We know that density for a student-t distribution is given as

$$\frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})} \left(\frac{\lambda}{\pi\nu}\right)^{\frac{1}{2}} \left[1+\frac{\lambda(x-\mu)^2}{\nu}\right]^{-\frac{\nu+1}{2}}$$

with $\text{E}(X) = \mu$, $\text{var}(X) = \frac{1}{\lambda}\frac{\nu}{\nu-2}$

where the three parameters are shape factor $\nu$, location $\mu$ and dispersion $\lambda$.

Now if I standardize X to Z as $\frac{X-\text{E}(X)}{\text{Std}(X)}$, what is the distribution of $Z$. If I standardize my input data to $Z$ as above, what is the density function for $f_Z(z)$?

I came across this question in Stephen Taylor's Asset Dynamics book. I am a bit confused after reading this question. As I understand, $\frac{X-\mu}{\sigma}$ where $\sigma=1/\sqrt(\lambda)$ will give the standardized t distribution with center 0 and disperson 1. But not sure what $Z$ above will look like.

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    $\begingroup$ Please check your statements. In the first line what is $V$? By definition a Student T has zero mean and its SD is determined by $\nu$, so where do $\mu$ and $\lambda$ come from? What, in the first two lines, is the relationship between $X$ and $Z$? Usually Student's T is used when mean and SD are being estimated from data, but your "standardization" presupposes you know $\mu$ and "$\text{std}(X)$". How? And why is there even a question here, when by definition the "log likelihood ... for $Z$" is the logarithm of the density, which you have explicitly written down? $\endgroup$
    – whuber
    Commented Mar 18, 2011 at 19:35
  • $\begingroup$ I have revised the question. My question is how to find the distribution of $Z$ and $f_Z(z)$ in the above case? Does this make sense now? $\endgroup$
    – user862
    Commented Mar 18, 2011 at 23:29

2 Answers 2


Assume $\nu \gt 2$ so that this distribution actually has a mean and standard deviation (otherwise you cannot standardize it). By direct calculation, its mean equals $\mu$ and its variance equals $\nu / (\lambda (\nu-2))$. Standardizing it, by construction, creates a distribution of the same shape but zero mean and unit standard deviation. Thus, for the standardized distribution, the formula is the same but we must have $\mu = 0$ and $\nu / (\lambda (\nu-2)) = 1$; that is, $\lambda = \nu/(\nu-2)$. Whence, plugging these values into the formula,

$$f_Z(z) = \frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})} \frac{1}{\sqrt{\pi(\nu-2)}} \left[1+\frac{z^2}{\nu-2}\right]^{-\frac{\nu+1}{2}}.$$

This assumes $\mathbb{E}[X] = \mu$ and $\mathbb{E}[(X-\mu)^2]$ are known in advance, not estimated from data.


If you're referring to the density function it's the exact same thing. You have simply shifted the distribution and normalized by the standard deviation.


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