PRICING OF RAINBOW OPTIONS: GAME THEORETIC APPROACH

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International Game Theory Review, Vol. 9, No. 2 (2007) 215–242 c World Scientific Publishing Company


PRICING OF RAINBOW OPTIONS: GAME THEORETIC APPROACH

Z. HUCKI Department of Coputing and Mathematics Nottingham Trent University, Nottingham NG1 4BU United Kingdom [email protected] V. N. KOLOKOLTSOV Department of Statistics, University of Warwick, CV4 7AL, UK and The Moscow Institute of Economics, Russia [email protected]

The general approach for the pricing of rainbow (or colored) options with fixed transaction costs is developed from the game theoretic point of view. The evolution of the underlying common stocks is considered in discrete time. The main result consists in the explicit calculation of the hedge price for a variety of the rainbow options including option delivering the best of J risky assets and cash, calls on the maximum of J risky assets and the multiple-strike options. The results obtained can be also used in the framework of real options. Keywords: Coloured (or rainbow) options; explicit formulas for hedge; transaction costs; interval model; submodular functions.

1. Model One can distinguish two main types of options: European and American. In case of European call options buyer is not engaged in financial activity and waits for the maturity date n of the options. The contradicting interests of the investor and the buyer give optimization problem of minmax nature. In this paper we will discuss mostly European options (See subsection 3.4 bellow). In our model, financial market is dealing with several securities: the risk-free bonds (or bank account) and J common stocks, J = 1, 2, . . . . In this paper we are concerned with the case J > 1, which means that we are discussing the so-called rainbow options. For a given J these options are called J-color rainbow options. The prices of the units of these securities, Bk and Ski , i ∈ {1, 2, . . . , J} respectively, change in discrete moments of time k = 1, 2, . . . according to the recurrent equations Bk+1 = ρBk , where the ρ ≥ 1 is an interest rate which remains unchanged over i i = ξk+1 Ski , where ξki , i ∈ {1, 2, . . . , J} are unknown sequences taking time, and Sk+1 values in some fixed intervals Mi = [di , ui ] ⊂ R. It is worth noting difference 215

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between the model considered and a colored version of the classical CRR model. In the latter a sequence ξki take values in the two point set {di , ui }, and it is supposed to be random with some given distribution. In our model any value in the interval [di , ui ] is allowed and no probabilistic assumption is made. The type of an option is specified by a given premium function f of J variables. The following are the standard examples Rubinstein (1995): option delivering the best of J risky assets and cash f (S 1 , S 2 , . . . , S J ) = max(S 1 , S 2 , . . . , S J , K),

(1)

calls on the maximum of J risky assets f (S 1 , S 2 , . . . , S J ) = max(0, max(S 1 , S 2 , . . . , S J ) − K),

(2)

multiple-strike options f (S 1 , S 2 , . . . , S J ) = max(0, S 1 − K1 , S 2 − K2 , . . . , S J − KJ ),

(3)

portfolio options f (S 1 , S 2 , . . . , S J ) = max(0, n1 S 1 + n2 S 2 + · · · + nJ S J − K),

(4)

and spread options f (S 1 , S 2 ) = max(0, (S 2 − S 1 ) − K).

(5)

Here, the S 1 , S 2 , . . . , S J represent the expiration date values of the underlying assets, and K, K1 , . . . , KJ represent the strike prices. In Rubinstein (1995), J = 2, but we shall be especially interested in the new features of the model arising under the assumption J > 2. The corresponding results for J = 2 were obtained in Kolokoltsov (1998). Let us stress that the assumptions on f that we shall make later on will include the first three options listed above. For the discussion of the last option we refer to Margrabe (1978) and Kulatilaka and Trigeorgis (1994). Let Xk denote the capital of the investor at the time k = 1, 2, . . . The investor is supposed to control the growth of his capital in the following way. At time n − 1 the investor determines his portfolio by choosing the numbers γni of common stocks to be held. Then one can write   J J

i i Xn−1 = γni Sn−1 + Xn−1 − γni Sn−1 , i=1

i=1

where the sum in bracket corresponds to the part of his capital laid on the bank account. The control parameter γn can take all real values, i.e. short selling and borrowing are allowed. The value ξn becomes known in the moment n and thus the capital at the moment n becomes   J J

i i i i i γn ξn Sn−1 + ρ Xn−1 − γn Sn−1 , Xn = i=1

i=1

if no transaction costs are taken into consideration.

