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I know that the reciprocal normal distribution does not have a defined mean or variance. But I wonder whether it is possible to get an expression for the mean when raising the reciprocal to a power greater than 1. As I understand it, the mean of the reciprocal normal does not exist because the tails are too heavy for the integral to be evaluated. Intuitively, I think that raising the reciprocal to a power greater than 1 would put less weight on the tails when evaluating the integral, and therefore potentially make the integral converge. In particular, I am looking at $\epsilon \approx 1.5$, in case the particular value of $\epsilon$ makes a difference. As I understand it, the question is whether the following integral can be solved:

$$ \int_{-\infty}^{\infty} \frac{1}{x^\epsilon \sigma \sqrt{2\pi}} \exp\left(\frac{-1}{2} \left(\frac{x-\mu}{\sigma}\right)^2\right)\mathrm dx$$

K.defaoite
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zandergordan
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  • In the case that $\mu=0$ this can be evaluated in terms of the Gamma function, using the well known Mellin transform of the Gaussian $$\int_{\mathbb R}x^{s-1}\exp(-x^2/2)\mathrm dx=2^{-1+s/2}(1-\mathrm e^{\mathrm i \pi s})\Gamma(s/2) \ \Re(s)>0$$ But if $\mu\neq 0 $ it's not so simple, and can only be written in terms of special functions (hypergeometrics). – K.defaoite Apr 13 '23 at 18:07
  • If $Y=(1/X)^\epsilon$, then to get less heavy tails I think you need to use smaller $\epsilon$, not greater. – leonbloy Apr 13 '23 at 18:49

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Notice first that $x^\epsilon$ for arbitrary $\epsilon \in \mathbb{R}$ and negative real $x$ is not very well defined (or at least is tricky).

If we consider instead $Y=1/|X|^{\epsilon}$. For large $y$, we have

$$P(Y>y)=P(|X| < y^{\beta}) \approx \gamma \, y^{\beta} $$

with $\beta = -1/\epsilon$ and $\gamma = \sqrt{\frac{2}{\pi \sigma^2}}\exp{(-\frac{\mu^2}{2\sigma^2})} $ which implies

$$f_Y(y) \approx \gamma \beta\, y^{\beta-1}$$

In order to have a finite variance, we need $1-\beta > 3$, or $\epsilon <1/2$

leonbloy
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