Eliza Gregory

2022-10-11

Why is the expectation of an exponential function:
$\mathbb{E}\left[\mathrm{exp}\left(Ax\right)\right]=\mathrm{exp}\left(\left(1/2\right){A}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}?$

Audrey Russell

Let $X\sim \mathcal{N}\left(0,1\right)$ and $a\in \mathbb{R}$. Then
$\begin{array}{rl}E\left[\mathrm{exp}\left(aX\right)\right]& ={\int }_{\mathbb{R}}\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-\frac{1}{2}{x}^{2}\right)\mathrm{exp}\left(ax\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x={\int }_{\mathbb{R}}\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-\frac{1}{2}\left(x-a{\right)}^{2}+\frac{1}{2}{a}^{2}\right)\\ & =\mathrm{exp}\left(\frac{1}{2}{a}^{2}\right){\int }_{\mathbb{R}}\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-\frac{1}{2}\left(x-a{\right)}^{2}\right)=\mathrm{exp}\left(\frac{1}{2}{a}^{2}\right)\end{array}$
because
$x↦\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-\frac{1}{2}\left(x-a{\right)}^{2}\right)$
is the density of an $\mathcal{N}\left(a,1\right)$ distribution.

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