How to obtain $e^{-\rho}e^{-b\mu+\frac{1}{2}b^2\sigma^2}$ from $e^{-\rho}E_t\{e^{-b\mu-b\sigma\epsilon}\}$ where $\epsilon$ is Normal $\sim(0,1)$, using the rule: If $X \sim N(\mu,\sigma^2)$ then for any constant k, $E\{e^{kW}\}=e^{k\mu+\frac{k^2}{2}\sigma^2}$?
I find "almost" the same expression, namely: $e^{-\rho}e^{-b\mu-\frac{1}{2}b^2\sigma^2}$. Any help to see what I am missing is appreciated.
I guess you made a trivial error. You need to evaluate $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{x^2}{2}-b\sigma x}dx$ The exponent can be rearranged to be $-\frac{1}{2}(x+b\sigma)^2+\frac{b^2\sigma^2}{2}$. From this the correct result follows. I suspect you made an erro in completing the square for the exponent.