3
$\begingroup$

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? $\frac{1}{2\pi\sigma^2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\big(-({(x-\mu_x)^2+(y-\mu_y)^2})/{2\sigma^2}\big) \,\mathrm{d}x\,\mathrm{d}y$

1 Answers 1

3

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.

  • 0
    Is 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