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Why is the expectation of an exponential function: $$\mathbb{E}[\exp(A x)] = \exp((1/2) A^2)\,?$$

I am struggling to find references that shows this, can anyone help me please?

If anyone could enlighten me it would be great!

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  • $\begingroup$ Expectation with respect to what probability density function? $\endgroup$
    – JohnD
    Dec 14, 2012 at 16:31
  • $\begingroup$ What is $A$ and $x$? $\endgroup$ Dec 14, 2012 at 16:34
  • $\begingroup$ A is a constant and x is a random variable that is assumed Gaussian $\endgroup$
    – Damian
    Dec 14, 2012 at 16:39
  • $\begingroup$ Yes: Expectation with respect to what probability density function $\endgroup$
    – Damian
    Dec 14, 2012 at 16:40
  • $\begingroup$ What is $E[x]$ and $\mathrm{Var}(x)$? $\endgroup$ Dec 14, 2012 at 16:46

1 Answer 1

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Let $X\sim\mathcal{N}(0,1)$ and $a\in\mathbb R$. Then $$ \begin{align*} E[\exp(aX)]&=\int_{\mathbb R}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}x^2\right)\exp(ax)\,\mathrm dx=\int_{\mathbb R}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}(x-a)^2+\frac{1}{2}a^2\right)\\ &=\exp\left(\frac{1}{2}a^2\right)\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}(x-a)^2\right)=\exp\left(\frac{1}{2}a^2\right) \end{align*} $$ because $$ x\mapsto \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}(x-a)^2\right) $$ is the density of an $\mathcal{N}(a,1)$ distribution.

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