I found that the polar coordinates of 2-dimensional Gaussian distribution with mean zero $$\frac{1}{2\pi\sigma^2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\big(-({x^2+y^2})/{2\sigma^2}\big) \,\mathrm{d}x\,\mathrm{d}y$$ is $$\frac{1}{\sigma^2}\int_{0}^{\infty}\exp\big(-r^2/{2\sigma^2}\big) \,r\mathrm{d}r$$ What if we consider non-zero mean, that is what would exactly be the following equation in polar coordinate system? $$\frac{1}{2\pi\sigma^2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\big(-({(x-\mu_x)^2+(y-\mu_y)^2})/{2\sigma^2}\big) \,\mathrm{d}x\,\mathrm{d}y$$
polar coordinates of Gaussian Distribution with non zero mean
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integration
normal-distribution
polar-coordinates
exponentiation
1 Answers
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If you mean a polar coordinate system with respect to the origin, then the result is a complicated mess. However, expressed in a polar coordinate system with respect to the point $(\mu_x,\mu_y)$, your third expression is again equal to your second expression, where $r$ now stands for the distance from the point $(\mu_x,\mu_y)$.
P.S.: I suggest to take more care to use terms precisely. These expressions are neither distributions, nor coordinates, nor equations; they're normalization integrals over distributions expressed in certain coordinates.
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0Thanks a lot for your reply. Actually I am looking for overlapping probability of two 2D Gaussians centred at $(\mu_x,\mu_y)$. For the amount of overlap I need to consider the centre point of both the Gaussian functions, since both normally distributed points are located at some distance from each other – 2011-07-04
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0Any suggestion, how can I find the overlapping probability of two 2D Gaussian functions??? – 2011-07-04
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0I don't know the term "overlapping probability" -- could you please define it? Also I'm not sure from what you wrote whether the two Gaussians are both centred at the same point $(\mu_x,\mu_y)$, or at different points. – 2011-07-04
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0Assuming that there are two 2D Gaussians at $(\mu_{x1},\mu_{y1})$ and $(\mu_{x2},\mu_{y2})$ respectively. Since the Gaussian functions are infinite, I am interested in their overlapping area. – 2011-07-04
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0Sorry, I don't see what could be meant by their "overlapping area". Their support is the entire plane, so you can't be referring to the area of overlap of their support. Please define what you mean. – 2011-07-04
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0In the following pdf, the first 5 pages show overlapping of two 2D Gaussians of same sigma. As shown in the images, I am interested in finding the overlapping region, as the probability of Gaussian becomes smaller and smaller as we move away from centre. http://brainimaging.waisman.wisc.edu/~oakes/PET_resolution_demo_2.pdf – 2011-07-04
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0@shaikh: There's no such thing as the overlapping region. As you wrote yourself, the support of the Gaussians is infinite; there are no finite areas to which they are restricted whose overlap you could calculate. You could define some (arbitrary) cutoff and then calculate the overlap of the areas where the Gaussian is above that cutoff. But instead of talking about areas, it would seem more natural to me to look at the integral over the product of the two Gaussian distributions. It so happens that the product of two Gaussians is again a Gaussian (with a different centre), ... – 2011-07-04
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0... so if you use this definition of "overlap", you can calculate the overlap just like a normalization integral for a single Gaussian. – 2011-07-04
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0Is it possible to do that using polar coordinates? I mean if I want to use polar coordinates for your mentioned product of Gaussians then how can I incorporate means $(\mu_{x1},\mu_{y1})$ and $(\mu_{x2},\mu_{y2})$ into the polar equation to have different centers. – 2011-07-04