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Assume $R$ is Gaussian random variable. I read this from a paper,$\newcommand{var}{\text{Var}}$

$\mathbb{E}[1-\exp\{-\lambda R\}] = 1-\exp\{-\lambda \mathbb{E}(R) + \frac{\lambda^2}{2}\var(R)\}$

and,$\newcommand{argmax}{\text{arg max}}$

$\argmax_{a} 1-\exp\{-\lambda \mathbb{E}(R) + \frac{\lambda^2}{2}Var(R)\} = \argmax_a \mathbb{E}(R) - \frac{\lambda}{2}\var(R)$

where $a$ is a parameter in $R$.

I didn't figure out the derivation steps. Could you please give me some hints on how to get this? Thanks!

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    @Dilip I guess it does not matter in which way $R$ is parametrized by $a$.2012-02-08

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For the first formula, it was used that $\mathsf E[1-x] = 1-\mathsf Ex$, and then the formula for the MGT of the Gaussian random variable (see examples in the linked article) was applied. For the second, you have that$\newcommand{argmax}{\operatorname{arg max}}$ $$ f(a) = \lambda\mathsf E(R_a)-\frac{\lambda^2}2\mathsf V(R_a) $$ is just some function of $a$. Moreover, the function $$ g(y) = 1-\mathrm e^{-y} $$ is strictly increasing for all $y\in\mathbb R$. That easily implies $$ \argmax_a g(f(a)) = \argmax_a f(a) $$

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    @Shuai: thanks, I've fixed in in the other way2012-02-08