I'm wondering about the monotonic quantile transformation in case of a t distribution. First, for explanation, lets consider a normal distribution:

\begin{align*} f(l | \mu , \sigma ^2)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(l-\mu)^2}{2 \sigma ^2}} \end{align*}

A quantile at $\alpha$ can be calculated by

$Quantile_\alpha=\mu + \sigma \Phi^{-1}(\alpha)$

so e.g. lets say $\mu = -0.0000615$ and $\sigma=0,0145$, the 0,95 quantile of the normal distribution is 1,644854 so


So now for the t distribution, I use the following notation:

\begin{align*} f(l|\nu ,\mu ,\beta) = \frac{\Gamma (\frac{\nu+1}{2})}{\Gamma (\frac{\nu}{2}) \sqrt{\pi \nu} \beta} \left(1+\frac{1}{\nu}\left(\frac{l - \mu}{\beta}\right)^2 \right)^{\text{$-\frac{1+\nu}{2}$}} \end{align*}

now, I have this from a book: quantile

ignore the "VaR", it only sais, we focus on the right side of the distribution.

And now I am not really sure about the quantile computation anymore. I think my $\beta$ is the $\sigma^2$ right?

So is this correct: $Quantile_{\alpha}^{t}=\mu + \beta t_\nu^{-1}(\alpha) $

or this: $Quantile_{\alpha}^{t}=\mu + \sqrt{\beta} t_\nu^{-1}(\alpha) $

As an example, I have the following values:



$\nu = 3.34331$

and therefore $t_{3.34331}^{-1}(0.95)=2.258039$

now, is





Mh, I am not sure, since I am confused about the $\sigma^2$ they use, but 0.2232954 CANNOT be empirical correct, so it MUST be the 0.021? Where is my mistake?

the book: Alexander J. McNeil, Rüdiger Frey, Paul Embrechts: Quantitative Risk Management 2005 page 40.


You have to use the scale parameter (whatever you denote it, $\sigma$, $\beta$, or $\spadesuit$).

In your density $f(\cdot; \mu,\beta,\nu)$, $\beta$ is the scale parameter and its value has to be plugged into the "monotonic" transformation.

P.S. By the terminology you use, I guess you also asked Quantile of a non standardized t distribution?

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  • $\begingroup$ so please tell me directly @user1 is 0.02105319 correct? or 0.223? $\endgroup$ – Stat Tistician Mar 28 '13 at 16:22
  • 1
    $\begingroup$ The correct result is $ 0.02105321$ which uses the value of the scale parameter $\beta$. You can check this by running qnst(0.95,−0.0012827,0.009891722,3.34331). $\endgroup$ – user1 Mar 28 '13 at 16:23

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