The effect of intermittent renewables on the electricity price variance

Abstract

The dominating view in the literature is that renewable electricity production increases the price variance on spot markets for electricity. In this paper, we critically review this hypothesis. Using a static market model, we identify the variance of the infeed from intermittent electricity sources (IES) and the shape of the industry supply curve as two pivotal factors influencing the electricity price variance. The model predicts that the overall effect of IES infeed depends on the produced amount: while small to moderate quantities of IES tend to decrease the price variance, large quantities have the opposite effect. In the second part of the paper, we test these predictions using data from Germany, where investments in IES have been massive in the recent years. The results of this econometric analysis largely conform to the predictions from the theoretical model. Our findings suggest that subsidy schemes for IES capacities should be complemented by policy measures supporting variance absorbing technologies such as smart-grids, energy storage, or grid interconnections to ensure the build-up of sufficient capacities in time.

Appendix

The proof of Proposition 1 requires the following proposition:

Proposition 2

Let $X$ be a positive random variable which takes values on an interval $[a,b]$. Let $h: \mathbb {R}^+ \times \mathbb {R}\rightarrow \mathbb {R}$ be a function such that the following hold:

• 1. $h(X,\cdot )$ is twice differentiable

• 2. $h(\cdot ,\epsilon )$ is increasing $\forall \epsilon \in [a,b]$, and

• 3. $\frac{\partial }{\partial \epsilon }h(\cdot ,\epsilon )$ is decreasing $\forall \epsilon \in [a,b]$.

Then for $\epsilon _1,\epsilon _2 \in [a,b]$ with $\epsilon _1 < \epsilon _2$ it holds true that $Var[h(X,\epsilon _2)] \le Var[h(X,\epsilon _1)].$

Proof

The proof is analogue to the proof of (See and Chen 2008, Theorem3.1). It suffices to show that $Var[h(X,\epsilon )]$ is a decreasing function of $\epsilon$. We have

$$\begin{aligned} \frac{\partial }{\partial \epsilon } {\text {Var}}(h(X, \epsilon ))= & {} \frac{\partial }{\partial \epsilon } \mathbb {E}(h(X,\epsilon )^2) - \frac{\partial }{\partial \epsilon } (\mathbb {E}(h(X,\epsilon ))^2 \\= & {} \mathbb {E}\left[ \frac{\partial }{\partial \epsilon } h(X,\epsilon )^2\right] - 2\mathbb {E}(h(X,\epsilon )) \frac{\partial }{\partial \epsilon } (\mathbb {E}(h(X,\epsilon )) \\= & {} \mathbb {E}\left[ 2 h(X,\epsilon ) \frac{\partial }{\partial \epsilon } h(X,\epsilon ) \right] - 2\mathbb {E}(h(X,\epsilon )) (\mathbb {E}( \frac{\partial }{\partial \epsilon } h(X,\epsilon )) \\\le & {} 0. \end{aligned}$$

The last inequality follows from the fact that for a random variable $X$, and for continuous functions $f,g$ on $\mathbb {R}$, $\mathbb {E}[f(X)g(X)] \le \mathbb {E}[f(X)] \mathbb {E}[g(X)]$ if $f$ is monotonically increasing and $g$ is monotonically decreasing (see, e.g., Gurland 1967). The interchange of the differential operator and the expectation is clearly justified by the assumptions we have made. $\square$

Proof

(Proof of Proposition 1) The proof is analogue to (See and Chen 2008, Theorem 3.3). We sketch the important steps.

Define $h(u,\epsilon ) = g \left( F_X^{-1}(u) + \epsilon [ F_Y^{-1}(u) - F_X^{-1}(u) ] \right)$, with $\epsilon \in [0,1]$. First, note that $h(\cdot ,\epsilon )$ is an increasing function $\forall \epsilon \in [0,1]$. Second, note that $\frac{\partial }{\partial \epsilon }h(\cdot ,\epsilon )$ is a decreasing function since

$$\begin{aligned} \frac{\partial }{\partial u}\frac{\partial }{\partial \epsilon }h(\cdot ,\epsilon )&= \frac{\partial }{\partial u} \left\{ [ F_Y^{-1}(u) - F_X^{-1}(u)] \ g^{\prime } \left( F_X^{-1}(u) + \epsilon [ F_Y^{-1}(u) - F_X^{-1}(u) ] \right) \right\} \\&= [(F_Y^{-1}{})^{\prime } -( F_X^{-1}{})^{\prime } ] \ g^{\prime } \left( F_X^{-1}(u) + \epsilon [ F_Y^{-1}(u) - F_X^{-1}(u) ] \right) \\&+ \epsilon (F_X^{-1}{})^{'} \ [F_Y^{-1} - F_X^{-1} ] \ g^{\prime \prime } \left( F_X^{-1}(u) + \epsilon [ F_Y^{-1}(u) - F_X^{-1}(u) ] \right) \\&+ (1 \!- \epsilon ) (F_X^{-1}{})^{'} \ [F_Y^{-1} \!- F_X^{-1} ] \ g^{\prime \prime } \left( F_X^{-1}(u) + \epsilon [ F_Y^{-1}(u) - F_X^{-1}(u) ] \right) \\&\le 0 \end{aligned}$$

For the last inequality note that the first term is negative due to condition (2) and $g$ being increasing ($g^\prime >0$), while the second and third terms are negative due to condition (1), convexity of $g$ ($g^{\prime \prime }>0$), and properties of distribution functions, $(F_X^{-1})^{\prime } \ge 0$ and $(F_Y^{-1})^{\prime } \ge 0$.

Now one can apply Proposition 2 and find

$$\begin{aligned} \mathrm{Var}[g(Y)] = \mathrm{Var}[h(u,1)] \le \mathrm{Var}[h(u,0)] = \mathrm{Var}[g(X)]. \end{aligned}$$

$\square$

The following lemma links inequality (1) to the variance of the random variables.

Lemma 1

If for two random variables $X$ and $Y$

$$\begin{aligned}F^{-1}_Y(\alpha ) - F^{-1}_X(\alpha )\end{aligned}$$

is monotonically decreasing, then ${\text {Var}}(Y) \le {\text {Var}}(X)$.

Proof

Notice that for $\alpha >\alpha '$, we have

$$\begin{aligned}F^{-1}_Y(\alpha ) - F^{-1}_X(\alpha ) \le F^{-1}_Y(\alpha ') - F^{-1}_X(\alpha ')\end{aligned}$$

and, therefore,

$$\begin{aligned}F^{-1}_Y(\alpha ) - F^{-1}_Y(\alpha ') \le F^{-1}_X(\alpha ) - F^{-1}_X(\alpha '),\end{aligned}$$

i.e., the inter-quantile range of $Y$ is smaller than that of $X$ for all quantiles $\alpha$, $\alpha '$.

Setting $\alpha _1 = F_X(\mathbb {E}(X))$ and $\alpha _2 = F_Y(\mathbb {E}(Y))$, yields

$$\begin{aligned} {\text {Var(X)}}= & {} \int _{-\infty }^\infty (x-\mathbb {E}(X))^2 f_X(x) dx = \int _0^1 (F_X^{-1}(\alpha )-F^{-1}_X(\alpha _1))^2 du \\\ge & {} \int _0^1 (F_Y^{-1}(\alpha )-F_Y^{-1}(\alpha _1))^2 du \ge \int _0^1 (F_Y^{-1}(\alpha )-F_Y^{-1}(\alpha _2))^2 du \\= & {} {\text {Var}}(Y), \end{aligned}$$

where the last inequality follows from

$$\begin{aligned}\mathbb {E}(Y) = F^{-1}_Y(\alpha _2) = \arg \min \left\{ a: \mathbb {E}\left[ (Y-a)^2\right] \right\} .\end{aligned}$$

$\square$