Price formation models Diogo A. Gomes

Transcription

1 models Diogo A. Gomes

2 Here, we are interested in the price formation in electricity markets where: a large number of agents owns storage devices that can be charged and later supply the grid with electricity; agents seek to maximize profit by trading electricity at a price ϖ(t), which is set by a supply versus demand balance condition.

3 Our model comprises three quantities of interest: a price ϖ C([0, T ]) a value function u C(R [0, T ]) a path describing the statistical distribution of the agents, m C([0, T ], P).

4 Problem Given ɛ 0, a Hamiltonian, H : R R R, H C, an energy production rate Q : [0, T ] R, Q C ([0, T ]), a terminal cost ū : R R, ū C (R) and an initial probability distribution m P Cc (R), find u : R [0, T ] R, m C([0, T ], P), and ϖ : [0, T ] R solving u t + H(x, ϖ(t) + u x ) = ɛu xx m t (D p H(x, ϖ(t) + u x )m) x = ɛm xx Ω D ph(x, ϖ(t) + u x )dm = Q(t), and satisfying the initial-terminal conditions { u(x, T ) = ū(x), m(x, 0) = m(x). (1)

5 u(x, t) is the value function for an agent whose charge is x at time t. u is a (continuous) viscosity of the first equation; if ɛ > 0, parabolic regularity gives additional regularity. m(x, t) determines the distribution of the energy storage of the agents at time t. m is a weak solution of the second equation. the spot price, ϖ(t), is selected so supply Q(t) is balances, the condition imposed by the last equation.

6 Main Result Theorem Under the assumptions below, there exists a solution (u, m, ϖ) where u is a viscosity solution of the first equation, Lipschitz and semiconcave in x, and differentiable almost everywhere with respect to m, m C([0, T ], P), and ϖ is Lipschitz continuous. Moreover, if ɛ > 0 this solution is unique. If ɛ = 0, under additional convexity assumptions on V and the terminal cost, there is a unique solution (u, m, ϖ). Moreover, u is differentiable in x for every x, and u xx and m are bounded.

7 A mean-field model for price formation Model derivation I To simplify, we set ɛ = 0. Each consumer has a storage device that is connected to the network, for example, an electric car battery. Consumers trade electricity, charging the batteries when the price is low and selling electricity to the market when the price is high.

8 A mean-field model for price formation A typical consumer has a battery with charge x(t). This charge changes according to ẋ(t) = α(t). Each consumer seeks to select α to minimize the cost.

9 A mean-field model for price formation This cost is determined by the Lagrangian l(α, x, t) = l 0 (α, x) + ϖ(t)α(t). For example, we often take l 0 (α, x, t) = c 2 α2 (t) + V (x). (2)

10 A mean-field model for price formation The singular case where V (x) = { 0 if 0 x 1 + otherwise, corresponds to the case where the battery charges satisfies 0 x 1.

11 A mean-field model for price formation Each consumer minimizes the functional J(x, t, α) = T t l(α(s), x(t), t)ds + ū(x(t )), where ū is the terminal cost and α A t, where A t is the set of bounded measurable functions α : [t, T ] A R.

12 A mean-field model for price formation The value function, u, is the infimum of J over all controls in A t ; that is, u(x, t) = inf α A t J(x, t, α). The Hamiltonian, H, for the preceding control problem is H(x, p) = sup ( pa l 0 (x, a)). a A For example, for l 0 as in (2), we have H(x, p) = p2 2c + V (x).

13 A mean-field model for price formation From standard optimal control theory, u is a viscosity solution of { u t + H(x, ϖ(t) + u x ) = 0 For l 0 as before: u(x, T ) = ū(x). u t + 1 2c (u x + ϖ(t)) 2 V (x) = 0. At points of differentiability of u, α (t) = D p H(x, ϖ(t) + u x (x(t), t)).

