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

  • $\begingroup$ 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 $1$? $\endgroup$ Jan 21, 2014 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, 2014 at 17:31
  • 2
    $\begingroup$ 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, 2014 at 17:38
  • $\begingroup$ Which question, specifically? $\endgroup$
    – Sam
    Jan 21, 2014 at 17:39
  • $\begingroup$ Sorry didnt realise tey got no answers except self answers. $\endgroup$
    – Lost1
    Jan 21, 2014 at 17:45

1 Answer 1


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.



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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    $\begingroup$ 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.) $\endgroup$ Oct 31, 2017 at 0:35
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    $\begingroup$ Why do we need the absolute sign here $\dfrac{1}{|x|}$? $\endgroup$
    – TrungDung
    Feb 11, 2022 at 14:21
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    $\begingroup$ Check the link in the post for criteria for mean to exist. $\endgroup$
    – Lost1
    Feb 12, 2022 at 15:06

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