Imagine that I measure 9 times a certain physical properties (which I know its exact value is 11) and the outcomes are:

$$X_i = 10, 11, 11, 10, 11, 12, 10, 11, 12$$

If I analyse $X_i$ as a random variable I can compute it's expected value and variance as it follows:

$$\mathbb{E}(X_i) = \frac{1}{9} \sum X_i = 11$$ $$Var(X_i) = \frac{1}{9} \sum (X_i-11)^2 =0.5$$

Hence the measurement can be written as $X = 10.9 \pm \sqrt{0.5}$

If I define a second random variable:

$$\bar X_j =\frac{1}{3} \sum_{\textrm{group of 3 $X_i$}} X_i$$

This means that:

$$X_i = 10.7, 11, 11$$ $$\mathbb{E}(\bar X_i) = \frac{1}{3} \sum \bar X_i = 10.9$$ $$Var(\bar X_i) = \frac{1}{3} \sum (\bar X_i-11)^2 = 0.05$$ Hence the measurement can be written as $X = 10.9 \pm \sqrt{0.05}$

The fact that the variance is smaller in the second case, follows directly by the sum of iid and goes like $1/\sqrt{N}$ where $N$ is the number of averaged value to obtain $\bar X_i$.

From a mathematical point of view it seems all clear but from a physical point of view what does it mean? If I consider averaged values the error I make on my measurement is smaller? The uncertainty of my measurement depends on how I treat my data?

  • $\begingroup$ This analysis is a particularly clever way to motivate the concept of standard error, q.v.. $\endgroup$
    – whuber
    Commented Oct 3, 2016 at 14:49
  • $\begingroup$ Note that what you are computing are sample statistics rather than random variable mean and variance. The former depend on the realized values of the random variables (the 9 numbers you use) while the latter are properties of the random variable itself. Therefore, you should divide by $n-1$ when computing variances (i.e., by 8 and by 2 respectively). $\endgroup$
    – A.G.
    Commented Oct 3, 2016 at 15:48

1 Answer 1


Very generally speaking, you are repeatedly decreasing the uncertainty of your derived values by averaging, which you already saw in your formulas. As an intuitive example, let's say, you are glueing sticks (let's say three at a time) with different lengths end to end together and then saw them into three equal length sticks. When you do this with all your sticks, the lengths of the resulting sticks will be closer together than the original population. If you do this repeatedly you are amplfying the effect. Note that this will only work if you combine new sticks everytime. This is because only new sticks add new information about stick lengths to the process. Glueing and cutting the same three sticks repeatedly does nothing more than doing it once, it's an idempotent operation.

  • $\begingroup$ Thank you for your reply, but what I can't see is how can the standard deviation account for statistical errors (which I assume have some physical origin), if they depend on how I play with my data. Imagine that I want to present an experiment showing the measurement above. What standard deviation should I use? I guess it doesn't matter as soon as everybody knows which one I am using right? And in any case a guy which makes the measurement of X will observe a spread $\sqrt{0.5}$ on its measurement. Right? $\endgroup$
    – Worldsheep
    Commented Oct 3, 2016 at 15:13
  • $\begingroup$ @Worldsheep. If you look at your data you will see that the spread of measurement error is from $-1$ ($X_1=10$) to $+1$ ($X_9=12$). $\endgroup$
    – A.G.
    Commented Oct 3, 2016 at 15:54

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