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A Nakagami fading distribution $X$ with parameter $m$ is given by the following $$ X\sim f(x;\,m,1) = \frac{2m^m}{\Gamma(m)}x^{2m-1}\exp\left(-mx^2\right)$$

Then the function $S:=|X|^2$ is Gamma distributed $$S\sim G(m,1/m)$$

In Wireless Communications, an example of a fading distribution on channel gains, is sometimes assumed to be Nakagami distributed. The case with

a- $m=\infty$ is referred to a case with no fading at all.

b- $m=1$ is referred to as Rayleigh fading.

Assuming I am interested in finding the Cumulative Density Function (CDF)of $S$

$$\mathbb{P}(S \geq T)$$

where $T$ is a non-negative constant. What happens to the two cases a and b above?

What I know

The CDF of Gamma distributed random variable is $$\frac{\gamma\left(m, mx\right)}{\Gamma(m)}$$

Then what is the CDF of the above cases a,b..?

Thanks

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  • $\begingroup$ For $m=1$ it is rather easy: $P(X \le x)=P(X^2 \le x^2) = \frac{\gamma\left(1, x^2\right)}{\Gamma(1)}$ and you can simplify this further to a closed function involving an exponential. (another -> $\endgroup$ – Henry Jan 8 '15 at 21:31
  • $\begingroup$ Yes I agree, thanks. Do you have any idea when $m=\infty$? $\endgroup$ – Henry Jan 8 '15 at 21:47
  • $\begingroup$ I would guess for large $m$ you would have $S$ approximately normally distributed with mean $m^2$ and variance $m^3$. Even though you are interested in $\sqrt{S}$, I would have thought this might suggest the limit when $m \to \infty$ was almost surely infinite with a CDF of $0$ for all positive finite $x$, and "no fading at all" might also suggest this. $\endgroup$ – Henry Jan 8 '15 at 22:01

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