1
$\begingroup$

There seem to be two canonical definitions for the standard deviation. $$\sigma_n = \langle(n - \langle n\rangle)^2\rangle^{1/2}$$ and $$\sigma_n = \sqrt{\langle n^2\rangle - \langle n\rangle^2}$$

How do we reconcile these two? I've tried expanding $\langle n \rangle = \sum n \cdot P(n)$ to no avail.

$\endgroup$
  • $\begingroup$ The first one is the variance en.wikipedia.org/wiki/Variance , whose square root is the standard deviation $\endgroup$ – Pierpaolo Vivo Feb 13 '16 at 8:30
  • $\begingroup$ Ah, that's my bad. Forgot to add the exponent. Edited! $\endgroup$ – Andrew H. Feb 13 '16 at 8:32
  • $\begingroup$ I would say that the first one is canonical and the second a derivation. $\endgroup$ – copper.hat Feb 13 '16 at 8:47
1
$\begingroup$

$$ \sigma_n = \langle(n - \langle n\rangle)^2\rangle^{1/2}=\langle (n^2+\langle n\rangle^2-2 n \langle n\rangle)\rangle^{1/2} $$ and then use linearity of the average to conclude $$ \sigma_n=(\langle n^2\rangle+\langle n\rangle^2-2\langle n\rangle^2)^{1/2}\ . $$

$\endgroup$
0
$\begingroup$

With $\bar{x} = Ex$, we have $E (x -\bar{x})^2 = E (x^2 -2x \bar{x} + \bar{x}^2) = E x^2 - 2 \bar{x}^2+ \bar{x}^2 = E x^2 - \bar{x}^2$.

$\endgroup$
0
$\begingroup$

Here is the proof. \begin{aligned} \sigma_n = \sqrt(\sum_i(x_i-\mu)^2) & =\sqrt(\sum_i x_i^2 +\sum_i\mu^2-2\sum_ix_i\mu) = \sqrt(\sum_ix_i^2-\sum_i\mu^2)\quad (since \sum_i x_i = n\mu$) \end{aligned}

$\endgroup$
0
$\begingroup$

$\langle(n - \langle n\rangle)^2\rangle^{1/2} = \langle n^2 - 2n \langle n\rangle+\langle n\rangle^2\rangle^{1/2} = \left(\langle n^2\rangle - 2\langle n \rangle \langle n\rangle+\langle n\rangle^2\right)^{1/2}= \left(\langle n^2\rangle - \langle n\rangle^2\right)^{1/2}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.