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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
    Mar 18 '11 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
    Mar 18 '11 at 23:29
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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.

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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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