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If n is the maturity date and hence the investor is obliged to pay the premium f to the buyer, the income of the investor equals Xn − f (Sn1 , Sn2 , . . . , SnJ ). There are several types of transaction costs. The proportional transaction costs are dealt with in Grannan and Swidle (1996). They have the form TC =

J


i κ|γni − γn−1 |Sni ,

i=1

where the constant κ is between 0 and 1. In other models (see Olsder (2000)) the transaction costs are considered to be the function of the numbers of stock i ) only. (γni , γn−1 We shall consider the so-called fixed transaction costs, considered, for example, in Morton and Pliska (1995). It means that whenever a trade is made, the investor pays a transaction cost equal to a fixed fraction 1 − β of his entire portfolio value, where β ∈ (0, 1) is some constant. J J i i becomes β i=1 γni ξni Sn−1 when So, the value of the portfolio i=1 γni ξni Sn−1 the transaction costs are met. The capital of the investor in the time n becomes:   J J

i i γni ξni Sn−1 + ρ Xn−1 − γni Sn−1 (6) Xn = β i=1

i=1

To obtain the optimal strategies for an investor we will use the game theoretic approach. This game is played on the space of (J + 1) nonnegative numbers X, S 1 , . . . , S J with the final income specified by the function G(X, S 1 , . . . , S J ) = X − f (S 1 , . . . , S J )

(7)

The strategy of the investor is by definition any sequences of vectors (γ1 , . . . , γn ) (with γj = (γj1 , . . . , γjJ )) such that each γji could be chosen using the whole previous i information: the sequences X0 , . . . , Xj−1 and S0i , . . . , Sj−1 (for every stock i = 1, 2, . . . , J). It is supposed that the investor, selling an option by the price C = X0 should organize the evolution of his capital in a way that would allow him to pay to the buyer some premium f (Sn1 , Sn2 , . . . , SnJ ) depending on the prices Sn1 , Sn2 , . . . , SnJ , in the prescribed moment n (maturity day). The strategy γ1i , . . . , γni , i = 1, . . . , J of the investor is called a hedge, if for any sequence (ξ1 , . . . , ξn ) (with ξj = (ξj1 , . . . , ξjJ )) the investor is able to meet his obligations, i.e. G(Xn , Sn1 , . . . , SnJ ) ≥ 0. The minimal value of the capital X0 for which the hedge exists is called the hedging price Ch of an option. As we can not control parameters ξni (for i = 1, 2, . . . , J), we will assume the worst possibility for the investor, i.e. that the nature minimises his income by means of ξni and the investor is looking for the maximum over γni .

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If the income is specified by a function G then the guaranteed income of the investor in one step of the game with the initial conditions X, S 1 , . . . , S J is BG(X, S 1 , . . . , S J )  = max min G ρX + β γ

ξ

J


γ iξiS i − ρ

i=1

J


 γ iS i, ξ1S 1, . . . , ξJ S J

(8)

i=1

From the theory of dynamic multistep games (see Bellman (1957)) it follows that the guaranteed income of the investor in the n step game with the initial conditions X0 , S01 , . . . , S0J is given by the formula Bn G(X0 , S01 , . . . , S0J ).

(9)

In our model G is given by (7). As the class of function G of the form ρk X − g(S 1 , . . . , S J ) is clearly invariant under the action of B it follows that in our model the guaranteed income in the n step game equals ρn X0 − (B n f )(S01 , . . . , S0J ), where the reduced Bellman operator is defined as:  (Bf )(S 1 , . . . , S J ) = min max f (ξ 1 S 1 , ξ 2 S 2 , . . . , ξ J S J ) − γ

ξ

J


 γ i S i (βξ i − ρ) .

(10)

i=1

The minimal value of X0 for which this income is not negative (and which by definition is the hedge price Ch ) is therefore given by Chn =

1 n (B f )(S01 , . . . , S0J ). ρn

(11)

This formula is the basic conclusion of the application of the game theory to the option pricing and can be used as a starting point for the construction of various schemes of numerical calculations. Our main result in this paper is analytical. It consists in a rather mysterious possibility to calculate the operator (10) explicitly on the class of submodular convex functions. For other achievements in the interval model for options we refer to the recent review, Bernard (2005). 2. Results We will denote vectors by bold letters, i.e. z = (z1 , z2 , . . . , zJ ), and to simplify writing we shall denote ξz = (ξ 1 z1 , . . . , ξ J zJ ) (which is not a usual scalar product). The Bellman operator (10) could be written like (Bf )(z) = min(Bf )(z, γ), γ

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where (Bf )(z, γ) = max f (ξ 1 z1 , . . . , ξ J zJ ) − ξ

J


γ i zi (βξ i − ρ).

