1. ## independent variables

IF$\displaystyle X \;and\; Y$ are independent continuous random variables with $\displaystyle identical \;but\; unknown$ distributions.how can I calculate $\displaystyle P(X>Y)$.

I tried using this concept from the independence and conditional distribution:

$\displaystyle P(X>Y)= E[P(X>Y)|Y] = \displaystyle \int P(X>Y) \; f(y)\;dy = \displaystyle \int (1-F_{X}(y))\; f(y)\;dy$

is this correct?

2. $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^xf(x)f(y)dydx$

$\displaystyle =\int_{-\infty}^{\infty}f(x)\int_{-\infty}^xf(y)dydx$

$\displaystyle =\int_{-\infty}^{\infty}f(x)F(x)dx$

$\displaystyle ={[F(x)]^2\over 2}\Bigr|^{\infty}_{-\infty}$

=.5-0=.5

3. Thank you so much. I have one more thing to ask. The second integral has limit going from -infinity to x. Is it because we are finding P(X>Y)?

4. because y is bounded above by x, this is just basic calculus

5. Originally Posted by chutiya
IF$\displaystyle X \;and\; Y$ are independent continuous random variables with $\displaystyle identical \;but\; unknown$ distributions.how can I calculate $\displaystyle P(X>Y)$.

I tried using this concept from the independence and conditional distribution:

$\displaystyle P(X>Y)= E[P(X>Y)|Y] = \displaystyle \int P(X>Y) \; f(y)\;dy = \displaystyle \int (1-F_{X}(y))\; f(y)\;dy$

is this correct?
$\displaystyle 1=P(X>Y)+P(Y>X)+P(X=Y)$

but $\displaystyle P(X>Y)=P(Y>X)$ by symmetry and for any continuous distribution $\displaystyle P(X=Y)=0$

So the result follows without doing any integrals.

CB

6. Originally Posted by matheagle
because y is bounded above by x, this is just basic calculus

thank you.. i am feeling so dumb!

lastly, i am getting confused on the third step of your second post..

isn't $\displaystyle \int_{-\infty}^{\infty}}f(x)F(x)dx ={[F(x)]^2}\Bigr|^{\infty}_{-\infty}$?

7. F(infty)=1
F(-infty)=0
those are really limits, but it's after 2AM

8. Originally Posted by chutiya
thank you.. i am feeling so dumb!

lastly, i am getting confused on the third step of your second post..

isn't $\displaystyle \int_{-\infty}^{\infty}}f(x)F(x)dx ={[F(x)]^2}\Bigr|^{\infty}_{-\infty}$?
NO

Calc 1, let u=F(x), then du/dx=F'(x)=f(x)

AND $\displaystyle \int udu=u^2/2+c$