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I ran across the following statement on Wikipeia at the following location

"chi-square distribution is a transformation of Pareto distribution"

I have looked for this transformation but cannot seem to find it. I have also looked here but it doesn't list it. I am curious if there is a transformation or if this may be a typo.

Any help would be greatly appreciated!

StubbornAtom
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dsmalenb
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1 Answers1

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You can get from any continuous distribution to a uniform distribution on $(0,1)$ by the probability integral transform. Take it from there.

Say $X$ has a Pareto distribution with pdf $$f(x)=\frac{a\theta^a}{x^{a+1}}1_{x>\theta>0}\quad,\,a>0$$

So cdf of $X$ is $$F(x)=1-\left(\frac{\theta}{x}\right)^a\quad,\,x>\theta$$

Then by probability integral transform, $$F(X)=1-\left(\frac{\theta}{X}\right)^a\sim U(0,1)$$

Or equivalently, $$Y=\left(\frac{\theta}{X}\right)^a\sim U(0,1)$$

And finally, $$-2\ln Y=2a\ln\left(\frac{X}{\theta}\right)\sim \chi^2_2$$

StubbornAtom
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