup$ – Denson Jul 25 '18 at 10:05
  • $\begingroup$ Formally speaking, you are right. But the densitiy function is continuous, so it doesn't matter if you include/exclude points or even curves in/from the event, it does not affect the result. $\endgroup$ – A. Pongrácz Jul 25 '18 at 11:39

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