# Inverting the Fourier Transform for a Fisher distribution

The characteristic function of Fisher $\mathcal{F}(1,\alpha)$ distribution is: $$C(t)=\frac{\Gamma \left(\frac{\alpha +1}{2}\right) U\left(\frac{1}{2},1-\frac{\alpha }{2},-i t \alpha \right)}{\Gamma \left(\frac{\alpha }{2}\right)}$$ where $U$ is the confluent hypergeometric function. I am trying to solve the inverse Fourier transform $\mathcal {F} _ {t,x}^{-1}$ of the $n$-convolution to recover the density of a variable $x$, that is: $$\mathcal {F} _ {t,x}^{-1} \left(C(t)^n\right)$$ with the purpose of getting the distribution of the sum of $n$ Fisher-distributed random variables. I wonder if someone has any idea as it seems to be very difficult to solve. I tried values of $\alpha=3$ and $n=2$ to no avail. Note: for $n=2$ by convolution I get the pdf of the average (not sum):

$$\frac{3 \left(12 \left(x^2+3\right) \left(5 x^2-3\right) x^2+9 \left(20 x^4+27 x^2+9\right) \log \left(\frac{4 x^2}{3}+1\right)+2 \sqrt{3} \left(x^2+15\right) \left(4 x^2+3\right) x^3 \tan ^{-1}\left(\frac{2 x}{\sqrt{3}}\right)\right)}{\pi ^2 x^3 \left(x^2+3\right)^3 \left(4 x^2+3\right)}$$,

where $x$ is an average of 2 variables. I know it is unwieldy but would love to get an idea of the approximation of the basin distribution.

• is this question alive? May 30, 2015 at 8:35
• Yes, it is still open.
– Nero
May 30, 2015 at 11:56
• I assume you are under some symbolic package right? May 30, 2015 at 17:56
• Dec 3, 2015 at 19:35
• Nero, response to this open letter? stats.stackexchange.com/a/36051
– BCLC
Oct 31, 2020 at 9:51

There are other ways to approach the problem without using the characteristic function. One would expect the convolution distribution to be something like an F-distribution in shape. One might try an approximation like $aF(n,k)$ for the $n$-convolution, and then choose $a$ and $k$ to make the first two moments of the distribution correct. This is easy given the known mean and variance of the F-distribution.
If $\alpha$ is large, then the convolution converges to a chisquare distribution on $n$ degrees of freedom. This is equivalent to choosing $a=n$ and $k=\infty$ in the above approximation, showing that the simple approximation is accurate for large $\alpha$.