Assume that $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$. How do I derive $\mathbb E(e^X)$? Note I'm not asking for a mere answer, but an approach how to solve it. I end up messing with integrals, and change of variables, and it's not working.
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1$\newcommand{\E}{\Bbb{E}}$Have you tried completing the square in the integral, to reduce it to a Gaussian integral? By the way, first you can just work out $M(t):=\E\left[e^{tZ}\right]$ for $t\in\Bbb{R}$, where $Z\sim \mathcal{N}(0,1)$. Once you work this out, you can use that $\E\left[e^X\right] = e^\mu M(\sigma)$ (because $X$ is equal in distribution to $\mu+\sigma Z$). This makes the integral that you need to calculate nicer. – Minus One-Twelfth Jun 18 '19 at 12:22
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You can also see this page: Calculating moment generating function with normal distribution. – Minus One-Twelfth Jun 18 '19 at 12:29
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As always, $$\mathbb E[f(X)]=\int_{\mathbb R} f(x)\mu_X(\,\mathrm d x),$$ where $\mu_X$ is the measure on $\mathbb R$ induced by $X$. In the case where $X\sim \mathcal N(\mu,\sigma ^2)$, $$\mathbb E[f(X)]=\frac{1}{\sigma \sqrt{2\pi}}\int_{\mathbb R}f(x)e^{-\frac{(x-\mu)^2}{2\sigma ^2}}\,\mathrm d x.$$
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