How is Nakagami function derived from Gamma function?

Question

So for example if I have ∼Gamma(,)

$$\ f(y) = \frac{1}{^Γ()}y^{k-1}e^{-y/}$$

Now according to wiki and other sources, to find Nakagami distribution we just have to =√ and we get this Nakagami distributed variable like this.

$$\ f(x) = \frac{2}{\Gamma(k)\theta^k}x^{2k-1} e^{-x^2/\theta}$$

So it looks like if we bring $f(y)^{1/2}$ we should get $f(x)$. But is there some special way of how to bring gamma functions to a power. I just don't see how this transformation happens. Can anybody share the proof or derivation process please.

Answer 1

It is very simple.

Let's start from

$$f_Y(y)=\frac{1}{\theta^k\Gamma(k)}y^{k-1}e^{-y/\theta}$$

Let's set

$$x=\sqrt{y}$$

That is

$$y=x^2$$

and its derivative is $y'=2x$

Immediately you get

$$f_X(x)=\frac{1}{\theta^k\Gamma(k)}x^{2k-2}e^{-x^2/\theta}2x=\frac{2}{\theta^k\Gamma(k)}x^{2k-1}e^{-x^2/\theta}$$

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Fundamenta Tranformation Theorem

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Remember that, if the transformation $y=g(x)$ is monotone, the density of Y can be derived by simply transforming the density of X with the following formula

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$