I regard a rather curious discrete distribution $X$ on $(1,2,...)$. Its weights are given by $P(X=i)=P_{NIG}(i)-P_{NIG}(i-1)$ where $P_{NIG}$ denotes the cumulative distribution function of the Normal-inverse Gaussian distribution. This CDF is not explicitly known, the probability density function is \begin{equation} f_{NIG}(x)=\frac{\alpha\delta K_1 \left(\alpha\sqrt{\delta^2 + (x - \mu)^2}\right)}{\pi \sqrt{\delta^2 + (x - \mu)^2}} \; e^{\delta \gamma + \beta (x - \mu)} \end{equation} with $K_j$ as the modified Bessel Function of the third kind. Now I want to fit this distribution to a dataset of discrete values. These dataset consists of two parts 1 and 2. In one part I assume a known $\beta_1$ in the other I assume a known $\beta_2$. Furthermore, assume $\mu_1=\mu_2=0$ in both parts. Thus, I want to estimate $\alpha=\alpha_1=\alpha_2$ and $\delta=\delta_1=\delta_2$. Due to the difficult analytic expression maximum likelihood or the method of moments are not directly feasible. I tried some numerical approximations for the first moment in R via optim() which did not work (did not converge).

I also tried to derive the characteristic function of the distribution $X$. A problem for the direct computation of the CF are the probability weights of $X$ which are only defined through integrals. I always end up in infinite sums of analytically possibly not solvable integrals. Moreover, I could not see an easy connection between the CF of a NIG-distributed random variable and the CF of my distribution $X$.

Is there a completely other way to fit this distribution to the data?


  • $\begingroup$ Out of curiosity, have you tried to work with the characteristic function instead of the density? I have worked with the NIG distribution in the past and I remember that it's characteristic function is quite a nice formula. Furthermore, characteristic function are often much easier to work with. Above that, there is a beautiful relationship between the moments and the derivative(s) of a characteristic function. $\endgroup$ – Cedric Cavents Mar 24 '16 at 20:08
  • $\begingroup$ Yes, I also tried this (explained in the edited paragraph above). $\endgroup$ – mikebrian18 Mar 25 '16 at 8:01

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