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This is a part of an exercise that I'm doing, in Durrett's Probability book.

Let $X$ be a r.v which is not constant. Let $\phi(\theta)=E\exp(\theta X)<\infty$ for $\theta\in(-\delta,\delta),$ and let $\psi(\theta)=\log \phi(\theta).$ Prove that $\psi$ is strictly convex.

I wanted to write $\psi$'' but it's not always well defined, because $\phi'$ is not always well defined. To calculate $\phi'$, we derive inside the expectation, so $\phi'(\theta)=E(X\exp(\theta X))$, but nothing garantees that $E(X\exp(\theta X))$ is finite.

I also tried to write the classic definition of convex functions $\psi(\lambda \theta_1+(1-\lambda)\theta_2)<\lambda\psi(\theta_1)+(1-\lambda)\psi(\theta_2),$ but it doesn't work either.

I hope that someone can help me solve it. Thanks!

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Durrett also assumes that $X$ is not constant. You should be careful to put in all the conditions of the exercise. – Byron Schmuland Jun 22 '12 at 2:50
You might want to expand on the reason why $\psi''$ could be not always well defined. – Did Jun 22 '12 at 5:54
Yes I've forgotten the condition that says $X$ is not constant. Thank you for pointing out. – Iew Jun 22 '12 at 16:43
I hope that someone can help me solve it. They do, but for that, you could explain what prevents, in some cases, $\psi''$ to be well defined. – Did Jun 22 '12 at 16:51
And this reason is wrong: if $E(\exp(\theta X))$ is finite for every $|\theta|\lt\delta$, then $E(X\exp(\theta X))$ exists for every $|\theta|\lt\delta$. And, for that matter, for every nonnegative $n$, $E(|X|^n\exp(\theta X))$ is finite for every $|\theta|\lt\delta$. – Did Jun 22 '12 at 19:25

I don't think that statement is actually true...

Take the simplest case, where X is a constant 0. Then $\phi(\theta)$ is $1$ for all $\theta$, so $\psi(\theta)$ is a constant $0$, which is not strictly convex. Check if the exercise adds other conditions you forgot.

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I have added the condition $X$ is not constant. Sorry for that. – Iew Jun 22 '12 at 16:44

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