MGF of a RV that is the ratio between a complex Gaussian and a Chi-squared RVs

Given the following p.d.f., which is the p.d.f. of the real and imaginary parts of a random variable that is the ratio between a complex Gaussian and a Chi-squared RVs:

$$\begin{equation*} f_U(u)=\exp\Big\{{-\frac{1}{4 u^2}}\Big\} \,\frac{\left(8 n u^2-1\right) I_n\left(\frac{1}{4 u^2}\right)+I_{n+1}\left(\frac{1}{4 u^2}\right)}{4 |u|^3}, \end{equation*}$$ where $$n$$ is a constant integer and $$I_n(z)$$ is the modified Bessel function of the first kind.

I'd like to find its Moment-generating function or at least a closed form expression for its mean and variance.

UPDATE 17/02/2019 Further information on this p.d.f. and how it was obtained can be found at: Distribution of ratio between complex Gaussian and Chi-square R.V.s

• are the numerator and denominator in this ratio of random variables independent? you might want to give more details on how you arrived at $f_U(u)$, now it is basically an exercise in integration, which might not be very motivating. – Carlo Beenakker Feb 17 at 18:19
• @CarloBeenakker, I've just updated the question to include more information on this p.d.f and how it was obtained. – Felipe Augusto de Figueiredo Feb 17 at 19:04
• @CarloBeenakker, I'm looking for the central moments. Do you think it can found based on the moments of $f_R (r)$, given in mathoverflow.net/questions/290092/… ? – Felipe Augusto de Figueiredo Feb 17 at 22:34

As discussed in the MO question linked to in the OP, the variable $$u$$ is defined as $$u=R^{1/2}\cos\phi$$, where $$\phi$$ is uniformly distributed in $$(0,2\pi)$$ and $$R=\xi_{2}/(\xi_2+\xi_{2n-2})^2$$ is composed from independent chi-squared distributions with $$2$$, respectively, $$2n-2$$ degrees of freedom. So all odd moments of $$u$$ vanish, while the even moments are given by $$\mu_{2p}\equiv\mathbb{E}(u^{2p})= \frac{(n-p-1)!}{(n+p-1)!}\frac{\Gamma(p+1/2)}{2^p\sqrt\pi}.$$ We can then reconstruct the moment generating function, which is a hypergeometric function: $$F(x)=\sum_{p=0}^\infty x^{2p}\mu_{2p}={}_2\!F_2\left[\left\{\tfrac{1}{2},1\right\},\{1-n,n\},-\tfrac{1}{2}x^2\right].$$
• well, moments of $\cos\phi$ are elementary integrals, and even moments of $R$ were calculated in the MO question you linked to. – Carlo Beenakker Feb 18 at 12:44