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(This question is more about solving the integral than what the integral represents.)

The homodyned K distribution has the following probability density function (PDF): $$ f(E | A, \nu, b) = \frac{2 E b^\nu}{\Gamma (\nu)} \int_0^\infty x^{\nu -2} \exp(-b x) \exp(-(A^2 + E^2)/x) I_0 (2EA/x) \, dx, $$ where $I_0$ is the modified Bessel function of the first kind. I am interested in solving for the complimentary cumulative distribution function (CCDF), or "tail probabilty" $\bar F(E_T | A, \nu, b)$: $$ \bar F(E_T | A, \nu, b) = \int_{E_T}^\infty f(E | A, \nu, b) \, dE \\ = \int_{E_T}^\infty \frac{2 E b^\nu}{\Gamma (\nu)} \int_0^\infty x^{\nu -2} \exp(-b x) \exp(-(A^2 + E^2)/x) I_0 (2EA/x) \, dx \, dE. $$ A fast approximate solution for $\bar F$ is needed for various values of A because this will be running inside an optimization routine, so each evaluation of the objective function must be fast. Right now I know of no closed form solution to either integral, and naively using a 2D numerical quadrature routine is too slow for my application.

Is there a way to simplify either integral?

A relatively large error is acceptable (i.e. up to 5%). At the very least, being able to reduce the CCDF from a double integral to a single integral would be acceptable.

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  • $\begingroup$ Have you tried the integral definition of the Bessel function? $$I_0 \left(\frac{2A}{x}E \right)=\frac{1}{\pi} \int_0^\pi \exp \left(\frac{2A}{x}E \cos t \right) dt$$ You can then easily integrate w.r.t. $E$ in terms of error function. You will still be left with a double integral, however it may be nicer numerically than the original one. $\endgroup$ – Yuriy S Oct 8 '18 at 23:43

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