a straightforward calculation of the $R$-transform with the help of Remark \[rem:F\_lambda\_sym\].

\[GIGPoissConv\] Let $X$ and $Y$ be free, $X$ free GIG distributed $\mu(\alpha,\beta,-\lambda)$ and $Y$ free Poisson distributed $\nu(1/\alpha,\lambda)$ respectively, for $\alpha,\beta,\lambda>0$. Then $X+Y$ is free GIG distributed $\mu(\alpha,\beta,\lambda)$.

We also quote another result from [@Szp17 Remark 2.2] which is again the free analogue of a property of classical GIG distribution. The proof is a simple calculation of the density.

\[GIGInv\] If $X$ has the free GIG distribution $\mu(\alpha,\beta,\lambda)$ then $X^{-1}$ has the free GIG distribution $\mu(\beta,\alpha,-\lambda)$.

The two propositions above imply some distributional properties of fGIG distribution. In the Section 4 we will study characterization of the fGIG distribution related to these properties.

\[rem:prop\]

1.  Fix $\lambda,\alpha>0$. If $X$ has fGIG distribution $\mu(\alpha,\alpha,-\lambda)$ and $Y$ has the free Poisson distribution $\nu(1/\alpha,\lambda)$ and $X,Y$ are free then $X\stackrel{d}{=}(X+Y)^{-1}$.

    Indeed by Proposition \[GIGPoissConv\] we get that $X+Y$ has fGIG distribution $\mu(\alpha,\alpha,\lambda)$ and now Proposition \[GIGInv\] implies that $(X+Y)^{-1}$ has the distribution $\mu(\alpha,\alpha,-\lambda)$.

2.  One can easily generalize the above observation. Take $\alpha,\beta,\lambda>0$, and $X,Y_1,Y_2$ free, such that $X$ has fGIG distribution $\mu(\alpha,\beta,-\lambda)$, $Y_1$ is free Poisson distributed $\nu(1/\beta,\lambda)$ and $Y_2$ is distributed $\nu(1/\alpha,\lambda)$, then $X\stackrel{d}{=}(Y_1+(Y_2+X)^{-1})^{-1}$.

    Similarly as before we have 