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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.

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    $\begingroup$ Do you mean that these independent random variables all have that F distribution? $\endgroup$ – Michael Hardy May 23 '15 at 22:43
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    $\begingroup$ In what context did this question arise? Just an exercise? Somes of F-distributed random variables arise naturally in applied problems, but I haven't seen it done in an case where they're independent, as far as I know. $\endgroup$ – Michael Hardy May 23 '15 at 22:49
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    $\begingroup$ I am looking at the distribution of the variance of Student-T distributed independent random variables. The squares for n=1 have the above distribution. $\endgroup$ – Nero May 23 '15 at 22:57
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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}

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  • $\begingroup$ Yet the beta prime distribution is reported as infinitely divisible, see Steutel and van Harn, 2003 (Appendix B) and before from Govaerts, 1979. Is there any contradiction that the form with n=2 does not seems to be a beta prime distribution. $\endgroup$ – Bentoy13 Jan 10 at 15:23

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