Consider a random variable $X$ such that for all $\lambda\in R$, $E\left(e^{\lambda(X-\mu)}\right)\leq e^{0.5 \lambda^2\sigma^2}$. How do I show that $E(X)=\mu$?

What I have done is that, by $1+\lambda X\leq e^{\lambda X}$, we have $E(1+\lambda X)\leq E(e^{\lambda X})\leq e^{0.5\lambda^2\sigma^2+\lambda\mu}\implies$

$\lambda E(X)\leq e^{0.5\lambda^2\sigma^2+\lambda\mu}-1\implies E(X)\leq\frac{e^{0.5\lambda^2\sigma^2+\lambda\mu}-1}{\lambda},\;\forall \lambda>0$.

Then by L'Hopital's rule, we have $\lim_{\lambda\to 0}\frac{e^{0.5\lambda^2\sigma^2+\lambda\mu}-1}{\lambda}=\mu$. Thus, we have $E(X)\leq \mu$.

Next, since $1+\lambda X\leq e^{\lambda X}$ still holds for $\lambda<0$, the previous inequality $E(X)\leq\frac{e^{0.5\lambda^2\sigma^2+\lambda\mu}-1}{\lambda},\forall \lambda>0$ becomes $E(X)\geq\frac{e^{0.5\lambda^2\sigma^2+\lambda\mu}-1}{\lambda},\;\forall \lambda<0$. Thus, we have $E(X)\geq\mu$.

Thus, $E(X)=\mu$.

Is my proof correct?


1 Answer 1


For fixed $\sigma^2, \mu \in \mathbb R$:

  • the inequality $\mathop{\mathbb E}\left[e^{\lambda\left(X-\mu\right)}\right]\leq e^{0.5 \lambda^2\sigma^2}$ for all $\lambda \in \mathbb R$ implies existence of the moment-generating function of $X$ and hence existence (and finiteness) of $\mathop{\mathbb E}\left[X\right]$;

and you have shown that

  • if $\mathop{\mathbb E}\left[e^{\lambda\left(X-\mu\right)}\right]\leq e^{0.5 \lambda^2\sigma^2}$ for all $\lambda \in \mathbb R_{>0}$, then $\mathop{\mathbb E}\left[X\right] \leq \mu$ for all $\lambda \in \mathbb R_{>0}$
  • if $\mathop{\mathbb E}\left[e^{\lambda\left(X-\mu\right)}\right]\leq e^{0.5 \lambda^2\sigma^2}$ for all $\lambda \in \mathbb R_{<0}$, then $\mathop{\mathbb E}\left[X\right] \geq \mu$ for all $\lambda \in \mathbb R_{<0}$.

Can you conclude what you want to show from these three points?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.