13
$\begingroup$

I am currently working through the Feynman Lectures, chapter 6: Probability. I have reached his problem of the "random walk".

After deriving this and getting some root mean square, wouldn't this just be the same as finding the standard deviation? The standard deviation is the root of the mean of the squared data. Isn't that also just the root mean square?

Also, what exactly are the implications of the root mean square, what does it even mean in regards to our problem?

http://www.feynmanlectures.caltech.edu/I_06.html

$\endgroup$
6
  • $\begingroup$ Yes, RMS=STD. Could you clarify your last question? $\endgroup$
    – A.S.
    Commented Nov 19, 2015 at 22:22
  • $\begingroup$ @A.S. Okay, that makes a lot more sense. My last question was just, what exactly is the root mean square, why do we use it? $\endgroup$ Commented Nov 19, 2015 at 22:26
  • $\begingroup$ It is one of the measures of how much around the mean the quantity is dispersed (that's why sometimes its square is sometimes called "dispersion"). For a constant quantity, RMS is zero, for example. It is used everywhere mostly because variance (which is STD^2) is mathematically easily tractable: $var(X+Y)=var(X)+var(Y)$ if $X$ and $Y$ are independent (or even just uncorrelated). $\endgroup$
    – A.S.
    Commented Nov 19, 2015 at 22:29
  • $\begingroup$ Another nice property of variance is that $var(X-c)$ is minimized when $c=E(X)$. $\endgroup$
    – A.S.
    Commented Nov 19, 2015 at 22:37
  • 3
    $\begingroup$ RMS is not the same as standard deviation, as another user pointed out. Standard deviation accounts for the deviation of individual data points from the mean, whereas RMS accounts for the absolute magnitude of those data points as well. Only when the mean is zero are RMS and standard deviation the same. $\endgroup$ Commented Oct 10, 2017 at 20:01

1 Answer 1

20
$\begingroup$

in the case of standard deviation, the mean is removed out from obsevations, but in root mean square the mean is not removed. however in the case of noise where the mean is zero, the two concept are the same. I hope that this is the difference. see: http://www.madsci.org/posts/archives/2004-11/1100200293.Ph.r.html

$\endgroup$
3
  • $\begingroup$ Welcome to Math.SE and thanks for tackling an open Question. It is desirable to quote from or summarize what a Reader will find at the specified link when this supplies the main content of your Answer, so that an informed decision can be made whether to follow your link. Please review How to write a good Answer. $\endgroup$
    – hardmath
    Commented Oct 13, 2016 at 14:57
  • $\begingroup$ Upvoting the answer to counter the erroneous comment that "this does not provide an answer to the question." $\endgroup$ Commented Jan 8, 2017 at 0:42
  • $\begingroup$ When do I use the sd and when rms? $\endgroup$
    – Ben
    Commented Jan 30, 2023 at 9:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .