I have quite a general question about Variance-gamma distribution. I am interested in how to estimate it's parameters given a set of training points?

I tried to find the answer in the internet, but surprisingly managed to find only a couple of relative links:

  1. VarianceGamma R package (and it's manual)
  2. The paper which is referenced in R package manual.

I am not an expert in R, but what I saw in R package was (as I understood) maximum likelihood estimation. They perform some iterative optimisation with different methods, starting with Skew Laplace to initialise the estimator.

In the paper they first estimate mean, variance, skewness, kurtosis through moments. Then they assume that asymmetry parameter $\theta$ is small, and set $\theta^2 = \theta^3 = 0$. And after all they solve combined equations to get all the parameter estimators. That's what I understood.

So my questions are:

  1. How exactly do they perform optimisation in R package? Maybe you know the place where it is described?
  2. What is the difference in the two methods? Which one is better to use in real life (well, maximum likelihood is better, but it is much more difficult to implement and the performance is not as good)? Are the assumptions about small $\theta$ strict in the first method?
  3. Where can I read more about Variance-Gamma parameter estimation?

I would be really grateful for every relative replies, papers and links. Thank you!

  • $\begingroup$ The approach you mention seems to be a form of the method of moments with restrictions on the higher order moments. Is there some reason that maximum likelihood does not work for this parametric model? $\endgroup$ Commented Jun 8, 2012 at 10:14
  • $\begingroup$ Well, it works (R package just prove that). The thing is that I don't know how they perform maximum likelihood. The only thing I understood is that they do it iteratively. I am almost sure that there is no exact analytical solution for this parametric model. $\endgroup$ Commented Jun 8, 2012 at 10:19
  • $\begingroup$ If the maximum likelihood estimates can be obtained an have the usual optimality properties, I don't understand why method of moments estimator would be suggested and why the higher order moments were set to 0. $\endgroup$ Commented Jun 8, 2012 at 10:36
  • $\begingroup$ Oh, I now got your point. Higher order moments are not set to zero, $\theta$ is an asymmetry parameter. In order to use method of moments, we have to assume that $\theta$ is small, otherwise we will not be able to solve equations. And about the optimality properties of ML, that is a question interesting to me as well. It works, but I don't know about theoretical background of ML in this case. $\endgroup$ Commented Jun 8, 2012 at 10:50
  • $\begingroup$ I would have thought that as the Variance-Gamma distribution is a variance-mean mixture of normal distributions that MLE would be most easily done via the EM algorithm (at least for some of the parameters). $\endgroup$ Commented Jul 12, 2013 at 2:10

1 Answer 1


1. In pp. 16 they mention that the package uses "BFGS" or "Nelder-Mead". The second one is the default option. Please, take a look at these links to see their differences. Optimisation is actually made on the likelihood function. This is, the fitted parameters in the output of vgFit are actually the maximum likelihood estimators.

2. It is difficult to tell which optimisation method is better in general. You can instead compare different methods and see if the results coincide using the command optim and the command vgFit. Next, I present a code for maximising the likelihood function, you can choose between 6 different optimisation methods.


# Simulate 100 observations from a variance-gamma distribution with parameters (0,1,0,1)
data = rvg(100, vgC = 0, sigma = 1, theta = 0, nu = 1)

# -log-likelihood function using the Variance-Gamma package
ll = function(par){
if(par[2]>0&par[4]>0) return( - sum(log(dvg(data, vgC = par[1], sigma=par[2],   
   theta=par[3], nu = par[4]) )))
else return(Inf)}

# Direct maximisation/minimisation using the command optim
optim(c(0,1,0,1),ll,method = c("Nelder-Mead", "BFGS", "CG", "L-BFGS-B", "SANN", "Brent"))

# Maximisation using the command vgFit

The advantage of vgFit is that you do not need to specify the starting value for searching the optimum. It implements three different methods for doing so: "US", "SL" and "MoM". "SL" is the default method. I would not trust blindly on these methods, I would rather compare the results from vgFit and optim.

3. You can check the references in the manual. For example

Seneta, E. (2004). Fitting the variance-gamma model to financial data. J. Appl. Prob., 41A:177– 187.

Kotz, S, Kozubowski, T. J., and Podgórski, K. (2001). The Laplace Distribution and Generalizations. Birkhauser, Boston, 349 p.

D.B. Madan and E. Seneta (1990): The variance gamma (V.G.) model for share market returns, Journal of Business, 63, pp. 511–524.

  • 1
    $\begingroup$ +1. Regarding 2., a couple things you could quantify "which is best" with are a) computational speed and b) look at the sensitivity to starting values - restart the optimizer from various starting points to see where it ends up. The algorithm that provides stabler results (across many different starting positions) is better, at least in that sense. This may also give hints about whether the objective function has multiple modes, etc. $\endgroup$
    – Macro
    Commented Jun 8, 2012 at 13:46
  • 1
    $\begingroup$ Thanks a lot for your effort! Your code really helped me to realize how R package works (thanks again for your experiments)! And thanks for the third reference, I will read it carefully. Actually, only one question still exists: the comparison of ML estimation and the one based on method of moments (assuming that asymmetry parameter is small, and actually how small it should be). $\endgroup$ Commented Jun 8, 2012 at 13:50
  • $\begingroup$ @JohnSmith I do not have an answer to this question. Please, have a look at this discussion. $\endgroup$
    – user10525
    Commented Jun 8, 2012 at 14:41
  • 1
    $\begingroup$ @Procrastinator Thanks anyway! I actually think there is no exact answer, so I will just perform some experiments with real data and compare the results. $\endgroup$ Commented Jun 8, 2012 at 14:48

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