# [SOLVED] question on markov inequality and functions

• Oct 26th 2008, 08:19 PM
joop
[SOLVED] question on markov inequality and functions
I want to prove that $\displaystyle Pr( Z \ge a) \le {E(g(Z)) \over g(a)}$ given that $\displaystyle g(x) \ge 0 \forall x \in\mathbb{R}$ and that for any
$\displaystyle x_1, x_2 \mbox{ if }x_1 < x_2 \mbox{ then }g(x_1) < g(x_2)$.
Finally I was given the formula
$\displaystyle I_{[z \ge a]} = \left\{\begin{matrix} 1, & \mbox{if }z \ge a \\ 0, & \mbox{otherwise} \end{matrix}\right.$
and the inequality
$\displaystyle g(a)I_{[Z \ge a]} \le g(Z)I_{[Z \ge a]} \le g(Z)$
which I am supposed to use somehow.

Any help will be appreciated.
• Oct 26th 2008, 11:04 PM
NonCommAlg
Quote:

Originally Posted by joop
I want to prove that $\displaystyle Pr( Z \ge a) \le {E(g(Z)) \over g(a)}$ given that $\displaystyle g(x) \ge 0 \forall x \in\mathbb{R}$ and that for any $\displaystyle x_1, x_2 \mbox{ if }x_1 < x_2 \mbox{ then }g(x_1) < g(x_2)$.

you've already used the fact that g is increasing to show that: $\displaystyle g(Z) \geq g(a)I_{[Z \ge a]}.$ thus: $\displaystyle \text{E}(g(Z)) \geq \text{E}(g(a)I_{[Z \ge a]})=g(a) \text{E}(I_{[Z \ge a]})=g(a)\text{Pr}(Z \geq a).$

now since g(x) is a positive function, you can divide both sides of the inequality by g(a) to finish the proof.
• Oct 26th 2008, 11:16 PM
joop
Thank you very much! I was simply not seeing the equality:

$\displaystyle \text{E}(I_{[Z \ge a]})=\text{Pr}(Z \geq a)$