I have a non-central chi variable $r$ with the distribution, \begin{align} p(r) = \frac{r^3\lambda}{(\lambda r)^{3/2}}\exp\left[-0.5(r^2 + \lambda^2)\right]I_{1/2}(r\lambda) \end{align}

I'm looking for the following expectations. \begin{align} E\left[\frac{1}{r}\right]; \quad E\left[\frac{1}{r^2} \right] \end{align}

In other words, is there a closed-form expressions for the integrals of the form, \begin{align} I(m) = \int_{0}^{\infty} \frac{1}{r^m} \frac{r^3\lambda}{(\lambda r)^{3/2}}\exp\left[-0.5(r^2 + \lambda^2)\right]I_{1/2}(r\lambda). \end{align}


Edit :

$I_v(.)$ is the modified Bessel function of the first kind and order $v$.

  • $\begingroup$ You need to define your I(.) notation. $\endgroup$
    – wolfies
    Nov 6, 2022 at 16:23
  • $\begingroup$ @wolfies : Thanks, I did. $\endgroup$ Nov 6, 2022 at 17:51
  • $\begingroup$ For the first moment: stats.stackexchange.com/a/391865/8402 $\endgroup$ Nov 7, 2022 at 8:54
  • 1
    $\begingroup$ @StéphaneLaurent : Thank you for the link, but it refers to non-central $\chi^2$ distributed random variable. Mine is a non-central $\chi$ variable. $\endgroup$ Nov 8, 2022 at 23:07

1 Answer 1


The density simplifies to

$$p(r)=\frac{\sqrt{\frac{2}{\pi }} r e^{-(\lambda ^2+r^2)/2} \sinh (\lambda r)}{\lambda }$$

where $\sinh$ is the hyperbolic sine function. Integrating with Mathematica results in

$$E(1/r)=\frac{\text{erf}\left(\frac{\lambda }{\sqrt{2}}\right)}{\lambda }$$

$$E(1/r^2)=\frac{\sqrt{2} F\left(\frac{\lambda }{\sqrt{2}}\right)}{\lambda }$$

where $\text{erf}$ is the error function and $F$ is the Dawson F function.

Higher moments don't seem to exist.

  • $\begingroup$ Amazing, this is exactly what I was looking for. Thank you very much. $\endgroup$ Nov 6, 2022 at 17:30
  • $\begingroup$ Re "don't seem to exist:" $I_{1/2}(x)$ behaves like $x^{1/2}$ near the origin, whence $p(r)$ behaves like $r^2.$ It is immediate that all moments of order $-3$ or less diverge at the origin. $\endgroup$
    – whuber
    Nov 6, 2022 at 18:21
  • $\begingroup$ @whuber That helps explain it. I was just taking Mathematica's "does not converge" message at face value. Hence the hedge of "don't seem to exist." $\endgroup$
    – JimB
    Nov 6, 2022 at 18:25

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