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.


1 Answer 1


There is no closed-form density for a convolution of F-statistics, so trying to invert the characteristic function analytically is not likely to lead to anything useful.

In mathematical statistics, the tilted Edgeworth expansion (also known as the saddlepoint approximation) is a famous and often used technique for approximating a density function given the characteristic function. The saddlepoint approximation if often remarkably accurate. Ole Barndorff-Nielsen and David Cox wrote a textbook explaining this mathematical technique.

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 $\alpha$.


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