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If we consider a complex Gaussian random variable as $h\sim\mathcal{CN}(0,\gamma)$, where $\gamma$ is the variance. Is there any closed-form solution with $\gamma$ for $\mathbf{E}\left[\frac{1}{\lVert h\rVert_2^2}\right]$? where $\lVert\cdot\rVert^2$ denotes the norm-$2$ operation.

Really appreciate for any comments!

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    $\begingroup$ Seems to be just $+\infty$ . . . the double integral of $1/|z|^2$ diverges logarithmically near $z=0$. $\endgroup$
    – Noam D. Elkies
    Commented Dec 17, 2022 at 5:59
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    $\begingroup$ @NoamD.Elkies said it all, but just to write this explicitly: $$\mathbf{E}\left[\frac{1}{\|h\|_2^2}\right]=\int_{-\infty}^\infty dx \int_{-\infty}^\infty dy\,(\pi\gamma)^{-1}e^{-(x^2+y^2)/\gamma}(x^2+y^2)^{-1}$$ $$=(2/\gamma)\int_0^\infty dr\, e^{-r^2/\gamma} r^{-1}=\infty.$$ $\endgroup$
    – Carlo Beenakker
    Commented Dec 19, 2022 at 21:20
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    $\begingroup$ @MichaelHardy, since you edited || || to \| \|—definitely an improvement—you might like to know about \lVert \rVert, which is semantically preferable. (For example, $\|-x\|$ \|-x\| spaces poorly compared to $\lVert-x\rVert$ \lVert-x\rVert.) I have edited accordingly. $\endgroup$
    – LSpice
    Commented Dec 20, 2022 at 0:54
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    $\begingroup$ @LSpice is this equivalent to \left\| and \right\| ? $\endgroup$
    – Noam D. Elkies
    Commented Dec 20, 2022 at 2:17
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    $\begingroup$ @NoamD.Elkies, re, basically, but without sizing. (It's probably close to \mathopen\| \mathclose\|.) Compare $\lVert{\displaystyle\sum}\rVert$ \lVert{\displaystyle\sum}\rVert to $\left\|{\displaystyle\sum}\right\|$ \left\|{\displaystyle\sum}\right\|. $\endgroup$
    – LSpice
    Commented Dec 20, 2022 at 3:51

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