What is the probability density function for the tau distribution?
The tau-distribution is typically defined in terms of the Student's t-distribution as follows: $$\tau_\nu \sim \sqrt{\frac{\nu \, t_{\nu - 1}^2}{\nu - 1 + t_{\nu - 1}^2}}.$$
I would be interested to compute the mean and variance of this random variable. Because it seems daunting to directly compute expectations from this formula, I started looking for a Probability Density Function (PDF) for this distribution.
Unfortunately, I was not very successful because everyone seems to use the formulation above. That made me wonder whether this formula (implicitly) defines the PDF and I am too stupid to see it, or if there simply is no proper density function. Hence my question: what is the PDF for the tau distribution?
1 answer
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.
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