# Am I obtaining the Incorrect Equation for this Joint Probability density Function?

I am confused over how I am messing the equation up to solve this problem listed below

I was asked in Part C to find P(X > Y).

Therefore I made a double integral as such:

The answer to the problem is 15/16, but why I am I getting -1 by doing it this way?

Am I calculating the equation incorrectly?

• but for $y > 1$ the integral $\int_y^1 f$ becomes negative for positive $f$… – Gono Nov 14 '18 at 11:53

Your integration domain is just wrong, you don't consider the restriction $$X < 1$$, e.g. $$y = \frac{3}{2}$$ is part of your domain and for this the inner integral becomes negative. Nevertheless $$y = \frac{3}{2}$$ is not valid for the domain $$\{X > Y\}$$
So your double integral should look like $$\int_0^1 \int_y^1 \ldots$$ because $$X>Y$$ implies $$y < 1$$.
• No… I'm saying that you have to restrict the range of Y to the given restriction $$\{X > Y\}$$ And this domain can be written as $$0 < y < 1, y < x < 1$$ what then gets your integration domain. – Gono Nov 14 '18 at 12:06