14 A mean-field model for price formation The associated transport equation is: { m t (D p H(x, u x + ϖ(t))m) x = 0, m(x, 0) = m(x), where m is the initial distribution of the agents. Taking l 0 as before, we have m t 1 c (m(ϖ + u x)) x = 0.

15 A mean-field model for price formation Finally, we fix an energy production function Q(t) and require that the production balances demand: α (t)m(x, t)dx = Q(t); that is, R R D p H(x, u x + ϖ(t))m(x, t)dx = Q(t). This constraint determines the price, ϖ(t).

16 Main Assumptions Assumption The Hamiltonian H is the Legendre transform of a convex Lagrangian: H(x, p) = sup pα l 0 (α) V (x), (3) α R where l 0 C 2 (R) is a uniformly convex function and V C 2 (R) is bounded from below.

17 Main Assumptions Assumption The potential V in (3) and the terminal data ū are globally Lipschitz.

18 Main Assumptions Assumption The potential V in (3) and the terminal data ū satisfy for some positive constant C. D 2 xxv C, D 2 xxū C

19 Main Assumptions Assumption There exists a constant, C > 0, such that m xx, ū xx C.

20 Main Assumptions Assumption There exists θ > 0 such that D 2 pph(x, p) > θ for all x, p R. In addition, there exists C > 0 such that D 3 ppph C.

21 Main Assumptions Existence proof main steps Bounds for u Bounds for m Bounds for ϖ Fixed point argument

22 Existence of a solution Proposition u(x, t) is locally bounded and the map x u(x, t) is Lipschitz for 0 t T. Moreover, the Lipschitz bound on u does not depend on ϖ nor on ɛ.

23 Existence of a solution Proposition Then, x u(x, t) is semiconcave with a semiconcavity constant that does not depend on ɛ nor on ϖ.

24 Existence of a solution Proof Fix an optimal control α for (x, t): [ T u(x, t) = E Then, for any h R, we have [ T u(x±h, t) E Therefore, t t ] l 0 (α ) + ϖα + V (x )ds + ū(x(t ) ). ] l 0 (α ) + ϖα + V (x ± h)ds + ū(x(t ) ± h). u(x + h, t) 2u(x, t) + u(x h, t) Ch 2. Note that C does not depend on ϖ, only on T and on the semiconcavity estimates for V and ū.

25 Existence of a solution Proposition Suppose that ϖ n ϖ uniformly on [0, T ], then u n u locally uniformly and u n x u x almost everywhere.

26 Existence of a solution Proof The local uniform convergence of u n follows from the stability of viscosity solutions. Because u n is semiconcave and converges uniformly to u, u n x u x almost everywhere.

27 Existence of a solution Now, we examine the Fokker-Planck equation. { m t div(md p H(x, ϖ + u x )) = ɛ m, m(x, 0) = m(x). (4) Let P denote the set of probability measures on R with finite second-moment and endowed with the 1-Wasserstein distance.

28 Existence of a solution Proposition Suppose ɛ > 0. The Fokker-Planck equation has a solution m C([0, T ], P). Moreover, d 1 (m(t), m(t + h)) Ch 1/2. In addition, for any sequence ϖ n ϖ uniformly on [0, T ] and corresponing solutions u n and m n, we have m n m in C([0, T ], P 1 ).

29 Existence of a solution Proof The existence of a solution in C([0, T ], P 1 ) and the estimate in the statement are well known. The constant C can be chosen to depend only on ɛ 0 for all ɛ < ɛ 0, on the problem data, and on ϖ L. By the Ascoli-Arzela theorem, we have that m n m in C([0, T ], P 1 ) Because ɛ > 0, m n m in L 2. Moreover, the Fokker-Planck equation has a unique solution. Thus, it suffices to check that m is a solution. Because ux n u x, almost everywhere, by semiconcavity, we have for any ψ Cc T ψ x D p H(x, ϖ n + ux n )m n dxdt 0 R is convergent. Hence, m is a weak solution.

