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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.

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Yes. $P(X>t, X\geq Y) \leq P(X>t)$ because $P(A\cap B) \leq P(A)$. –  Patrick Li Oct 6 '12 at 17:03
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1 Answer

up vote 0 down vote accepted

Let $$ g(y) = \Pr(X>t\mid X\ge y\ \&\ Y=y) = \Pr(X>t \mid X\ge y)\quad\text{by indedpendence.} $$ Show that $\Pr(X>t\mid X\ge y)\ge \Pr(X>t)$ regardless of the value of $y$.

Then observe that $\Pr(X>t) = \mathbb E(g(Y))$ and go on from there.

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Thanks for putting me on the right track! –  ActuariallyImpaired Oct 6 '12 at 18:16
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