I have the following in a proof I am following:

  • For $\theta \in (0, \theta_+), E(e^{\theta X}) < \infty$
  • From this, it is derived that $E(X^{+}) < \infty$. I can prove this.
  • If $|h| < h_0$, then $|e^{hx}-1| \leq |hx|e^{h_0x}$. I can also prove this.

Then, the book makes the following statement:

$\lim_{h \to 0}\int \frac{e^{hx}-1}{h}e^{\theta x}dF(x) = \int xe^{\theta x}dF(x)$ which it claims can be infered from the dominated convergence theorem.

Now, my reasoning is the following:

For a given (small) $h$, $|\frac{e^{hx}-1}{h}e^{\theta x}| < |x|e^{\theta_0 x}$ for some $\theta_0 \in (\theta, \theta_+)$ so, to prove that the integrals converge as claimed (using the dominated convergence theorem), it must be the case that $E(|X|e^{\theta X}) < \infty$.

But I cannot convince myself that the last inequality holds.

Is that the case that it holds? Or maybe my reasoning is flawed and the proof goes along other line of reasoning?

Thank you very much for any suggestion. It is appreciated.


(Assuming $\theta_+=1$.)

It is false that $Ee^X<\infty$ implies $E|X|e^X<\infty$, or, for some particular value of $\theta$, that $Ee^{\theta X}<\infty$ implies $E|X|e^{\theta X}<\infty$. It is true that $Ee^{\theta X}<\infty$ for all $\theta\in(0,1)$, implies that $E|X|e^{\vartheta X}<\infty$ for each $\vartheta\in(0,1)$. To check this for some $\vartheta$, look at what you know for $\theta\in(\vartheta,1)$, and use the fact that $|X|=O(\exp((\theta-\vartheta)X)$.

  • $\begingroup$ Thank you! I have just changed the title to better reflect the spirit of my question, yet your answer accurately addresses it so I accepte it. Thank you very much. $\endgroup$ – Cristián Antuña May 6 at 13:35

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