I'm having a hard time seeing why $Ee^{\theta X} < \infty$ implies $E[\left|X\right| \max(1,e^{\theta X})] < \infty$ but according to page 44 here it's true: https://drive.google.com/file/d/0B_WFOD8f4jS5eUdiRm1TYWRMTjA/view

$\theta > 0$ here too.

All I can say is that $E[e^{\theta X}] < \infty$ implies $E[X] < \infty$ because $e^{\theta X} \geq 1 + X$ for all $X$ but that still doesn't mean the expectation of $E[\left| X e^{\theta X} \right|] < \infty.$ Is there something I'm missing here?


1 Answer 1


This is not what is claimed in the reference you provided.

Rather, the claim is that if there exists $\theta_1 > 0$ such that $E e^{\theta_1 X} < \infty$ then for $\theta_0 < \theta_1$, you get $E (|X| \max(1, e^{\theta_0 X})) < \infty$.

To see this, start by breaking up the integral into the regions where $X \ge 0$ and where $X < 0$. You have shown in your question that you know how to deal with the $X < 0$ integral, so consider the $X \ge 0$ integral, which is $E (X e^{\theta_0 X})$. The reason this integral converges is that (since $\theta_0 < \theta_1$ by assumption) $X e^{\theta_0 X}$ grows much more slowly than $e^{\theta_1 X} = e^{(\theta_1 - \theta_0) X} e^{\theta_0 X}$, and the latter is integrable by assumption.


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