Given a set of $n$ values, the error associated with their average will be

$$\text{standard deviation}/\sqrt{n}.$$

But if the values themselves have an uncertainty attached to them, such as $100\pm 1,$ $110 \pm 1, \ldots$ is the previous formula correct?

  • 3
    $\begingroup$ for independent sources of noise (with variances $\sigma_1^2$ and $\sigma_2^2$) you could add the variances and then the error would be $N^{-1/2}\sqrt{\sigma_1^2+\sigma_2^2}$. $\endgroup$ May 9 at 21:19

1 Answer 1


One interpretation of your situation is that you have obtained a random sample $Y_1, \ldots, Y_n$ from a distribution with a finite variance, say $\sigma^2,$ but you have observed the values $X_i = Y_i + E_i$ where the variances of the "errors" $E_i$ are known. Let's refer to these error variances as $\sigma_i^2.$

One interpretation of the "error associated with the average" is that you are looking for the variance of the arithmetic mean of the $X_i.$ This mean is defined as

$$\bar X = \frac{1}{n}\sum_{i=1}^n X_i = \frac{1}{n}\sum_{i=1}^n \left(Y_i + E_i\right) = \bar Y + \bar E.$$

Assuming all the values $Y_i$ and all the errors $E_j$ are uncorrelated, the laws of variances say

$$\operatorname{Var}(\bar X) = \frac{1}{n^2}\sum_{i=1}^n (\operatorname{Var}(Y_i) + \operatorname{Var}(E_i)) = \frac{\sigma^2}{n} + \frac{1}{n^2}\sum_{i=1}^n \sigma_i^2.$$

Your formula for the "error" of $\bar X$ is the square root of the right hand side, neglecting the observational error variances $\sigma_i^2.$

At this point we're stuck, because we almost never assume we know $\sigma^2.$ Usually we are trying to use the observations $X_i$ to estimate it. But we can still say some things from this result, including

  1. The error of $\bar X$ exceeds the error of $\bar Y$ (which is $\sigma/\sqrt{n}$).

  2. Since you know or assume you know the $\sigma_i^2,$ if you can somehow obtain an estimate of $\sigma^2,$ then you can adjust it to estimate $\operatorname{Var}(\bar X).$

One subtlety is that with observations reported in the form "$X_i \pm \tau_i,$" as in the question, it is possible that the $\tau_i$ equal the $\sigma_i,$ or $2\sigma_i,$ or $1.32 \sigma_i,$ or $1.645\sigma_i,$ or maybe even something else--I have listed only the commonest conventions. You have to rely on your source of these data to tell you what it means.


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