# function of expectation

• Apr 17th 2013, 08:51 AM
Juju
function of expectation
Hello,

I have to show the following assertion:
$\displaystyle \sup\limits_{a \in [0,1]} E[log(1+a(exp(Z)-1))]=0$
I know that
$\displaystyle E[Z]\leq 0$ and $\displaystyle \sup\limits_{a \in [0,1]} E[log(1+a(exp(Z)-1))]\geq 0$
and for $\displaystyle a=1$ we have $\displaystyle \sup\limits_{a \in [0,1]} E[log(1+a(exp(Z)-1))]=E[Z]$

Unfortunately I have no idea how to show it.
Does anyone have a hint?
• Apr 17th 2013, 09:18 PM
chiro
Re: function of expectation
Hey Juju.

One idea I have is to use Jensens Inequality since you are dealing with a convex function log(x):

Jensen's inequality - Wikipedia, the free encyclopedia
• Apr 17th 2013, 11:06 PM
Juju
Re: function of expectation
Thanks Chiro.

But I don't think that Jensen's inequality helps (or at least I don't know how):

$\displaystyle E[\log(1+\alpha(\exp(Z)-1))]\geq \log E[1+\alpha(\exp(Z)-1)]\geq \log(1+\alpha(\exp(E(Z))-1))\geq 0$

since $\displaystyle (1+\alpha(\exp(E(Z))-1)) \leq 1.$ And that is what I already know.

Or did I anything wrong?
• Apr 17th 2013, 11:56 PM
chiro
Re: function of expectation
Doesn't that mean that log of that term is <= 0 though?
• Apr 18th 2013, 12:45 AM
Juju
Re: function of expectation
Of course you are right. But anyway I think that I cannot conclude that
$\displaystyle E[\log(1+\alpha(e^Z-1))]=0$

since the inequality does not imply
$\displaystyle E\log(1+\alpha(e^Z-1))\leq 0.$

Am I right?
• Apr 18th 2013, 01:37 AM
chiro
Re: function of expectation
It is just an inequality: If you know that the value is zero then it is zero.

The inequality just gives a possible range without knowing any more information: now you can add your other piece of information to reduce the inequality even further.
• Apr 18th 2013, 01:52 AM
Juju
Re: function of expectation
Quote:

It is just an inequality: If you know that the value is zero then it is zero.
?? But I do not know that the value is zero. This is what I want to show.

But I think I can show it quite easily; just using
$\displaystyle E[\log(1+\alpha(e^Z-1))]\leq E[\alpha(e^Z-1)]$

$\displaystyle (\log(1+x))\leq x)$

Thank you.