I want to prove thatgiven that
and that for any
.
Finally I was given the formula
![I_{[z \ge a]} = \left\{\begin{matrix} 1, & \mbox{if }z \ge a \\ 0, & \mbox{otherwise} \end{matrix}\right.](http://latex.codecogs.com/png.latex? I_{[z \ge a]} = \left\{\begin{matrix} 1, & \mbox{if }z \ge a \\ 0, & \mbox{otherwise} \end{matrix}\right.)
and the inequality
which I am supposed to use somehow.
Any help will be appreciated.
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I want to prove that
Finally I was given the formula
and the inequality
which I am supposed to use somehow.
Any help will be appreciated.
you've already used the fact that g is increasing to show that:Quote:
Originally Posted by joop
I want to prove that
thus:
![\text{E}(g(Z)) \geq \text{E}(g(a)I_{[Z \ge a]})=g(a) \text{E}(I_{[Z \ge a]})=g(a)\text{Pr}(Z \geq a).](http://latex.codecogs.com/png.latex?\text{E}(g(Z)) \geq \text{E}(g(a)I_{[Z \ge a]})=g(a) \text{E}(I_{[Z \ge a]})=g(a)\text{Pr}(Z \geq a).)
now since g(x) is a positive function, you can divide both sides of the inequality by g(a) to finish the proof.
Thank you very much! I was simply not seeing the equality: