If I have a distribution of data, X, representing N samples taken during one measurement, then the mean square of X is $\bar{X^2} = \langle X^2 \rangle$, the variance of $X^2$, $\mathrm{var}(X^2)$ is $\langle\langle X^4 \rangle – \langle X^2 \rangle \rangle$, and the standard error of the mean square is $\sqrt{\frac{\mathrm{var}(X^2)}{N}}$.
Thus, the standard error of the mean square represents one standard deviation of the distribution that would be produced by repeating the measurement (taking N samples each time), assuming that $X^2$ is normally distributed with the variance of the original measurement.
What is the standard error of the quantity $\sqrt{\langle X^2 \rangle}$ (the standard error of the root mean square?)?
A rephrasing: assume Y is normally distributed with mean $\mu$ and variance $\sigma^2$. Define Z = $\sqrt{\frac{1}{N}\sum_1^N Y_i}$ What is the variance of Z?
Best Answer
If $Y$ is normally distributed with mean $\mu$ and variance $\sigma^2$, $U = \frac{1}{N} \sum_1^N Y_i$ is normally distributed with mean $\mu$ and variance $\sigma^2/N$. Note that $P\{U < 0\} > 0$, so $Z$ will have complex values. Anyway, $E[|Z|] = \sqrt{\frac{N}{2\pi \sigma^2}} \int_{-\infty}^\infty \exp(-N(t-\mu)^2/(2 \sigma^2)) \sqrt{|t|}\, dt$. I doubt there is a closed form for this integral. $E[|Z|^2] = E|U| = \sqrt{\frac{2}{\pi N}} \sigma \exp(-\mu^2 N/(2 \sigma^2)) + \mu\, {\rm erf}(\mu \sqrt{N/2}/\sigma)$, and ${\rm Var}(Z) = E[|Z|^2] - E[|Z|]^2$.