**Appendix A**

Consider a random process *X*(*t*), which is sampled at times *t<sup>i</sup>* , each sample *X<sup>i</sup>* = *X*(*ti*) is a random variable. We will assume that the process is stationary in the sense that each random variable has the same expectation value:

$$E[X\_i] = E\left[X\_j\right] = \overline{X} \quad \forall \; i, j \tag{A1}$$

and standard deviation:

$$
\sigma\_{\mathbf{x}\_{\mathbf{i}}} = \sigma\_{\mathbf{x}\_{\mathbf{j}}} = \sigma\_{\mathbf{x}} \quad \forall \, \mathbf{i}\_{\mathbf{i}} \mathbf{j} \tag{A2}
$$

We will also assume that the random variables are independent. Thus the covariance satisfies *Cxix<sup>j</sup>* = 0 , ∀ *i* 6= *j*. Let us now average *n* samples as:

$$X\_d = \frac{1}{n} \sum\_{i=1}^n X\_i. \tag{A3}$$

This implies that the expectation value of the average will be the same as that of the samples:

$$E[X\_a] = E\left[\frac{1}{n}\sum\_{i=1}^n X\_i\right] = \frac{1}{n}\sum\_{i=1}^n E[X\_i] = \frac{1}{n}\sum\_{i=1}^n \overline{X} = \overline{X} \tag{A4}$$

However, the standard deviation of *X<sup>a</sup>* will be much smaller, since:

$$\begin{split} \sigma\_{\mathbf{X}\_{d}}^{2} &= \operatorname{E} \left[ (\mathbf{X}\_{d} - \overline{\mathbf{X}\_{d}})^{2} \right] = \frac{1}{n^{2}} \operatorname{E} \left[ \sum\_{i,j}^{n} (\mathbf{X}\_{i} - \overline{\mathbf{X}\_{i}})(\mathbf{X}\_{j} - \overline{\mathbf{X}\_{j}}) \right] \\ &= \frac{1}{n^{2}} \sum\_{i,j}^{n} \mathbf{C}\_{\mathbf{x}\_{i} \mathbf{x}\_{j}} = \frac{1}{n^{2}} \sum\_{i}^{n} \mathbf{C}\_{\mathbf{x}\_{i} \mathbf{x}\_{i}} = \frac{1}{n^{2}} \sum\_{i}^{n} \sigma\_{\mathbf{X}\_{i}}^{2} = \frac{1}{n^{2}} \sum\_{i}^{n} \sigma\_{\mathbf{X}}^{2} = \frac{\sigma\_{\mathbf{X}}^{2}}{n} \end{split} \tag{A5}$$

Hence:

$$
\sigma\_{\mathbf{X}\_d} = \frac{\sigma\_{\mathbf{X}}}{\sqrt{n}}.\tag{A6}
$$

Additionally, for a large amount of samples:

$$\lim\_{n \to \infty} \sigma\_{\mathbf{X}\_{4}} = \lim\_{n \to \infty} \frac{\sigma\_{\mathbf{X}}}{\sqrt{n}} = 0. \tag{A7}$$
