# Why is the RMS value taken to calculate uncertainty in random errors

My Text Book mentions the use of RMS (Root Mean Square) to calculate the value of uncertainty.

"Random errors are handled using statistical analysis. Assume that a large number ($$N$$) of measurements are taken of a quantity $$Q$$ giving values $$Q_1, Q_2, Q_3,…Q_N.$$ Let $$Q$$ be the mean value of these measurements. $$\langle Q\rangle = \sum_{i=1}^n \frac{Q_i}{N}$$ and let $$d$$ be the deviation in the measurements, $$d=\sqrt\frac{\sum_{i=1}^n(Q-\langle Q\rangle)^2}{N}$$ The result of the measurement is quoted (assuming systematic errors have been eliminated) as,

$$Q = \langle Q\rangle +d$$ My question is, how did we arrive at the expression for $$d$$ in the first place?

• d is basically the standard deviation of the set of $Q$'s if that is your question. Oct 9, 2021 at 5:22
• Yes, why exactly do we take the standard deviation only? Sorry, I do not have much knowledge of statistics. Oct 9, 2021 at 7:51

When the errors that contribute to fluctuations in a measured value are normal in their distribution, an infinite number of measurements will yield a Gaussian distribution. The mean $$\mu$$ and standard deviation $$\sigma$$ (square root of the variance) characterize the distribution and its uncertainty.

For a finite number of measurements on values $$V_j$$ where contributions to error in $$V_j$$ remain normal in their distribution, we use the average $$\langle V \rangle$$ and standard uncertainty $$\Delta V$$ of the distribution to define the expected value and its overall uncertainty. This is the report that you present.

The average approaches the mean $$\langle V \rangle \rightarrow \mu$$ and the standard uncertainty approaches the standard deviation $$\Delta V \rightarrow \sigma$$ as the number of measurements we make approach infinity. We can characterize the approach using the standard uncertainty of the mean $$S_m$$. Further considerations lead to approaches such as the t-test for the confidence of the measured average relative to the true mean.

When the errors that contribute to $$V_j$$ are not normal in their distribution, additional complexities arise. Also, when we have to combine various values using an equation, we have to consider how we must propagate the uncertainties in the measured values. Both topics are expansions of your starting question.

A thorough reference to appreciate uncertainties or errors in measurements can be found in the Guide to Uncertainties in Measurements (GUM).

Perhaps two examples will clarify things

1. Data 1,2,3

The mean is $$\langle Q\rangle = \sum_{i=1}^n \frac{Q_i}{N} = \frac{6}{3} = 2$$

The standard deviation formula is a way to calculate a 'measure of spread' i.e. an indication of how spread out the data is. This can also be an indication of the errors in the measurement.

Since some data is lower than the mean and some higher, the square is included before the square root to make all the differences from the mean positive, otherwise we might be using a formula that gives a spread of zero.

$$d=\sqrt\frac{\sum_{i=1}^n(Q-\langle Q\rangle )^2}{N} =\sqrt\frac{(-1)^2+0^2+(1)^2}{3} = 0.8165$$

1. Data 10,20,30

The mean is now $$20$$, and $$d$$ works out to be $$8.165$$

The second set of data is more spread out, or the error in the measurements is higher.

In each case the 'true' value of the thing being measured is likely to be in the range $$Q = \langle Q\rangle \pm d$$

In the second example, if a theory predicted that $$Q=40$$ and measurements had given $$20 \pm 8.165$$ the probability that the $$40$$ could be true, can then be found using a normal distribution and finding the probability that a measurement would be $$\frac{20}{8.165}$$ i.e. $$2.45$$ standard deviations or $$2.45d$$ from the mean.

If the true standard deviation isn't known, but is estimated from the data, there is a further adjustment, but that's a technicality...

The purpose of taking any sort of average is to come up with a number that represents the whole data set--one that is, in some sense, close to most of the points. A natural way to calculate how well a number summarizes a data set is to sum the distance from that point to all other points: $$D_1 = \sum_i |Q_i - \hat{q}|$$ where $$D_1$$ is the total distance, $$Q_i$$ is each data point, and $$\hat{q}$$ is the candidate summary point. The best choice of $$\hat{q}$$ is the one that minimizes $$D_1$$. This minimum turns out to be the median of the data set, where $$\hat{q}$$ is less than or equal to at least half of the data points and greater than or equal to the other half (more simply, sort the data points and take the middle one). The associated spread of the data set around the median is the $$D_1/N$$: $$S = \frac{1}{N}\sum_i |Q_i - \hat{q}|.$$ We divide by $$N$$ so taking a bigger sample doesn't make the spread bigger.

One quirk of this choice of distance is that the magnitude of extreme points don't affect the median. The median of $$\{1, 2, 3, 4, 5\}$$ is $$3.$$ The median of $$\{1, 2, 3, 40, 500\}$$ is also $$3.$$ For some purposes, this is fine [1], but for others we want extreme values to affect the average.

To make sure extreme points affect the average, we choose another metric: the squared distance. $$D_2 = \sum_i \left(Q_i - \bar{q}\right)^2$$ Squaring the distance magnifies the influence of values that are far from $$\hat{q}$$. The minimum of $$D_2$$ occurs when $$\bar{q}$$ is the mean of the data set. $$\bar{q} = \frac{1}{N}\sum_i Q_i.$$ We once again define the spread as $$D_2/N.$$ $$\sigma^2 = \frac{1}{N}\sum_i \left(Q_i - \bar{q}\right)^2$$ But, this quantity has units of $$Q^2,$$ which often isn't easily comparable to $$Q$$ and makes it impossible to write $$Q = \bar{q} \pm \sigma^2.$$ So, we define the spread around the mean as the square root of this value: $$\sigma = \sqrt{\sigma^2} = \sqrt{\frac{1}{N}\sum_i \left(Q_i - \bar{q}\right)^2}.$$ Now we can write $$Q = \bar{q} \pm \sigma.$$

In summary, the formula for the deviation comes from our choice of how to calculate the quality of the average value of a data set. That choice of quality calculation then determines how the average itself is calculated.

[1] One example of using the median instead of the mean is in economic data. Calculating the mean wealth of a population can be misleading due to a few extremely rich persons dragging the mean upwards. If a sample of 10 has one person with a million dollars and nine with nothing, the mean wealth is \$100,000. This sounds like well-off sample. The median, however, is \$0, which gives a better picture of how well-off these people are.