30 Existence of a solution Proposition Let ɛ > 0. Then T 0 R D 2 pphu 2 xxmdxdt C.

31 Existence of a solution Proof We diffferentiate the Hamilton-Jacobi twice with respect to x, multiply by m, and integrate by parts using the Fokker-Planck equation.

32 Existence of a solution Price-supply relation We observe that there exists a unique ϑ 0 such that D p H(x, ϑ 0 + u x (x, 0)) mdx = Q(0). R Moreover, ϑ 0 is bounded by a constant that depends only on the problem data.

33 Existence of a solution Price-supply relation Next, we differentiate D p H(x, ϖ + u x )mdx = Q(t) R R to get the identity ϖ DppHmdx 2 [ ] + D 2 pp Hu xt m + D p Hm t dx = Q. R

34 Existence of a solution Price-supply relation Now, we use the equations to get the identity DppHu 2 xt m + D p Hm t = DppH 2 ( ɛ u x + D p Hu xx + D x H) m R R + D p H (ɛ m + (md p H) x ). R R Because DxpH 2 = 0, we have DppHu 2 xt m + D p Hm t = DppHD 2 x Hm + ɛdppphu 3 xxm. 2 R

35 Existence of a solution Price tracking ODE Accordingly, ϖ DppHm 2 = Q R R ( D 2 pp HD x H + ɛdppphu 3 xx 2 ) m.

36 Existence of a solution Given ϖ, we solve the Hamilton-Jacobi equation, then the Fokker-Planck equation, and define ϑ = Q R D2 pp H(x,ϖ+ux )Dx H(x,ϖ+ux )m+ɛd3 ppp H(x,ϖ+ux )u2 xx m ϑ(0) = ϑ 0, R D2 pph(x,ϖ+ux )m (5) where ϑ 0 is as before. Then, (u, m, ϖ) solves (1) if ϖ solves (5).

37 Existence of a solution Proposition Consider the setting of Problem 1 with ɛ > 0. Suppose that ϖ n ϖ uniformly in C([0, T ]). Let u n, m n, and ϑ n be the corresponding solutions. Then, ϑ n converges to ϑ, uniformly in C([0, T ]), where ϑ solves (5). there exists a constant C that depends only on the problem data but not on ϖ such that ϑ W 1, ([0,T ]) C.

38 Existence of a solution Proof The bounds in W 1, ([0, T ]) for ϑ follow from the assumptions and the Lispchitz bounds for u. The uniform convergence of ϖ n ϖ gives the convergence of u n x u x, almost everywhere and m n m in C([0, T ], P). Because D 2 pph is bounded from below, we have the convergence of the right-hand side of (5) as follows, for any ψ C([0, T ]), T 0 ψ ϑ n ds T 0 ψ ϑds. Because the family ϑ n is equicontinuous, any subsequence has a further convergent subsequence that must converge to ϑ. Thus, ϑ n ϑ, uniformly.

39 Existence of a solution Proof of Theorem 1 - existence for ɛ > 0 Let ɛ > 0. The map ϖ ϑ determined by the Hamilton-Jacobi, Fokker-Planck and (5) is continuous in C([0, T ]), bounded, and compact due to the W 1, bound for ϖ. Thus, by Schauder s fixed-point theorem, it has a fixed point.

40 Existence of a solution Proof of Theorem 1 - existence for ɛ = 0 Now, we examine the case ɛ = 0. The key difficulty is the continuity of the map ϖ m in the case ɛ = 0. To overcome this difficulty, we use the vanishing viscosity method. The key idea is to fix (u ɛ, m ɛ, ϖ ɛ ) solve (1) with ɛ > 0 and justify the limit ɛ 0.

41 Existence of a solution Proof of Theorem 1 - existence for ɛ = 0 By the above, ϖ ɛ and u ɛ are uniformly locally bounded and Lipschitz. Therefore, as ɛ 0, extracting a subsequence if necessary, ϖ ɛ ϖ and u ɛ u where u is a viscosity solution. The key difficulty is to show that m solves the transport equation.

42 Existence of a solution Proof of Theorem 1 - existence for ɛ = 0 We introduce a phase-space measure µ ɛ as follows T 0 R 2 ψ(x, p, t)dµ ɛ (x, p, t) = T 0 R ψ(x, ϖ ɛ + u ɛ x, t)m ɛ dxdt for all ψ C b (R R [0, T ]). Because m ɛ C([0, T ], P) with a modulus of continuity that is uniform in ɛ, as ɛ 0, we have µ ɛ µ; that is T 0 R 2 ψdµ ɛ T 0 R 2 ψdµ.

43 Existence of a solution Proof of Theorem 1 - existence for ɛ = 0 Moreover, T 0 ψ t D p H(x, p)d x ψdµ R 2 = ψ(x, T )m(x, T )dx ψ(x, 0) m(x)dx. R It just remains to show that p = ϖ + u x, µ-almost everywhere. R

44 Existence of a solution Proof of Theorem 1 - existence for ɛ = 0 Next, we fix δ > 0 and consider a standard mollifier η δ. We define v δ = η δ u. We note that D 2 v δ C δ 2. Then, using the uniform convexity of the Hamiltonian, we get v δ t + γη δ u x v δ x 2 + H(x, ϖ + v δ x ) O(δ).

45 Existence of a solution Proof of Theorem 1 - existence for ɛ = 0 Therefore, w = v δ u ɛ satisfies w t + D p H(x, ϖ ɛ + u ɛ x)w x ɛw xx + γη δ u x v δ x 2 + γ ϖ + v δ x ϖ ɛ u ɛ x 2 O(δ) + O( ɛ δ 2 ). Integrating with respect to m ɛ, we conclude that T 0 R 2 γη δ u x v δ x 2 +γ ϖ+v δ x p 2 dµ ɛ O(δ)+O( ɛ δ 2 )+ v δ u ɛ L.

46 Existence of a solution Proof of Theorem 1 - existence for ɛ = 0 Next, we let ɛ 0, to get T γ η δ u x vx δ 2 + ϖ + vx δ p 2 dµ O(δ). 0 R 2 Finally, we let δ 0 and conclude that m-almost every point is a point of approximate continuity of u x. Therefore, vx δ u x almost everywhere. Hence, p = ϖ + u x µ-almost everywhere. Therefore, we obtain T (ψ t D p H(x, ϖ + u x )D x ψ) dµ 0 R 2 T = (ψ t D p H(x, ϖ + u x )D x ψ) mdxdt 0 R = ψ(x, T )m(x, T )dx ψ(x, 0) m(x)dx, R which gives that m solves (4) with ɛ = 0. R

47 Existence of a solution Proposition Suppose that ɛ = 0 and that the potential, V, and the terminal cost, ū, are convex. Let ϖ be a Lipschitz function. Then, u is differentiable in x for every x R. Moreover, u xx is bounded.

48 Existence of a solution Proof By a direct inspection of the variational problem using the additional convexity assumptions, we see that u(x, t) is convex in x Moreover, u is semiconcave in x. This gives the bound for u xx and the differentiability of u in x.

49 Existence of a solution Corollary Suppose that ɛ = 0 and that the potential, V, and the terminal cost, ū, are convex. Then, there exists a solution (u, m, ϖ) with u differentiable in x for every x and u xx bounded. Moreover, m is also bounded.

50 Existence of a solution Proof of Theorem 1 - additional regularity for ɛ = 0 Additional regularity for the case ɛ = 0 follows the preceding corollary.

