Talk:Nakagami distribution

Remove the claim that m \geq 0.5

The Nakagami density function is defined for any m > 0, as can easily be verified by integration. Kolar et. al in "Estimator Comparison of the Nakagami-m Parameter and Its Application in Echocardiography" claim that "In radio channels modeling, the parameter m is constrained so that m ≥ 1/2 [2]. But this condition may be violated in ultrasound applications [6], [7]." It appears reasonable to remove the constrain that m \geq 1/2 and replace it with m>0, as that isn't subject-matter specific, then add a paragraph about why m>=1/2 is a constraint in radio channels modelling.

old

The legend of the first graph shows "omega = 1 = 2". 1 = 2 is rubbish. 131.215.45.223 (talk) 21:44, 28 April 2008 (UTC)

From M.Nakagamis 1960 paper, page 10, the MGF is as follows: $E\left[X^{k}\right]={\frac {\Gamma (\mu +k/2)}{\Gamma (\mu )\mu ^{k/2}}}\omega ^{k/2}$— Preceding unsigned comment added by 80.3.31.37 (talk) 13:12, 16 September 2010

formula for the median (table on the right)

The formula for the median in the table on the right "median = square root (ω)" cannot be correct. If so, all curves with the same ω in fig. 2 (Cumulative distribution function) should cross at the same point: cdf(median) = 0,5 I am not so familiar with the Nakagami distribution, therefore I don't know the correct formula. Hartmut Voigt --217.93.248.136 (talk) 12:14, 22 February 2011 (UTC)

(I came here to make a similar comment) Yes, the formula for the median cannot be correct, as it would imply nice closed forms for the median of the chi-square distribution, which are not known to exist. (c.f. chi-square distribution) Shabbychef (talk) 23:09, 3 December 2014 (UTC)

On-line reference

There is difficulty with the online reference to work by Laurenson. This depends on browsers being able to deal with the XBM format which, acoording to XBM is not supported by IE or Mozilla. Is there an alternative? Melcombe (talk) 16:32, 22 July 2011 (UTC)

m<0.5

It seems that in some applications, 0<m<0.5 is possible as well. Maybe this should be mentioned? See eg.:

Dumane, V.A.; Shankar, P.M.; , "Use of frequency diversity and Nakagami statistics in ultrasonic tissue characterization," Ultrasonics, Ferroelectrics and Frequency Control, IEEE Transactions on , vol.48, no.4, pp.1139-1146, July 2001 doi: 10.1109/58.935733

VladimirSlavik (talk) 14:05, 15 November 2012 (UTC)

related distributions

The Nakagami-m and -q distributions are only approximately equal (exactly equal in special cases), and the parameter relationship is more involved than claimed here.

Separately, the Nakagami-m includes as special cases the one-sided Gaussian and the Rayleigh distributions.

All of this can be found in ref 5, Nakagami, M. (1960). — Preceding unsigned comment added by 105.184.227.147 (talk) 19:15, 27 August 2017 (UTC)

Agreed. $m=0.5$ corresponds exactly to a halfnormal distribution, and $m=1$ corresponds exactly to a Rayleigh distribution scaled by $1/{\sqrt {2}}$. For $0.5<m<1$ the Nakagami-m is only approximately equivalent to the Nakagami-q with

$$\Omega ={\alpha +\beta \over 2}$$

$$m={(\alpha +\beta )^{2} \over (\alpha +\beta )^{2}+(\alpha -\beta )^{2}}.$$

These are equations (58) from Nakagami(1960). The language is a litle difficult, but Nakagami refers to errors being "negligible for our present purposes". Dozy lizard (talk) 11:07, 23 December 2022 (UTC)