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 Cumulative distribution function (CDF).

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

I've tried to solve with Mathematica:

ii = 1/Pi Integrate[w^(-1 - n) Exp[(-b*v)/(a*w)] 1/Sqrt[w - u^2], {w,u^2, \[Infinity]}, Assumptions -> (u > 0 && n > 2 && 1 > v > 0 && a > 0 && b > 0)]
jj = (b^n)/((n - 2)! a^n) Integrate[v^n (1 - v)^(n - 2) ii, {v, 0, 1}, Assumptions -> (u > 0 && n > 2 && a > 0 && b > 0)]
Integrate[jj, {u, -Infinity, t}, Assumptions -> (u > 0 && n > 2 && a > 0 && b > 0)]

but it gives out a solution with this Hypergeometric function. I wonder if there is a tricky or some other technique to find a simpler equation for the CDF.

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    $\begingroup$ I am surprised that you get something as simple as Mathematica's hypergeometric answer. I don't think that you can reasonably ask for anything better than that. $\endgroup$ – Neil Strickland Feb 19 '19 at 9:12
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    $\begingroup$ Not sure if helpful, but your Bessel functions should simplify, $(8 n u^2-1) \, I_n(\frac{1}{4 u^2}) + I_{n+1}(\frac{1}{4 u^2}) = I_{n-1}(\frac{1}{4 u^2}) - I_n(\frac{1}{4 u^2})$. Also, the form is very close to $(\partial_z + n/z) \, \exp\{-z\} \, I_n(z)$, so there might just be an expression? $\endgroup$ – student Feb 19 '19 at 10:21
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    $\begingroup$ ... which of course means the whole thing is essentially $z^{-n} \partial_z \, z^n \exp\{-z\} \, I_n(z)$. $\endgroup$ – student Feb 19 '19 at 10:31
  • $\begingroup$ @student, what does $\partial_{z}$ mean? $\endgroup$ – Felipe Augusto de Figueiredo Feb 19 '19 at 11:54
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    $\begingroup$ I was merely pointing out that it is not entirely unreasonable to expect a closed form since the function is close to a derivative itself. But I had not considered that the PDF is a second derivative of the CDF for a complex random variable. So I doubt you can do much better than the accepted answer, unless the polynomials turn out to be of something nice. $\endgroup$ – student Feb 19 '19 at 21:50

The cumulative distribution function you are seeking is $$F_n(x)=2\int_0^{x}\exp\left({-\frac{1}{4 u^2}}\right) \,\frac{\left(8 n u^2-1\right) I_n\left(\frac{1}{4 u^2}\right)+I_{n+1}\left(\frac{1}{4 u^2}\right)}{4 u^3}\, du.$$ It has the general form $$F_n(x)=e^{-1/z}\bigl(A_n(z)I_0(1/z)+B_n(z)I_1(1/z)\bigr),$$ with $A_n(z)$ a polynomial in $z=4x^2$ of degree $\max(0,n-3)$ and $B_n(z)$ a polynomial in $z$ of degree $\max(0,n-2)$. For $n\in\{1,2,\ldots 10\}$ I find $$\{A_n\}=\left\{1,1,3,3-8 z,48 z^2-8 z+5,-384 z^3+48 z^2-32 z+5,3840 z^4-384 z^3+336 z^2-32 z+7,-46080 z^5+3840 z^4-4224 z^3+336 z^2-80 z+7,645120 z^6-46080 z^5+61440 z^4-4224 z^3+1296 z^2-80 z+9,-10321920 z^7+645120 z^6-1013760 z^5+61440 z^4-23424 z^3+1296 z^2-160 z+9\right\},$$ $$\{B_n\}=\left\{0,2,2-4 z,16 z^2-4 z+4,-96 z^3+16 z^2-20 z+4,768 z^4-96 z^3+160 z^2-20 z+6,-7680 z^5+768 z^4-1632 z^3+160 z^2-56 z+6,92160 z^6-7680 z^5+19968 z^4-1632 z^3+736 z^2-56 z+8,-1290240 z^7+92160 z^6-284160 z^5+19968 z^4-11232 z^3+736 z^2-120 z+8,20643840 z^8-1290240 z^7+4608000 z^6-284160 z^5+192768 z^4-11232 z^3+2336 z^2-120 z+10\right\}.$$ (I checked that $\lim_{x\rightarrow\infty}F_n(x)=1$, as it should be.)

Perhaps these are known sequences of polynomials, I don't know.

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  • $\begingroup$ Dear Carlo, do you know what $\partial_x$ means? Thanks. $\endgroup$ – Felipe Augusto de Figueiredo Feb 19 '19 at 18:43
  • $\begingroup$ certainly, $\partial_x$ is an abbreviation for $\partial/\partial x$ (so partial derivative with respect to $x$) $\endgroup$ – Carlo Beenakker Feb 19 '19 at 20:41

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