What is the probability density function for the tau distribution?

I managed to find a reference to page 241 of the textbook "Mathematical method of Statistics" from H. Cramer, where the PDF of the tau distribution should be defined as

$$p(\tau \mathbin{;} \nu) = \frac{\Gamma(\nu / 2)}{\Gamma\bigl((\nu - 1) / 2\bigr) \sqrt{\nu \pi}} \Big(1 - \frac{\tau^2}{\nu}\Big)^{(\nu - 3) / 2},$$

with support in $\bigl[-\sqrt{\nu}, \sqrt{\nu}\bigr]$ and $\nu > 0$.

I also realised that we can use the change of variables formula to transform the formulation from the question into this PDF: Let $g(t) = t \sqrt{\frac{\nu}{t^2 + \nu - 1}}$, then we have $$\begin{align} p(\tau \mathbin{;} \nu) &= p_t\bigl(g^{-1}(\tau) \mathbin{;} \nu - 1\bigr) |g'(\tau)|^{-1} \\ &= p_t\biggl(\pm \tau \sqrt{\frac{\nu - 1}{\nu - \tau^2}} \mathbin{;} \nu - 1\biggr) \nu \sqrt{\frac{\nu - 1}{(\nu - \tau^2)^3}} \\ &= \frac{\Gamma(\nu / 2)}{\Gamma\bigl((\nu - 1) / 2\bigr) \sqrt{(\nu - 1) \pi}} \Big(1 + \frac{\nu - 1}{\nu - \tau^2} \frac{\tau^2}{\nu - 1}\Big)^{-\nu / 2} \nu \sqrt{\frac{\nu - 1}{(\nu - \tau^2)^3}} \\ &= \frac{\Gamma(\nu / 2)}{\Gamma\bigl((\nu - 1) / 2\bigr) \sqrt{\pi}} \Big(\frac{\nu}{\nu - \tau^2}\Big)^{-\nu / 2} \sqrt{\frac{\nu^2}{(\nu - \tau^2)^3}} \\ &= \frac{\Gamma(\nu / 2)}{\Gamma\bigl((\nu - 1) / 2\bigr) \sqrt{\nu \pi}} \Big(\frac{\nu - \tau^2}{\nu}\Big)^{(\nu - 3) / 2}. \end{align}$$

PS: Mean and variance of this distribution are 0 and 1, respectively.

It is well worth knowing how to do what I show you how to do below. The books in which you find formulas were not brought down from Heaven by an archangel. They are derived by humans, and you are one of those.

First, notice that (assuming $\nu>1$) as $t^2_{\nu-1}$ gets bigger, so does $\tau_\nu \sim \sqrt{\frac{\nu \, t_{\nu - 1}^2}{\nu - 1 + t_{\nu - 1}^2}},$ and also $\tau_\nu<\nu.$ That can be seen as follows: $$ \frac{aw}{a-1+w} = a - \frac{a(a-1)}{a-1+w} $$ and this increases as $w$ increases, but always remains less than $a.$

Hence we have $\tau_\nu\le\text{something}$ if and only if $t_{\nu-1}^2\le \text{something}.$ A bit of algebra tells us what $\text{“something”}$ is: Suppose $0\le u < \nu.$ Then $$ \tau_\nu \le u $$

$$ \sqrt{\frac{\nu t^2_{\nu-1}}{\nu-1+t^2_{\nu-1}}} \le u $$$$ \text{iff } \nu - \frac{\nu(\nu-1)}{\nu-1+t^2_{\nu-1}} \le u^2 $$$$ t^2 \le \frac{(\nu-1)u^2}{\nu - u^2} $$$$ |t| \le \sqrt{ \frac{(\nu-1)u^2}{\nu - u^2} }. $$

Therefore $$ f_{\tau_\nu}(u) = \frac d{du} \Pr(\tau_\nu \le u) $$

$$ = \frac d{du} \int_{-\sqrt{(\nu-1)u^2/(\nu-u^2)}}^{ +\sqrt{(\nu-1)u^2/(\nu-u^2)} } f(s) ~ ds $$

where $f$ is the density of Student's t-distribution $$ = 2 \frac d{du} \int_0^{ +\sqrt{(\nu-1)u^2/(\nu-u^2)} } f(s) ~ ds $$ because $f$ is an even function and we are integrating over and interval that is symmetric about $0$ $$ = 2f \left( \sqrt{(\nu-1)u^2/(\nu-u^2)} \right) \cdot \frac d{du} \sqrt{(\nu-1)u^2/(\nu-u^2)}. $$

Evaluate this derivative, and plug the argument to the function $f$ into the density of Student's t-distribution, then do the routine simplifications.