# expectation derivations on a Gaussian random variable

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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Since $R$ is a Gaussian random variable, what does "$a$ is a parameter in $R$" mean? That $a$ is the mean $\mu$ or the variance $\sigma^2$ (these are the only two parameters defining a Gaussian distribution) or some function of $\mu$ and $\sigma^2$? – Dilip Sarwate Feb 8 '12 at 11:49
@Dilip I guess it does not matter in which way $R$ is parametrized by $a$. – Ilya Feb 8 '12 at 11:55

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 $$f(a) = \lambda\mathsf E(R_a)-\frac{\lambda^2}2\mathsf V(R_a)$$ 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 $$\argmax_a g(f(a)) = \argmax_a f(a)$$
I would like to revise a bit: $$g(y) = 1-e^{-\lambda y}$$ and it's strictly increasing for all positive $\lambda$. – shuaiyuancn Feb 8 '12 at 14:05