Thread: function of expectation

1. function of expectation

Hello,

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

Unfortunately I have no idea how to show it.
Does anyone have a hint?

2. 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

3. Re: function of expectation

Thanks Chiro.

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

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

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

Or did I anything wrong?

4. Re: function of expectation

Doesn't that mean that log of that term is <= 0 though?

5. Re: function of expectation

Of course you are right. But anyway I think that I cannot conclude that
$E[\log(1+\alpha(e^Z-1))]=0$

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

Am I right?

6. 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.

7. Re: function of expectation

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
$E[\log(1+\alpha(e^Z-1))]\leq E[\alpha(e^Z-1)]$

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

Thank you.