51 Uniqueness We set and Ω T = R [0, T ], D =(C (Ω T ) C([0, T ], P)) (C (Ω T ) W 1, (Ω T )) C ([0, T ]), D + ={(m, u, ϖ) D s.t. m > 0}, D b ={(m, u, ϖ) D s.t. m(x, 0) = m(x), u(x, T ) = ū(x)}, D b + =D b D +, Then, we define A : D+ b D as m m m A u =A 1 u + A 2 u ϖ ϖ ϖ u t + ɛu H(x, Du + ϖ) xx = m t ɛm xx + div(md p H(x, ϖ + u x )). 0 md p H(x, ϖ + u x )dx + Q(t) Ω

52 Uniqueness Furthermore, for w = (m, u, ϖ), w = ( m, ũ, ϖ) D, we set w, w = Ω T (m m + uũ) dxdt + T 0 ϖ ϖdt. Then, A is a monotone operator if A[w] A[ w], w w 0 for all w, w D+. b

53 Uniqueness Proposition Suppose the map p H(x, p) is convex. Then A is a monotone operator.

54 Uniqueness Proof Let w = (m, u, ϖ), w = ( m, ũ, ϖ) D+. b Then A 1 [w] A 1 [ w], w w = ((u ũ) t + ɛ (u ũ))(m m) Ω T + ((m m) t ɛ (m m))(u ũ) Ω T =0, because u ũ and m m vanish at t = 0, T.

55 Uniqueness Proof Furthermore, A 2 [w] A 2 [ w], w w ( ) = m H(ũ x + ϖ) H(u x + ϖ) (ũ x + ϖ u x ϖ)d ph(u x + ϖ) Ω T + ( ) m H(u x + ϖ) H(ũ x + ϖ) (u x + ϖ ũ x ϖ)d ph(ũ x + ϖ) Ω T 0, by the convexity of p H(x, p).

56 Uniqueness Proof Combining the previous inequalities, we conclude that A[w] A[ w], w w = A 1 [w] A 1 [ w], w w + A 2 [w] A 2 [ w], w w 0.

57 Uniqueness Proof of main result, uniqueness Let (m, u, ϖ) and ( m, ũ, ϖ) solve Problem 1. Note that m and m are absolutely continuous and strictly positive. The monotonicity gives T 0 R Therefore, ϖ + u x = ϖ + ũ x a.e.. Hence, ϖ + u x ϖ ũ x 2 ( m + m) = 0. u t = ũ t, almost everywhere and, thus, u = ũ. Finally, the uniqueness of the Fokker-Planck equation, for ɛ > 0 or for the transport equation, when ɛ = 0 give m = m.

58 A linear-quadratic model Linear-quadratic model I We set l(t, α) = c 2 α2 + αϖ(t), where c is a constant that accounts for the usage-depreciation of the battery. The corresponding MFG is (ϖ(t)+ux )2 u t + 2c = 0 m t 1 c (m(ϖ(t) + u x)) x = 0 1 c R (ϖ(t) + u x)mdx = Q(t).

59 A linear-quadratic model The stored energy by each agent follows optimal trajectories that solve the Euler Lagrange equation: cẍ + ϖ = 0. Integrating the previous equation in time, we get where θ is time independent. ẋ(t) = 1 (θ ϖ(t)), (6) c

60 A linear-quadratic model Next, differentiating the Hamilton-Jacobi equation: (u x ) t + (u x + ϖ) u xx c = 0. Using the previous equation, taking into account the transport equation, give d u x mdx = u xt m + u x m t = u xt m + 1 dt R R R c u x (m(ϖ + u x )) x = 1 (ϖ + u x )u xx m u xx m(ϖ + u x )dx = 0. c R

61 A linear-quadratic model Thus, the supply vs demand balance condition becomes Q(t) = 1 (u x + ϖ)mdx = 1 (Θ ϖ), c c where R Θ = u x mdx (7) R is constant. From the above, we obtain the following linear price-supply relation ϖ = Θ cq(t). (8)

62 A linear-quadratic model Integrating (6) in time and taking into account the linear price-supply relation (8): x(t ) = x(t)+ 1 c T t (θ ϖ(s))ds = x+ T t c Accordingly, u is given by the optimization problem u(x, t) = inf θ ( + ū x + T t T (θ Θ)+ Q(s)ds. t [ (θ Θ + cq(s)) ] 2c c (θ Θ + cq(s))(θ cq(s)) ), (θ Θ) (T t) + K(t) c where K(t) = T t Q(s)ds.

