I'm having a bit of trouble proving this inequality:
$\Pr(X>t\mid X\ge Y) \ge \Pr(X>t)$
Where $t\in \Re^+$ and $X,Y$ are positive independent random variables.
I started off in the following way,
$\frac{\Pr(X>t,X\ge Y)}{\Pr(X\ge Y)}\ge \Pr(X>t,X\ge Y) = \int_t^\infty \int_0^x f_Y(y)f_X(x) \, dy \, dx=\int_t^\infty f_X(x) F_Y(x) \, dx$
From here I have no idea how to show it's bigger than $\int_t^\infty f_X(x) \, dx$, in fact I'm pretty sure that it isn't... Which leads me to believe my approach is wrong. I would greatly appreciate some help.
Thank you.