Where $X$ follows an F Ratio distribution F$(1,\alpha)$ with pdf:
$$
f(x)= \frac{\alpha ^{\alpha /2} (\alpha +x)^{\frac{1}{2} (-\alpha -1)}}{\sqrt{x} B\left(\frac{1}{2},\frac{\alpha }{2}\right)},\; x\in [0,\infty).$$
Looking for the distribution of the $n$-summed independent F Ratio-distributed variables $Y= \sum_{1 \leq i \leq n}X_i$, with $\alpha>2$.
I tried to work with the $n$-convoluted characteristic function:
$$\chi_n(t)=\left( \frac{\Gamma \left(\frac{\alpha +1}{2}\right) U\left(\frac{1}{2},1-\frac{\alpha }{2},-i t \alpha \right)}{\Gamma \left(\frac{\alpha }{2}\right)}\right)^n$$
(where $U(.,.,.)$ is the confluent hypergeometric function with integral representation $U(a,b,z)=\frac{1}{a \Gamma }\int _0^{\infty } t^{a-1} (t+1)^{-a+b-1} e^{t (-z)} \mathrm{d} t$ ) and was unable to go anywhere. I can pull the moments from $\chi(t)$ (which turn out to be rapidly infinite at higher orders) but I am interested in the density.
With gratitude.
Best Answer
Note that \begin{align} &F(x) = \int\limits_0^xf(\xi)\,\mathrm d\xi = \frac{\alpha ^{\frac\alpha2}}{B\left(\frac12,\frac{\alpha }{2}\right)}\int\limits_0^x \xi^{-\frac12}(\alpha +\xi)^{\frac12 (-\alpha -1)} \,\mathrm d\xi,\\[4pt] &t=\dfrac{\xi}{\alpha + \xi},\quad \xi=\dfrac{\alpha t}{1-t},\quad \alpha+\xi=\dfrac\alpha{1-t},\quad \mathrm d\xi = \dfrac\alpha{(1-t)^2}\mathrm dt,\\[4pt] &F(x) = \frac{\alpha ^{\frac\alpha2}}{B\left(\frac12,\frac\alpha2\right)} \int\limits_0^{\frac x{\alpha+x}} \left(\dfrac{\alpha t}{1-t}\right)^{-\frac12}\left(\dfrac\alpha{1-t}\right)^{\frac12 (-\alpha -1)}\dfrac\alpha{(1-t)^2}\mathrm dt\\ &\qquad = \frac1{B\left(\frac12,\frac\alpha2\right)} \int\limits_0^\frac x{\alpha+x} t^{-\frac12}(1-t)^{\frac\alpha2-1}\,\mathrm dt,\\[4pt] &F(x) = \frac{B_{\frac x{\alpha+x}}\left(\frac12,\frac\alpha2\right)}{B\left(\frac12,\frac\alpha2\right)} = I_{\frac x{\alpha+x}}\left(\frac12,\frac\alpha2\right),\tag1\\[4pt] \end{align} where $B_t(a,b)$ is incomplete beta function and $I_t(a,b)$ is regularized beta function.
If $\mathbf{n=1},$ then solution is trivial: $$f_1(y) = f(y).\tag2$$ If $\mathbf{n=2},$ then \begin{align} &f_2(y) = \int\limits_0^yf(\xi)f(y-\xi)\,\mathrm d\xi = f(y)\ast f_1(y),\tag3\\ &f_2(y) = \dfrac{\alpha^\alpha}{B^2\left(\frac12,\frac\alpha2\right)} \int\limits_0^y\dfrac{\left((\alpha+\xi)(\alpha+y-\xi)\right)^{-\frac{\alpha+1}2}}{\sqrt{\xi(y-\xi)}}\,\mathrm d\xi,\\[4pt] &\xi=\frac y2(1-\cos t),\quad d\xi=\dfrac y2\sin t\,\mathrm dt,\\[4pt] &f_2(y) = \dfrac{\alpha^\alpha}{B^2\left(\frac12,\frac\alpha2\right)} \int\limits_0^\pi\dfrac{\left(\left(\alpha+\frac y2\right)^2 - \left(\frac y2\sin t\right)^2\right)^{-\frac{\alpha+1}2}}{\frac y2\sin t}\dfrac y2\sin t\,\mathrm dt\\ &\qquad = \dfrac{\alpha^\alpha}{2B^2\left(\frac12,\frac\alpha2\right)} \int\limits_0^{2\pi}\left(\left(\alpha+\frac y2\right)^2 - \left(\frac y2\sin