**Appendix B**

The revenue *R* is the product of *En*, the energy produced, and *C*, the price of energy. For a small amount of energy *dEn*, *dR* is equal to:

$$d\mathbb{R} = \mathbb{C} \, dE \,\mathfrak{n} \tag{A8}$$

The gain *G* is defined as the revenue per unit time:

$$\mathcal{G} = \frac{dR}{dt} = \mathbb{C}\,\frac{dEn}{dt} = \mathbb{C}P\tag{A9}$$

in which *P* = *dEn dt* is the power. Thus the expectation of the gain is the correlation of the price and the power.

$$E[\mathbf{G}] = E[\mathbf{C}\mathbf{P}] = \mathcal{R}\_{\mathbf{C}\mathbf{P}}.\tag{A10}$$

Since the price is a random process, which is largely independent of the power and is dependent, among other things, on the price of other energy sources and administrative decisions, we will assume that *C* and *P* are independent random variables, hence:

$$E[G] = \overline{\mathbb{C}} \cdot \overline{P} \tag{A11}$$

in which *E*[*C*] = *C* is the average price and *E*[*P*] = *P* is the power average. The variance of the gain, as its standard deviation square, is thus:

$$
\sigma\_{\mathbf{G}}\mathbf{r}^2 = \mathbb{E}\left[\mathbf{G}^2\right] - \overline{\mathbf{G}}^2 = \mathbb{E}\left[\mathbf{C}^2\mathbf{P}^2\right] - \overline{\mathbf{C}}^2\overline{\mathbf{P}}^2 = \mathbb{E}\left[\mathbf{C}^2\right]\mathbb{E}[\mathbf{P}^2] - \overline{\mathbf{C}}^2\overline{\mathbf{P}}^2\tag{A12}
$$

However, since

$$E\left[\mathbb{C}^2\right] = \sigma\_{\mathbb{C}}\,^2 + \overline{\mathbb{C}}^2 \,, \ E\left[P^2\right] = \sigma\_P\,^2 + \overline{P}^2\tag{A13}$$

it follows that:

$$
\sigma\_{\mathbb{C}}\,^2 = \left(\sigma\_{\mathbb{C}}\,^2 + \overline{\mathbb{C}}^2\right) \left(\sigma\_{\mathbb{P}}\,^2 + \overline{\mathbb{P}}^2\right) - \overline{\mathbb{C}}^2\overline{\mathbb{P}}^2 = \sigma\_{\mathbb{C}}\,^2\sigma\_{\mathbb{P}}\,^2 + \overline{\mathbb{C}}^2\sigma\_{\mathbb{P}}\,^2 + \overline{\mathbb{P}}^2\sigma\_{\mathbb{C}}\,^2\tag{A14}$$

Hence:

$$\frac{\sigma\_{\mathbb{C}}^{2}}{\overline{\mathbb{C}}^{2}} = \frac{\sigma\_{\mathbb{C}}}{\overline{\mathbb{C}}^{2}} + \frac{\sigma\_{\mathbb{P}}}{\overline{\mathbb{P}}^{2}} \left(1 + \frac{\sigma\_{\mathbb{C}}}{\overline{\mathbb{C}}^{2}}\right) \tag{A15}$$

Thus, if *<sup>σ</sup><sup>P</sup>* 2 *P* <sup>2</sup> <sup>≪</sup> *<sup>σ</sup><sup>C</sup>* 2 *C* 2 , and since necessarily *<sup>σ</sup><sup>C</sup>* 2 *C* <sup>2</sup> < 1, we can write:

$$\frac{\sigma\_{\mathbb{C}}^{2}}{\overline{\mathbb{C}}^{2}} \cong \frac{\sigma\_{\mathbb{C}}^{2}}{\overline{\mathbb{C}}^{2}}.\tag{A16}$$

However, notice that power fluctuations are not small, it is only the power average fluctuations over many samples that are small (see Appendix A), hence a more sophisticated approach is needed. We notice that the typical time for the change of power is much shorter than the typical time for the change of price. For example, we can assume that price changes once a month while power is sampled every ten min. We define ∆*T* as a duration for which the price remains constant. Next, we calculate the revenue for the total duration *T*:

$$R\_T = \int\_0^{R\_T} dR = \int\_0^T \mathbb{C}Pdt\tag{A17}$$

We now divide the duration *T* to the subduration of intervals ∆*T*, and integrate *R* as follows:

$$R\_T = \sum\_{n=1}^{N} \int\_{T\_{n-1}}^{T\_n} \mathbb{C}Pdt\tag{A18}$$

In the above, *<sup>T</sup><sup>n</sup>* <sup>−</sup> *<sup>T</sup>n*−<sup>1</sup> <sup>=</sup> <sup>∆</sup>*T*, and *<sup>T</sup>*<sup>0</sup> <sup>=</sup> 0, *<sup>T</sup><sup>N</sup>* <sup>=</sup> *<sup>T</sup>*. Since *<sup>C</sup>* is constant for <sup>∆</sup>*<sup>T</sup>* intervals, we can write:

$$R\_T = \sum\_{n=1}^{N} \mathbb{C}\_n \int\_{T\_{n-1}}^{T\_n} Pdt\tag{A19}$$

Let us further divide the duration ∆*T* into subdurations ∆*t* = <sup>∆</sup>*<sup>T</sup> <sup>M</sup>* , in which *M* is a large enough number; we may now write:

$$\int\_{T\_{n-1}}^{T\_n} Pdt = \sum\_{i=1}^{M} \int\_{T\_{ni-1}}^{T\_{ni}} Pdt \tag{A20}$$

In the above, *<sup>T</sup>ni* <sup>−</sup> *<sup>T</sup>ni*−<sup>1</sup> <sup>=</sup> <sup>∆</sup>*t*, and *<sup>T</sup>n*<sup>0</sup> <sup>=</sup> *<sup>T</sup>n*−1, *<sup>T</sup>nM* <sup>=</sup> *<sup>T</sup>n*. If *<sup>M</sup>* is large enough, we may write approximately:

$$\int\_{T\_{n-1}}^{T\_n} Pdt \cong \sum\_{i=1}^{M} \Delta t P\_{\text{ni}} = \frac{\Delta T}{M} \sum\_{i=1}^{M} P\_{\text{ni}} = \Delta T P\_{\text{na}}.\tag{A21}$$

where *Pna* is an average in the sense of (A3), hence the fluctuations of *Pna* are small enough. The total revenue is therefore:

$$R\_T = \Delta T \sum\_{n=1}^{N} \mathbb{C}\_n P\_{\text{na}} = \Delta T \sum\_{n=1}^{N} \mathbb{G}\_{\text{na}} \tag{A22}$$

Thus the revenue is proportional to the sum of the average gains *Gna*, which, according to Equations (A14) and (A15), have fluctuations that are a result of the dominant price fluctuations and the negligible power fluctuations.
