confusion about radii

[ian19981210]

Hi, thanks for sharing this great work！I found that you calculate the radius of gaussian in forward.cu by float my_radius = ceil(3.f * sqrt(max(lambda1, lambda2))); but the radius should be 1/sqrt(lamda) in linear algebra.so why you don't use a reciprocal and why you multiple 3.0? thanks for your reply!!!

[Snosixtyboo]

Hi,

why do you think the radius should be the reciprocal? The eigenvalues of the covariance matrix are equivalent to the variance of an unrotated Gaussian. The square root of the variance gives the standard deviation. We extend the radius to 3 * the standard deviation of the Gaussian (99% confidence interval) to compute its "fuzzy" radius (strictly speaking, a Gaussian extends into infinity).

[Snosixtyboo]

Closed due to inactivity

[elenacliu]

@Snosixtyboo That's because the ellipse's major radius = $\frac{1}{\sqrt{\lambda1}}$, minor radius = $\frac{1}{\sqrt{\lambda2}}$. I have the same question as @ian19981210

[elenacliu]

I found an explanation here

[elenacliu]

Hello, I guess I know why.

In fact, the 2D gaussian distribution should be (I ignore the multiplier):

$$ G(x)=\exp(-\frac{1}{2}(x)^T \Sigma^{-1} (x)) $$

with its covariance $\Sigma$. In paper, $\Sigma=RSS^TR^T$ and the ellipse is represented by $x^T\Sigma^{-1} x=1$.

For example, I draw a picture where $\Sigma=\text{diag}([5,1])$. If the ellipse is represented by $x^T\Sigma x=1$, the direction of the ellipse will be wrong.

So the major radius should be $\max(\sqrt{\lambda_1}, \sqrt{\lambda_2})$, minor $\min(\sqrt{\lambda_1}, \sqrt{\lambda_2})$, bacause the ellipse is represented by the inverse of $\Sigma$.