63 A linear-quadratic model By setting µ = θ Θ, we get T [ (µ + cq(s)) 2 u(x, t) = inf + 1 ] µ t 2c c (µ + cq(s))(θ cq(s)) ds + ū (x + µ ) c (T t) + K(t). Thus, given Θ, we determine a function, u Θ, solving the preceding minimization problem: [ u Θ T t (x, t) = inf µ T ( µ 2c c (T t)θµ + Θ c Q(s) ) Q(s)ds t 2 ( +ū x + µ ) ] c (T t) + K(t).

64 A linear-quadratic model The optimality condition in the preceding minimization problem gives µ + ū x (x(t )) = Θ. (9) Thus, if ū is a convex function, there is a unique solution, µ(θ) for each given Θ. Given Θ, we obtain a solution, u Θ for the Hamilton-Jacobi equation. We use the resulting expression for u Θ in (7) at t = 0 to get Θ = ux Θ (x, 0)m 0 (x)dx. (10) R Solving the preceding equation, we obtain Θ and hence ϖ using the price-supply relation, (8).

65 A linear-quadratic model As an example, we consider the terminal cost ū(y) = γ 2 (y ζ)2. Solving(9), we obtain µ = γ(k(t) + x ζ) + Θ 1 + γ T t. c

66 A linear-quadratic model Accordingly, we have u Θ (x, t) = γ(k(t) + x ζ) 2 + (t T ) c Θ(2γ(K(t) + x ζ) + Θ) 2 ( 1 + γ T t ) c T Q 2 (s) + ΘK(t) c ds. 2 t

67 A linear-quadratic model Therefore, u x (x, t) = γ K(t) + x ζ (T t) c Θ 1 + γ T t c

68 A linear-quadratic model Using the previous expression for t = 0 in (10), we obtain the following equation for Θ Θ = γ K(0) + x ζ T c Θ 1 + γ T c where Thus, x = R xm 0 dx. Θ = γ(k(0) + x ζ).

69 A linear-quadratic model Therefore, using (8), we obtain ϖ = γ(k(0) + x ζ) cq. Finally, we use the above results and conclude that each agent dynamics is {ẋ = ( x x)γ 1+ T c γ + Q x(0) = x.

70 A linear-quadratic model In alternative, using ẋ(t) = ϖ + u x(x(t), t) c we have {ẋ(t) = ( x(t) x(t))γ x(0) = x, 1+ T t γ c + Q (11) where x(t) = xm(x, t)dx. R

71 A linear-quadratic model Averaging (11) with respect to m, we obtain x(t) = Q(t), which is simply the conservation of energy. Thus, the trajectory of an individual agent can be computed by solving {ẋ(t) = ( x(t) x(t))γ x(t) = Q. 1+ T t γ c + Q(t) The previous system is a closed system of ordinary differential equations that only involves Q and the parameters of the problem.

72 Real Data Daily energy consumption in the UK, during a twenty-four hour period. Power supply oscillation, Q(t) (in MW) time (hours) Normalized electricity production Q.

73 Real Data We compare our price-formation model with the MFG model by De Paola et al, where the energy price is a function of the aggregate consumption.thus, there may be an energy imbalance. we consider the state-independent quadratic cost model; thus, the price depends only on the constant that accounts for battery s wear and tear. We calibrate our model against the model by De Paola et al. using a least squares approach.

74 Real Data Price ($/MWh) Time (hours) 24-hour electricity price. In green, energy s price when no batteries are connected to the grid. In blue, price with batteries connected as predicted by the model by De Paola et al. In yellow, the price corresponding to our model.