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Given a random variable $x$ which is assumed to follow a Gaussian distribution $x \sim N( \mu, \sigma^2 )$ and $x$ is further known to be positive, I am interested in the following expectation value: $E\left[ \frac{1}{x} \right]$ .

In my case $\mu \gg 0$, which might allow to ignore that $x$ is always positive.

Does anyone knows about a collection of known expectation values under the normal distribution?

Many thanks in advance

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What do you mean by $x$ is further known to positive? Is this a truncated normal distribution? – Learner Mar 12 '13 at 10:40
In any case $E[1/x]=\infty$ . – Learner Mar 12 '13 at 10:42
Yes, it does. I chose $\mu = 10$ and $\sigma = 1$. $E\left[ \frac{1}{x} \right]$ than converges to 0.1010 – Matthias Mar 12 '13 at 13:30
@Did: I was replying to "I doubt that", which was in response to Matthias saying to joriki that the mean converged. The mean of $\frac1x$ is going to be taken over all samples where $\frac1x$ does not overflow. This leaves out a small, symmetric interval around $x=0$. This corresponds to what happens when taking the principal value integral. The smaller we take the symmetric interval, the closer the expected value comes to $0.10103161564918598872=\frac{\sqrt{2\pi}}{2}e^{-50}\,\mathrm{erfi}(5\sqrt2)$. – robjohn Mar 13 '13 at 0:35
@Did: Matthias does not claim that $\mathrm{E}\left(\frac1{|x|}\right)$ converges, he says that his samples (I assume via simulation) indicate that $\mathrm{E}\left(\frac1{x}\right)$ converges (no absolute values). Now, it did bother me that he said he rejected $x<0$. However, I then considered that $\int_\epsilon^1\frac{\mathrm{d}x}{x}\lt11356$ where $\epsilon=2^{-16382}$, which is the smallest positive extended precision number, and $e^{-50}\lt2\times10^{-22}$. Thus, the extended precision contribution of the singularity would be less than $1\times10^{-18}$. So much for simulations. – robjohn Mar 13 '13 at 15:35

We will use $$ \int_{-\infty}^\infty x^{2k}e^{-\frac{x^2}{2}}\,\mathrm{d}x=(2k-1)!!\sqrt\pi\tag{1} $$ If we take the principal value, we get the convergent series $$ \begin{align} &\mathrm{PV}\frac1{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty\frac1xe^{\large-\frac{(x-\mu)^2}{2\sigma^2}}\,\mathrm{d}x\\ &=\frac1{\sqrt{2\pi}\sigma}\int_0^\infty\frac1x\left(e^{\large-\frac{(\mu-x)^2}{2\sigma^2}}-e^{\large-\frac{(\mu+x)^2}{2\sigma^2}}\right)\,\mathrm{d}x\\ &=\frac1{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty\frac1x\sinh\left(\frac{\mu x}{\sigma^2}\right)e^{\large-\frac{\mu^2+x^2}{2\sigma^2}}\,\mathrm{d}x\\ &=\frac1{\sqrt{2\pi}\sigma}e^{\large-\frac{\mu^2}{2\sigma^2}}\int_{-\infty}^\infty\frac1x\sinh\left(\frac\mu\sigma x\right)e^{\large-\frac{x^2}{2}}\,\mathrm{d}x\tag{2}\\ &=e^{\large-\frac{\mu^2}{2\sigma^2}}\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k+1)!}\frac{\mu^{2k+1}}{\sigma^{2k+2}}\tag{3} \end{align} $$ We can also get an asymptotic expansion from $(2)$ using stationary phase: $$ \begin{align} &\frac1{\sqrt{2\pi}\sigma}e^{\large-\frac{\mu^2}{2\sigma^2}}\int_{-\infty}^\infty\frac1x\sinh\left(\frac\mu\sigma x\right)e^{\large-\frac{x^2}{2}}\,\mathrm{d}x\\ &=\mathrm{PV}\frac1{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty\frac12\left(\frac1{\frac\mu\sigma+x}+\frac1{\frac\mu\sigma-x}\right)e^{\large-\frac{x^2}{2}}\,\mathrm{d}x\\ &=\mathrm{PV}\frac1{\sqrt{2\pi}\mu}\int_{-\infty}^\infty\frac12\left(\frac1{1+\frac\sigma\mu x}+\frac1{1-\frac\sigma\mu x}\right)e^{\large-\frac{x^2}{2}}\,\mathrm{d}x\\ &\sim\frac1\mu\sum_{k=0}^\infty(2k-1)!!\frac{\sigma^{2k}}{\mu^{2k}}\tag{4} \end{align} $$ We can get a "closed form" in terms of $\mathrm{erfi}$ from $(2)$ $$ \begin{align} &\frac{\mathrm{d}}{\mathrm{d}\alpha}\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\frac1x\sinh\left(\alpha x\right)\,e^{\large-\frac{x^2}{2}}\,\mathrm{d}x\\ &=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\cosh\left(\alpha x\right)\,e^{\large-\frac{x^2}{2}}\,\mathrm{d}x\\ &=e^{\large\frac{\alpha^2}{2}}\tag{5} \end{align} $$ Therefore, $$ \begin{align} \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\frac1x\sinh\left(\alpha x\right)\,e^{\large-\frac{x^2}{2}}\,\mathrm{d}x &=\int_0^\alpha e^{\large\frac{t^2}{2}}\,\mathrm{d}t\\ &=\frac{\sqrt2}i\int_0^{i\alpha/\sqrt2} e^{\large-t^2}\,\mathrm{d}t\\ &=\frac{\sqrt{2\pi}}{2}\,\mathrm{erfi}(\alpha/\sqrt2)\tag{6} \end{align} $$ and plugging $(6)$ into $(2)$ yields $$ \mathrm{PV}\frac1{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty\frac1xe^{\large-\frac{(x-\mu)^2}{2\sigma^2}}\,\mathrm{d}x =\frac{\sqrt{2\pi}}{2\sigma}e^{\large-\frac{\mu^2}{2\sigma^2}}\mathrm{erfi}\left(\frac{\mu}{\sigma\sqrt2}\right)\tag{7} $$

Extended precision is not enough

It is mentioned in a comment that $x<0$ was rejected. This poses a theoretical problem. The computations above are carried out in principal value, that means that a small interval $[-\delta,\delta]$ is rejected, where $\delta\to0$. However, if $x\lt\delta$ is rejected, then, as $\delta\to0$, the the contribution to expected value from the singularity grows like $$ -\frac{\log(\delta)}{\sqrt{2\pi}}e^{-50}\tag{8} $$ Even using extended precision, where $\delta=2^{-16382}$, $(8)$ amounts to about $8.74\times10^{-19}$ which is pretty insignificant. However, as $\delta\to0$, $(8)\to\infty$.

Therefore, even extended precision arithmetic is insufficient to expose the problems with a simulation where $x\lt0$ is rejected.

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