# Meaning of sigma in a 2D sigma

1. Jul 29, 2015

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!

Last edited: Jul 29, 2015
2. Jul 29, 2015

### 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.

3. Jul 29, 2015

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?

4. Jul 29, 2015

### davidmoore63@y

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

5. Jul 30, 2015

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.

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6. Aug 4, 2015

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?

Last edited: Aug 4, 2015
7. Aug 6, 2015