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wikipedia says that

Also, the scaled inverse chi-squared distribution is presented as the distribution for the inverse of the mean of ν squared deviates, rather than the inverse of their sum. The two distributions thus have the relation that if $$ X \sim \text{Scale-inv-}\chi^2(\nu, \tau^2) \qquad \text{then} \qquad \dfrac{X}{\nu \tau^2} \sim \text{inv-}\chi^2(\nu) $$

The pdf of the scaled-inverse-chi-square distribution is
$$f(x, \nu, \tau^2) = C\cdot(\tau^2 \nu)^{\nu/2} \dfrac{\exp[-\nu\tau^2 (2x)^{-1}]}{x^{1 + \nu/2}} \qquad (1.)$$
and the pdf of the inverse-chi-square distribution is
$$f(x, \nu) = C \dfrac{\exp[-(2x)^{-1}]}{x^{1 + \nu/2}} \qquad (2.)$$
where $$C = (2^{\nu/2} \Gamma(\nu/2))^{-1}$$

If we assume $Q = (\sum_{i=1}^\nu Z_i^2)^{-1}$ where $Z_i$ are iid standard normal, then $Q \sim \text{inv-}\chi^2(\nu)$. Based on the quoted passage

Also, the scaled inverse chi-squared distribution is presented as the distribution for the inverse of the mean of ν squared deviates, rather than the inverse of their sum.

I thought that if we define $X := \nu Q$ then we should have $X \sim \text{Scale-inv-}\chi^2(\nu, \tau^2)$. But I don't understand where $\tau^2$ comes from, and if I try to calculate the pdf of $X$ based on the pdf of $Q$ I get:
$$Pr(X = x) = Pr(\nu Q = x) = Pr(Q = \nu^{-1}x) = /(2.)/ \\ C \dfrac{\exp[(2(\nu^{-1}x))^{-1}]}{(\nu^{-1}x)^{1 + \nu/2}} = \\ C \nu^{1 + \nu/2}\dfrac{\exp[\nu(2x)^{-1}]}{x^{1 + \nu/2}}$$

Which does not seem to be equal to (1.) for any value of $\tau^2$ (wherever $\tau$ comes from). I wonder where my calculation goes wrong and how to understand the quoted passages.

DancingIceCream
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  • Although it isn't clear what you're trying to ask, the general topic seems to be that of transforming PDFs. Perhaps that link (to a site search) will turn up helpful posts here on CV. https://stats.stackexchange.com/questions/14483 covers the intuition; for a mathematically rigorous account, see https://stats.stackexchange.com/a/154298/919 or search the Web for "Jacobian." – whuber Mar 31 '22 at 02:41
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    @whuber Thanks, that actually sorts out my question!

    Sorry for being unclear about the question. The question was essentially this:

    Prove that if $Q\sim \text{inv-}\chi^2(\nu)$ then $X = Q\nu \implies X\sim \text{scale-inv-}\chi^2(\nu, \tau^2)$.

    From the quoted wikipedia passage, I gathered that it should be possible to prove this. Only that I did not understand where $\tau$ was supposed to come from. However, from your links I understand that it's just a matter of a pdf transformation and realizing that in my example $Q\sim \text{inv-}\chi^2(\nu) \implies \tau=1$.

     – DancingIceCream Mar 31 '22 at 08:52

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