What Does Sigma Represent in a 2D Gaussian Distribution?

[TheCanadian]

Case 1: I have a 2D Gaussian: $Ae^{-[\frac { (x-x_o)^2 }{2 \sigma_x ^2} + \frac { (y-y_o)^2 }{2 \sigma_y ^2}]}$ where $\sigma_x \neq \sigma_y$ (at least not necessarily). Using this as my 2D Gaussian, would the normalization constant be $A = \frac {1}{2\pi (\sigma_x ^2 + \sigma_y ^2)}$? In this context, what does $\sigma = \sqrt{ \sigma_x ^2 + \sigma_y ^2}$ even mean?

Case 2: if $\sigma_x = \sigma_y = \sigma$ in all cases, then I have: $Ae^{-[\frac { (x-x_o)^2 + (y-y_o)^2 }{2 \sigma^2}]}$. Would the normalization constant be $A = \frac {1}{2\pi \sigma^2}$ in that case?

Also, in terms of physical significance, what is the difference between sigma in the first case and the second case (if any)?

Any help would be great!

 

[davidmoore63@y]

Case 1: I think the normalization constant is 1/(2pi sigma(x) sigma(y))
Case 2 I agree it is 1/(2pi (sigma(x))^2)
I am not sure what you mean by physical significance. You didn't introduce any physical context.

 

[TheCanadian]

Case 1: I think the normalization constant is 1/(2pi sigma(x) sigma(y))
Case 2 I agree it is 1/(2pi (sigma(x))^2)
I am not sure what you mean by physical significance. You didn't introduce any physical context.

Thank you for the response.

For example, if $z = Ae^{-[\frac { (x-x_o)^2 + (y-y_o)^2 }{2 \sigma^2}]}$ (where A is any constant), then if one integrates z over the space $[x_o - \sigma, x_o + \sigma]\times[y_o - \sigma, y_o + \sigma]$ the result is $0.68^2$. If the integration area is $[x_o - 2\sigma, x_o + 2\sigma]\times[y_o - 2\sigma, y_o +2 \sigma]$ the result is $0.95^2$. This will happen if $\sigma_x = \sigma_y$. I am wondering what will happen if $\sigma_x \neq \sigma_y$. Is there an analogous region over which one can find $0.68^2$ or $0.95^2$ of the datapoints for this type of Gaussian?

 

[davidmoore63@y]

The analogous region is [x0 - sigma(x), x0+sigma(x)] x [y0 - sigma(y), y0 + sigma [y]].

 

[TheCanadian]

The analogous region is [x0 - sigma(x), x0+sigma(x)] x [y0 - sigma(y), y0 + sigma [y]].

Now this may seem a little odd, but if I was to choose another arbitrary 2D-axis besides x and y, is there a way to find the sigma of the Gaussian on these axes just based on $\sigma_x$ and $\sigma_y$ and the angle these new set of axes make with original axes? For example, If I know $\sigma_x = 4$ and $\sigma_y = 5$ but I want to find 0.68 of all data in a region spanned by the axes going through x and y at 45 degrees and also -45 degrees, is there a method to do this? I've tried to describe my intent in the image--the Gaussian's characteristics are known on x and y, but now I want to know what the characteristics are on x' and y', where the angular different between x and x' and y and y' is known and also the same (i.e. x' any y' are also perpendicular to each other).

My main purpose of doing this is so that I know what the Gaussian's features are, regardless of which axis I view and analyze it from initially.

 

[TheCanadian]

The analogous region is [x0 - sigma(x), x0+sigma(x)] x [y0 - sigma(y), y0 + sigma [y]].

Now I have a slightly odd inquiry. If using this equation in cylindrical coordinates, the equation becomes: $z = Ae^{-[\frac { (rcos(\theta)-r_ocos(\theta_o))^2}{2 \sigma_x^2} + \frac {(rsin(\theta)-r_osin(\theta_o))^2 }{2 \sigma_y^2}]}$

Now, I want to introduce a rotation to the Gaussian so that it is no longer strictly oriented in one direction, but now it is instead oriented randomly throughout space. I want to randomly orient this Gaussian by $\theta_{rot}$ so that the new equation is: $z1 = Ae^{-[\frac { (rcos(\theta - \theta_{rot})-r_ocos(\theta_o))^2}{2 \sigma_x^2} + \frac {(rsin(\theta - \theta_{rot})-r_osin(\theta_o))^2 }{2 \sigma_y^2}]}$. This accomplishes what I want, but out of curiosity, is the mean position of the centroid is still given by $(r_o, \theta_o)$? For some odd reason, I keep computing the mean position's angle and I keep getting $\theta_o + \theta_{rot}$ instead of just $\theta_o$. It should be just $\theta_o$, right?

Or should the actual equation be: $z2 = Ae^{-[\frac { (rcos(\theta - \theta_{rot})-r_ocos(\theta_o - \theta_{rot}))^2}{2 \sigma_x^2} + \frac {(rsin(\theta- \theta_{rot})-r_osin(\theta_o - \theta_{rot}))^2 }{2 \sigma_y^2}]}$

Also, what exactly is the difference between z1 and z2?

 

[TheCanadian]

The analogous region is [x0 - sigma(x), x0+sigma(x)] x [y0 - sigma(y), y0 + sigma [y]].

Also, if one has a 2D Gaussian with a sigmax and sigmay predefined (e.g. sigmax=3, sigmay=1). Then if you rotate it either 30 degrees, 40 degrees, or 50 degrees, what would be the new method to compute the standard deviation of this function in the original x- and y-axis? Is there a "simple" method to do this or is integrating the function and comparing intervals the only way, despite knowing the sigmas on two perpendicular axes and the exact rotation of the function?