Consider the generalized hyperbolic distribution given by (from wikipedia): gh

So I now wanted to derive the standardized version, so mean zero variance one. I wanted to do the following:

  1. Set the mean to zero and the variance equal to one.
  2. Rewrite the mean equation, such that

    \begin{align*}\beta=-\mu\frac{\gamma K_\lambda (\delta \gamma)}{\delta K_{\lambda+1}(\delta\gamma)}\end{align*}

  3. Insert this expression into the variance equation.

  4. Rewrite the equation, such that $\mu =.....$
  5. Insert this expression into the pdf. This gives the formula for the standardized ghyp, which now only has the paramger $\lambda$, $\alpha$, $\beta$, $\delta$, since $\mu$ is fixed.

My question is now: Can I do this, I mean, is this correct?

My second important question is: I think, there are different ways to express the point, that the mean is zero and the variance equal to one. So instead of solving for $\beta$ and then $\mu$, I could have solved for different combinations of the parameters to get the mean zero and the variance equal to one. So my solution is just one possibility. Consider the following simplified example (arbitrary equations, just for illustration):

\begin{align*} \text{GH}= \mu + \alpha + 3 \beta + 4 \delta \end{align*}

The mean e.g. is given by and set to zero: \begin{align*} \text{mean}= 0 = \mu + \delta \beta \end{align*}

The variance e.g. is given by and set to one: \begin{align*} \text{variance} = 1 = \beta \delta +\delta /(\alpha-\beta) \end{align*}

There are now different ways of rewriting and solving. I can compute a solution, in which the final equation depends on $\alpha$, $\beta$, and $\delta$ and $\mu$ fixed, or, there is e.g. a solution in which the final equation depends only on $\mu$, and $\delta$ and $\alpha$ and $\delta$ is fixed. So both would fulfill the requirement of mean zero and variance one. I think by going the way I proposed in case of the GHYP above, there is the same problem, that this is only one solution, but there exist other combinations, in which the mean zero and variance one equation is still fulfilled.

So: What would you do in such a case?


Section 2.3.5 of the following manual


This question is a duplicate of Standardized generalized hyperbolic distribution

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  • $\begingroup$ no, it is not! The link does not give an answer to my question! $\endgroup$ – Stat Tistician May 1 '13 at 17:14

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