Can the Difficult Double Integral Be Simplified with Approximate Functions?

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    Double integral Integral
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SUMMARY

The discussion centers on the difficulty of simplifying the double integral defined as \iint_{(ax+\mu _{1})^{2}+(bx+cy+\mu _{2})^{2}\leqslant z}\frac{1}{2\pi \sigma ^{2}}e^{-(\frac{1}{2\sigma ^{2}})(x^{2}+y^{2})}dxdy, which represents the cumulative distribution function (C.D.F.) of a specific elliptical domain involving Gaussian random variables. The integrand can be evaluated using polar coordinates, simplifying the domain to x^{2}+y^{2}\leqslant z, but no analytical solution exists for the elliptical domain. The user seeks to simplify the integral using approximate functions like the error function (ERF) and Bessel functions of the first kind, but finds the resulting expressions complex and unwieldy.

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Jeff.Nevington
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Here is the beast
[tex]\iint_{(ax+\mu _{1})^{2}+(bx+cy+\mu _{2})^{2}\leqslant z}\frac{1}{2\pi \sigma ^{2}}e^{-(\frac{1}{2\sigma ^{2}})(x^{2}+y^{2})}dxdy[/tex]

The integral gives the C.D.F. of [itex](ax+\mu _{1})^{2}+(bx+cy+\mu _{2})^{2}\leqslant z[/itex] where [itex]x[/itex] and [itex]y[/itex] are identically distributed gaussian random variables with zero mean and unit variance.

The integrand can be easily evaluated with polar coordinates over the less complex domain [itex]x^{2}+y^{2}\leqslant z[/itex] (In this case it becomes chi-square with two degrees of freedom). I am quite certain however that over the ellipse-shaped domain that I require, there is no analytical solution. On the other hand it would greatly speed up the numerical solution if I could just get rid of one of the integrals and/or solve in terms of approximate functions like ERF and Bessel of the first kind.

Any ideas? Anyone seen anything similar to this before?
 
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Use coordinates matching the boundaries:
x'=ax+m1 and y'=bx+cy+m2
This will likely make your work easier.
 
Thanks for the reply maajdl,

I altered the coordinates in this way as you suggested before and I didn't feel any closer to a solution, the integrand became a massive long mess and I couldn't simplify it. If I find my notes I will post it up (took me a while to re-arrange it).

I think for now I will admit defeat on this. It doesn't take so long to evaluate numerically for its purpose.

Thanks again,
Jeff
 

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