I got a question concerning the standardization of a student t distribution. I see that the "plain vanilla" t distribution has density $f(x|\nu)=\frac{\Gamma(\frac{\nu+1}{2}) }{\sqrt{\pi\nu}\Gamma(\frac{\nu}{2})}\bigg(1+\frac{x^2}{\nu}\bigg)^{\frac{-(v+1)}{2}} $ with $\nu$ being the df parameter. This distribution has a variance of $\nu/(\nu-2)$, iff $\nu>2$.

If my data has e.g. $\nu = 6$ it follows that the variance of it is equal 1.5. If I want to standardize it, I have to transform the data like $z= \frac{x-\mu}{\sigma}$.

What I don't know is how this works if I want to apply the transformation to the density function?

I found the following definition: $f(x|\mu,\sigma^2,\nu) = \frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})} \frac{1}{\sqrt{\pi (\nu - 2)}} \frac{1}{\sigma} \bigg( 1 + \frac{1}{\nu - 2} \left( \frac{x - \mu}{\sigma}\right)^2 \bigg)^{-\frac{\nu + 1}{2}}$,

where $\sigma^2$ is the variance and $\mu$ is the mean.

But I don't know how the transition from the first equation to the second works. Why is there an adjustment of $\nu - 2$? I've read a lot of text books but unfortunately I couldn't find the answer. I really appreciate your help, thanks a lot in advance!

  • $\begingroup$ Please tell us what $\sigma$ is intended to mean in the definition you found. $\endgroup$ – whuber Mar 30 '15 at 16:56
  • $\begingroup$ Thanks for noticing. I edited my question. Btw, the source is Lambert/Laurent (2002). $\endgroup$ – Henry_Cnvxty Mar 30 '15 at 17:00
  • $\begingroup$ OK, so that means the original variance of $\nu/(\nu-2)$ has to be multiplied by $\sigma^2(\nu-2)/\nu$ in order to create a variance of $\sigma^2$. That's done by rescaling $x$ by the square root of that quantity; the rest is pure algebra. I have re-directed this thread to another that explains and illustrates the concepts. $\endgroup$ – whuber Mar 30 '15 at 17:04