I have samples $x_i$ of lets say a random variable $X$ (euclidean distances, $X=\sqrt{Y}$, where $Y$ is the squared distance) which I computed from squared distances samples $y_i$.
I can now calculate the sample mean $\bar{x}$ and sample variance $s^2$ as:
$$
\bar{x} = 1/n \sum_{i=1}^n x_i \\
s^2 = 1/(n-1) \sum_{i=1}^n (x_i- \bar{x})^2
$$
as described here: link
Is it possible to somehow compute the sample mean/variance for $Y$ (squared distances)
(more effiecient since no square root computation needs to be done)
and then transform it to the sample mean and sample variance of $X$ in an easy way?
Best Answer
If $X_1, \dots, X_n$ are iid normal with mean $\mu$ and SD $\sigma,$ then $Z_1, \dots, Z_n,$ where $Z_i = (X_i - \mu)/\sigma,$ are iid standard normal (with zero mean and unit variance).
Finally, $Q = \sum_{i=1}^n Z_1^2$ has a chi-squared distribution with $n$ degrees of freedom, and $E(Q) = n$ and $V(Q) = 2n.$ Of course, $Q$ is the squared distance from the origin in $n$-space, if each linear component is standard normal.
If this is on the right track for you, maybe look at Wikipedia on chi-squared distribution. Also the Rayleigh distribution has to do with the (Euclidean) distance between points in a plane.