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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?

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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?

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