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I'm currently reading "Concepts in Thermal Physics", and in the chapter on independent variables it has the following example:

If we have $n$ independent variables $X_i$, each with a mean $\left<X\right>$, and a variance $\sigma_X^2$, we can sum them to get the following:

$$\begin{split} Y & = \sum^n_iX_i \\ \left<Y\right> & = \sum^n_i\left<X_i\right> = n\left<X\right> \\ \sigma_Y^2 & = n\sigma_X^2 \end{split}$$

I understand the derivation of all this fine, however the following is then stated:

The results proved in this last example have some interesting applications. The first concerns experimental measurements. Imagine that a quantity $X$ is measured $n$ times, each time with an independent error, which we call $\sigma_X$. If you add up the results of the measurements to make $Y = \sum_iX_i$, then the rms error in $Y$ is only $\sqrt{n}$ times the rms error of a single $X$. Hence if you try and get a good estimate of $X$ by calculating $(\sum_iX_i)/n$, the error in this quantity is equal to $\sigma_X/ \sqrt{n}$.

I'm not entirely sure what they mean here by the root mean square error. Is that just another way of saying the standard deviation? If it is, in what sense can the above example lead to the statement that follows?

The only way I can personally see this making sense, is if they are modelling the error in a single measurement as the standard deviation of a probability distribution. This doesn't seem correct to me, is this actually what they are doing?

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2 Answers 2

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Regarding your first question about rms error:

Say the true value of $X$ is $\bar{X}$, and you measured $X_i$ (which on average should be $\bar{X}$).

The measurement error would be: $X_i - \bar{X}$.

The mean of the square of the errors would be $\langle (X_i - \bar{X})^2 \rangle $ which is exactly the variance.

The root of the mean of the squares is the square-root of the variance, meaning the standard deviation.


Second, after you had $n$ measurements you want to estimate $\bar{X}$, so you average your measurements and get $\langle X_i \rangle $. Of course, this cannot be equal precisely to $\bar{X}$ because all of these numbers are on a continuum. So how far off are you from the truth? The central limit theorem tells us that after taking enough measurement, no matter the distribution of $X_i$, your estimation will behave as a Gaussian with a mean of $\bar{X}$ and standard deviation of $\frac{\sigma}{\sqrt{n}}$, meaning the more you increase $n$, the narrower your gaussian will be and the closer your estimation will be to the truth. The intuition behind this is as @Physics Enthusiast answered.

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  • $\begingroup$ I see what you're saying, and thankyou for the Central Limit theory! But in this case the statement says each measurement has some error $\sigma_X$, but the error in the case of a single measurement is not a standard deviation of a distribution, it's a single value. Secondly I have seen the rms calculated thus: $$\sqrt{\frac{1}{n}\sum_i^n\sigma_{X_i}^2}$$ But it seems in the statement that all the $\sigma_{X_i}$ are the same, hence you would have: $$\sqrt{\sigma_{X}^2} = \sigma_{X}$$ Not the $\frac{\sigma_{X}}{\sqrt{n}}$ as expected. $\endgroup$
    – Connor
    Sep 21, 2021 at 8:29
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    $\begingroup$ @Connor Physicists like to use the word error in 2 different (but related) ways: sometimes they mean the true error $X_i - \bar{X}$ and sometimes they mean the standard deviation of the distribution of $X_i$, meaning, $\sigma_X$. About the second question, there is the rms of the errors in $X_i$, which is $\sigma_X$ as you calculated, and there is standard deviation of the average of the measurements, which has a distribution with a lower standard deviation than each one of the measurements - that's the point of this paragraph, the more measurements you take, your average will be more precise. $\endgroup$ Sep 21, 2021 at 8:46
  • $\begingroup$ Okay! Thanks, that makes a lot of sense. So is it fair to say the statement is slightly sloppy in defining terms in that case? Or are rms error and standard deviation interchangable? $\endgroup$
    – Connor
    Sep 21, 2021 at 10:35
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    $\begingroup$ Both statements are correct :) $\endgroup$ Sep 21, 2021 at 12:21
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Let's say you do $n$ measurements and add up the results of the measurements. The reason the rms error in the sum is not $n$ times the rms of a single error is that the errors are assumed to be independent ("each time with an independent error"), so they cancel out each other to some degree (i.e., one experiment might give a positive error, the next experiment might give a negative error, etc.).

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  • $\begingroup$ So you would say intuitively that repeating the experiment gives you a distribution of errors, from which you can get a standard deviation which is considered a reduced error? $\endgroup$
    – Connor
    Sep 21, 2021 at 10:41
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    $\begingroup$ I don't think I would phrase it that way, but I think your overall idea is correct. The key is to make sure you understand that there are two ways that the word error can be used (@Ofek Gillon explains this very well in the main response as well as the comment that starts with "Physicists like to use the word error ...") $\endgroup$ Sep 21, 2021 at 15:12

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