t\right)^2\right)^{-\frac{\alpha+1}2}\,\mathrm dt,\\ &z=e^{it},\quad dt = -i\dfrac{dz}z,\\ &f_2(y) = -i\dfrac{\alpha^\alpha}{B^2\left(\frac12,\frac\alpha2\right)} \oint\limits_{|z|=1}\left(\left(\alpha+\frac y2\right)^2 +\frac{y^2}4 \left(z-\frac 1z\right)^2\right)^{-\frac{\alpha+1}2}\,\dfrac{\mathrm dz}z,\\ \end{align} wherein \begin{align} &\left(\left(\alpha+\frac y2\right)^2 +\frac{y^2}4 \left(z-\frac 1z\right)^2\right)^{-\frac{\alpha+1}2} = \left(\alpha^2+\alpha y - \frac {y^2}4 +\frac{y^2}4 \left(z^2+\frac 1{z^2}\right)\right)^{-\frac{\alpha+1}2}\\[4pt] &\qquad = \left(\alpha^2+\alpha y - \frac {y^2}4\right)^{-\frac{\alpha+1}2}\left(1+\dfrac{y^2}{4\alpha^2+4\alpha y - y^2}\left(z^2+\frac 1{z^2}\right)\right)^{-\frac{\alpha+1}2}\\[4pt] & = \left(\alpha^2+\alpha y - \frac {y^2}4\right)^{-\frac{\alpha+1}2}\Bigg(1-\frac{\alpha+1}2\dfrac{y^2}{4\alpha^2+4\alpha y - y^2}\left(z^2+\frac 1{z^2}\right)\\[4pt] &+\dfrac1{2!}\frac{\alpha+1}2 \frac{\alpha+3}2\left(\dfrac{y^2}{4\alpha^2+4\alpha y - y^2}\right)^2\left(z^2+\frac 1{z^2}\right)^2\\[4pt] &-\dfrac1{3!}\frac{\alpha+1}2\frac{\alpha+3}2\frac{\alpha+5}2\left(\dfrac{y^2}{4\alpha^2+4\alpha y - y^2}\right)^3\left(z^2+\frac 1{z^2}\right)^3\\[4pt] &+\dfrac1{4!}\frac{\alpha+1}2\frac{\alpha+3}2\frac{\alpha+7}2\frac{\alpha+9}2\left(\dfrac{y^2}{4\alpha^2+4\alpha y - y^2}\right)^4\left(z^2+\frac 1{z^2}\right)^4 + \dots\Bigg)\\ &\qquad = c_0(y) + c_2(y)\left(z^2+\dfrac1{z^2}\right) + c_4(y)\left(z^4+\dfrac1{z^4}\right) + \dots,\\ &c_0(y) = \left(\alpha^2+\alpha y - \frac {y^2}4\right)^{-\frac{\alpha+1}2}\Bigg(1 +{\dfrac1{2!}\frac{\Gamma\left(\frac{\alpha+5}2\right)}{\Gamma\left(\frac{\alpha+1}2\right)}\binom21\left(\dfrac{y^2}{4\alpha^2+4\alpha y - y^2}\right)^2}\\[4pt] &+\dfrac1{4!}\frac{\Gamma\left(\frac{\alpha+9}2\right)}{\Gamma\left(\frac{\alpha+1}2\right)}\binom42\left(\dfrac{y^2}{4\alpha^2+4\alpha y - y^2}\right)^4 +\dfrac1{6!}\frac{\Gamma\left(\frac{\alpha+13}2\right)}{\Gamma\left(\frac{\alpha+1}2\right)}\binom63\left(\dfrac{y^2}{4\alpha^2+4\alpha y - y^2}\right)^6+\dots\Bigg),\\[4pt] &c_0(y) = \left(\alpha^2+\alpha y - \frac {y^2}4\right)^{-\frac{\alpha+1}2}\frac1{\Gamma\left(\frac{\alpha+1}2\right)}\sum\limits_{n=0}^\infty \dfrac{\Gamma\left(\frac{\alpha+4n+1}2\right)}{(n!)^2}\left(\dfrac{y^2}{4\alpha^2+4\alpha y - y^2}\right)^{2n},\tag4\\[4pt] &f_2(y) = 2\pi c_0(y)\dfrac{\alpha^\alpha}{B^2\left(\frac12,\frac\alpha2\right)}.\tag5\\[4pt] \end{align} If $\mathbf{n>2},$ then solution can be obtained as convolution in the forms of \begin{align} &f_n(y) = \int\limits_0^yf(y-\xi)f_{n-1}(\xi)\,\mathrm d\xi = f(y)\ast f_{n-1}(y),\tag6\\ \end{align} or \begin{align} &f(y)=\int\limits_0^y\int\limits_0^{y-\xi_1}\dots\int\limits_0^{y-\xi_1\dots-\xi_{n-2}}f(\xi_1)f(\xi_2)\dots f(\xi_{n-1})f(y-\xi_{n-1})\,\mathrm d\xi_1\,\mathrm d\xi_2\dots\,\mathrm d\xi_{n-1}.\tag7\\ \end{align}