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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.2012-06-22
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    You might want to expand on the reason why $\psi''$ could be *not always well defined*.2012-06-22
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    Yes I've forgotten the condition that says $X$ is not constant. Thank you for pointing out.2012-06-22
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    *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.2012-06-22
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    did, I have now written in my question the reason why $\psi''$ could not be always well defined.2012-06-22
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    @Sasha It's corrected now. It was an error of typing.2012-06-22
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    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 $|\theta|\lt\delta$.2012-06-22
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    Yes you're right, I've figured this out by Holder's inequality. Now the problem can be easily solved by calculating $\psi''$!2012-06-22
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    Indeed. (There are other, simpler, ways than Hölder but this is not terribly important.)2012-06-23

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