# From sample mean and variance of $X$ to $\sqrt{X}$

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:

$$\bar{x} = 1/n \sum_{i=1}^n x_i \\ s^2 = 1/(n-1) \sum_{i=1}^n (x_i- \bar{x})^2$$

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?

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