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?

  • $\begingroup$ The link mentions the variance of the sample mean, but not the sample variance. $\endgroup$ – drhab Jun 17 '15 at 9:39
  • $\begingroup$ jep, I corrected the question, thanks :-) $\endgroup$ – Gabriel Jun 17 '15 at 15:09
  • $\begingroup$ Link gone now. Trying to guess what is going on. See possible Answer. $\endgroup$ – BruceET Jun 17 '15 at 23:47

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.

  • $\begingroup$ Where did it say the distribution is normal? $\endgroup$ – Indominus Jun 20 '15 at 5:43
  • $\begingroup$ Question unstable. Link (when present) not helpful. Context, purpose unclear. Took a stab at trying to help.See my Comment above. Notice my answer begins with "If...". If @Indominus has something useful to offer, why just a comment? $\endgroup$ – BruceET Jun 20 '15 at 17:50

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