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

Given this Problem here

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

Therefore I made a double integral as such:

My Equation

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?

  • $\begingroup$ but for $y > 1$ the integral $\int_y^1 f$ becomes negative for positive $f$… $\endgroup$ – 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 $y < 1$.

  • $\begingroup$ Sorry to bother you again, but are you saying that I should use the range of X for the outer bound integral? This is because since 2 is not in the range of X, I should use the range 0< x< 1 $\endgroup$ – FoolishNumber Nov 14 '18 at 12:05
  • $\begingroup$ 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. $\endgroup$ – Gono Nov 14 '18 at 12:06
  • $\begingroup$ I see. I believe I understand now. Thank you so much! $\endgroup$ – FoolishNumber Nov 14 '18 at 12:09

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