# 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?

## 2 answers

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.

#### 0 comment threads

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.

## 1 comment thread