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I was wondering if anyone knows a good resource to learn about variance mixture models ? My interest is in particular the normal variance mean mixture.

I know what they mean with their definition of $Y=\mu+\beta X+\sigma\sqrt{X}Z$, with $Y$ the variance mixed distribution, X the mixing distribution and Z the standard normal distribution, which are independent of each other.

In an article on ArXiV the immediately suggest the generalized hyperbolic distribution as a solution. Now I was wondering how one would go from the prescription above for $Y$ to a formula for the distribution. Is it by guessing and hoping it works, or is there some kind of procedure ?

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The source of the generalized hyperbolic distribution is explained clearly in the wikipedia article - it arises by choosing $V$ (in their notation, $X$ in yours) to be generalized inverse Gaussian and then calculating the distribution of that mixture in your question; the wikipedia article indicates that this might be done by using

$$f(x) = \int_0^\infty \frac{1}{\sqrt{2 \pi \sigma^2 v}} \exp \left( \frac{-(x - \alpha - \beta v)^2}{2 \sigma^2 v} \right) g(v) \, dv$$

or by using MGFs, via

$$M(s) = \exp(\alpha s) \, M_g \left(\beta s + \frac12 \sigma^2 s^2 \right),$$

for example. I have not attempted to verify the specifics of either calculation in this case.

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  • $\begingroup$ My question was more of the form "If you say $Y=\mu+\beta X + \sigma \sqrt{X}Z$, then do they get to the form of f(x) ? Is it simply by postulating this form and testing wether it suffises that equation, or is there a deeper proces, somekind of "plan of attack" ? $\endgroup$ – Nick Oct 1 '13 at 17:27
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    $\begingroup$ I've explained examples of the kinds of mechanical procedure used to find what the mixture distribution is from the specified distribution of $X$ and $Z$ (in your notation) quite directly. Two possible lines of calculation in answer to the explicit question "I was wondering how one would go from the prescription above for Y to a formula for the distribution" are provided above. Specify $X$ and $Z$, and do some algebra. ... (ctd) $\endgroup$ – Glen_b -Reinstate Monica Oct 1 '13 at 21:50
  • $\begingroup$ (ctd)... If you're lucky or have chosen $f_X$ judiciously (e.g. perhaps by knowing related results) for special cases, it may be that for some distribution you can obtain an explicit expression. In the cases where explicit expressions can be obtained and those are distributions that are useful in practice, these tend to be used. $\endgroup$ – Glen_b -Reinstate Monica Oct 1 '13 at 21:55

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