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I'm currently running into the following inequality in a measure-theory related proof that are related to moment-generating functions and their derivatives.

Let $X \in \mathbb{R}_+$ and $h \in \mathbb{R}$. Show that:

$\cfrac{e^{hX} - 1}{h} \le X e^{hX}$

I'm fairly more conditions are required on $h$ for the inequality to hold (such as $h\ne 0$, though none were stated).

Other facts that I have are that may be relevant are: The quantity on the right is most likely the derivative of $e^{hX}$.

  • $X$ is actually a non-negative random variable.
  • $\mathbb{E}(e^{rX}) < \infty$ for $r\in[-\infty,s]$ where $s>0$.
  • $\mathbb{E}(X^k) < \infty$ for $k>0$.
  • $\mathbb{E}(X^k e^{rX}) < \infty$ for $k>0$ and $r\in[-\infty,s]$.
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3 Answers 3

up vote 2 down vote accepted

Let $y = h X$. It looks like you're trying to prove, for $y \ge 0$, $$ e^y-1 \le y e^y \Rightarrow f\left(y\right) = e^y\left(1-y\right) \le 1. $$ Since for $y \ge 0$ $$ f'\left(y\right) = -y e^y \le 0, $$ and $$ f\left(0\right) = 1, $$ the inequality holds.

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If you are assuming $X \in \mathbb{R_+}$ and $ h\in \mathbb{R} $ ,then the inequality does not hold. For instance, if we test it at the point $x=1,h=-1$, it does not hold

$$ Xe^{hX}=e^{-1}=.3678794412 ,\quad \frac{e^{hX} - 1}{h}= \frac{e^{-1} - 1}{(-1)}=0.6321205588 \,. $$

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Either by the convexity of the function $z\mapsto\mathrm e^{-z}$ or by a second order Taylor expansion of this function at zero, one sees that $\mathrm e^{-z}\geqslant1-z$ for every real number $z$, hence $\mathrm e^z-1\leqslant z\mathrm e^z$. In particular $\mathrm e^{hx}-1\leqslant hx\mathrm e^{hx}$ for every $x$ and $h$.

Thus, for $h\ne0$, $\dfrac{\mathrm e^{hx}-1}h\leqslant x\mathrm e^{hx}$ if and only if $h\gt0$ (and then this inequality holds for every real number $x$).

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