1
$\begingroup$

Assume $X \sim f_{X}(n,\mu, \sigma)$ is a Variance-Gamma random variable. The density function involves a modified Bessel function, therefore is not that trivial to handle.

I'm looking for $\mathbb{E}|X|$ (and, by extension, $Var(|X|)$)

Looking at this straightforwardly, it should hold that $\mathbb{E}|X| = \int_{-\infty}^{\infty} x dF_{|X|}$ where $F_{|X|}(t)=\mathbb{P}(|X|\leq t) = \mathbb{P}(-t \leq X \leq t)$.

Without deriving the direct expression for $F_{|X|}(t)$, can we arrive at $\mathbb{E}|X|$ through integration? Or would this be easier through characteristic functions?

$\endgroup$
1
  • $\begingroup$ The easiest way is probably to go through the mixture representation (i.e. condition, use known results for the normal distribution, then use the law of total expectation/variance). $\endgroup$ – Chris Haug Apr 20 at 0:21
1
$\begingroup$

This is just an extended comment.

You might want to try a symbolic computation program such as Mathematica, Maple, or MATLAB. Here are the results for the mean and variance using Mathematica. (And maybe there is some simplification available?)

dist = TransformedDistribution[Abs[x], 
   x \[Distributed] VarianceGammaDistribution[\[Lambda], \[Alpha], \[Beta], 0]];
mean = Mean[dist]

Mean

var = Variance[dist]

Variance

$\endgroup$
1
  • $\begingroup$ Wow, that's a bit more complicated expression than I initially expected. Thanks for the code, will play around with it. $\endgroup$ – runr Apr 20 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.