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If $X$ is a normal distributed with mean $\mu$ and variance $\sigma^2$. What would be the mean and variance of $Y = \dfrac{1}{X}$

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  • $\begingroup$ What makes you think that the mean and/or variance exist? Both are defined by improper integrals; are you sure these integrals actually converge? Have you tried it explicitly with mean $0$ and variance $1$? $\endgroup$ – John Hughes Jan 21 '14 at 17:26
  • $\begingroup$ I didnot do that mathematically. However, I ran this code in the MATLAB: a = randn(1000000,1); b = 1./a; The mean and variance of b does have some value. So I was interested in if there is a closed mathematical term, anyone has ever worked out. $\endgroup$ – Sam Jan 21 '14 at 17:31
  • $\begingroup$ Please accept answers to your other questions by pressing tick button. Many people do not like helping people who do not accept answers. $\endgroup$ – Lost1 Jan 21 '14 at 17:38
  • $\begingroup$ Which question, specifically? $\endgroup$ – Sam Jan 21 '14 at 17:39
  • $\begingroup$ Sorry didnt realise tey got no answers except self answers. $\endgroup$ – Lost1 Jan 21 '14 at 17:45
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Mean and variance do not exist. For the mean to exist, the integral

$\int^\infty_{-\infty} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\frac{1}{|x|} \text{d}x$

needs to be finite. This is clearly not the case.

Note it is necessary that mean exists for variance to exist.

See

http://en.wikipedia.org/wiki/Inverse_distribution#Reciprocal_normal_distribution

Note inverse gaussian is something completely different. It is connected to brownian motion hitting a level. I changed the title. The thing you are referring to is a reciprocal normal.

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  • 5
    $\begingroup$ The integral is not finite since, denoting by $p(x)$ the density of a standard normal, $$\begin{array}{rl}\displaystyle\int_{-\infty}^{\infty} \frac{1}{|x|} p(x)dx &=& 2 \int_0^{\infty} \frac{1}{x} p(x)dx \\ & = & 2 \int_0^1 \frac{1}{x} p(x) dx + 2\int_1^{\infty} \frac{1}{x}p(x)dx \\ & \ge & 2 p(1)\int_0^1 \frac{1}{x}dx + 2 \int_1^{\infty} \frac{1}{x}p(x)dx = \infty\end{array}$$ since $\int_0^1 \frac{1}{x}dx$ is divergent and $p(x)$ is decreasing on $[0,1]$, so bounded below by $p(1)$. (I assume the answerer knows this, I am writing it for future reference of anyone else slow like me.) $\endgroup$ – Chill2Macht Oct 31 '17 at 0:35

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