Show that $E(logY|X) > log X$ implies $E(Y|X) > X$ using Jensen's inequality

Question

I'm currently stuck in a demonstration involving Jensen's inequality :

Show that

$E(log Y|X) > log X$

implies

$E(Y|X) > X$

using Jensen's inequality.

The suggestion is :

1. Start with $E(Y|X) > X$
2. Replace Y with $exp(log Y)$
3. Use the Jensen's inequality

Doing 1-3, the inequality becomes

$E(exp(log Y)|X) > X$

and

$E(exp(log Y)|X) > exp(E(logY|X))$

I guess you have to log $exp(E(logY|X))$ and $X$ so you will have $E(logY|X))$ and $logX$ but I don't know how to link them.

Answer 1

The suggestion seems to be actually wrong. Correct solution is, for me :

$$E(logY|X)>logX$$

$$exp(E(logY|X) > X$$

By Jensen's Inequalities :

$$E(exp(logY)|X) > exp(E(logY|X) > X$$

$$E(exp(logY)|X) > X$$

$$E(Y|X) > X$$

QED