(12)

i=1

For a set I ⊂ {1, 2, . . . , J} let us denote by fI the value of f (ξ 1 z1, ξ 2 z2 , . . . , ξ J zJ ) / I. with ξ i = di for i ∈ I and ξi = ui for i ∈ Also, we shall write

γ i zi (βdi − ρ) − γ i zi (βui − ρ). BI (z, γ) = fI (z) − i∈I /

i∈I

For example, f{1,3} (z) = f (d1 z1 , u2 z2 , d3 z3 ) and B{1,3} (z, γ) = f{1,3} (z) − 1 γ z1 (βd1 − ρ) − γ 2 z2 (βu2 − ρ) − γ 3 z3 (βd3 − ρ). Using this notation, operator (12) for a convex function f can be written simply as Bf (z, γ) =

max

I⊂{1,2,...,J}

BI (z, γ).

(13)

2 Definition 1. Let f : R+ → R+ . The function f is called submodular if it satisfies the inequality:

f (z1 , ω2 ) + f (ω1 , z2 ) − f (z1 , z2 ) − f (ω1 , ω2 ) ≥ 0

(14)

for every z1 < ω1 and z2 < ω2 . d → R+ is submodular if it is submodular with respect to every Function f : R+ two variables. The set of all submodular convex functions we will denote by N S. Inequality (14) clearly implies f{1} (z) + f{2} (z) − f{1,2} (z) − f∅ (z) ≥ 0,

(15)

which will be frequently used in the paper. The properties of submodular and supermodular functions are discussed in various contexts in Altman, Gaujal and Hordijk (1999), Altman and Koole (1998), Yao (1995) and Pap (1995). d ), then the function f is submodular iff Remark 1. If function f ∈ C 2 (R+ 0 for all i = j.

∂2 f ∂zi ∂zj

≤

To simplify our formulas let us introduce special notation for the increments of functions. Namely, we shall denote by ∆j fI∪{j} (z) the difference / I. fI (z) − f{j}∪I (z) for any I ⊂ {1, 2, . . . , J} and j ∈ Let us denote by r the adjusted interest rate: ρ r= . β

We will suppose that 0 < di < r < ui for all i ∈ {1, 2, . . . , J}. (If this condition is not satisfied then the solution of our problem is trivial, because if r ≥ ui , say, then the corresponding bond i should not be taken into account.)

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Now, let us introduce the following coefficients:
uj − r , where I ⊂ {1, 2, . . . , J}. αI = 1 − uj − dj j∈I

For the case I = 3 the following analysis:  u1 − r α123 = 1 − u1 − d1  u1 − r α12 = 1 − u1 − d1  u1 − r α13 = 1 − u1 − d1  u2 − r α23 = 1 − u2 − d2

coefficients will play the major role in our u2 − r u3 − r − u2 − d2 u3 − d3  u2 − r − u2 − d2  u3 − r − u3 − d3  u3 − r − . u3 − d3



−

(16)

In this section we will formulate our main result that concerns the case when J = 3. It is divided in two theorems. The form of (Bf )(z) turns out to depend on the signs of coefficients (16). Theorem 1. Let f ∈ N S. (i) If α123 ≥ 0, then

 u1 − r f{1} (z) α123 f∅ (z) + u1 − d1  u2 − r u3 − r + f{2} (z) + f{3} (z) . u2 − d2 u3 − d3

1 (Bf )(z) = r

(ii) If α123 ≤ −1, then

 d1 − r f{2,3} (z) −(α123 + 1)f{1,2,3} (z) − u1− d1  d3 − r d2 − r f{1,3} (z) − f{1,2} (z) . − u2− d2 u3− d3

(17)

1 (Bf )(z) = r

(18)

Theorem 2. Let f ∈ N S and suppose that 0 ≥ α123 ≥ −1. (i) If α12 ≥ 0, α13 ≥ 0 and α23 ≥ 0, then (Bf )(z)

  (−α123 ) f{1,2} (z) + α13 f{2} (z) + α23 f{1} (z) +       1 = max (−α123 ) f{1,3} (z) + α12 f{3} (z) + α23 f{1} (z) +  r      (−α123 ) f{2,3} (z) + α12 f{3} (z) + α13 f{2} (z) +

 u3 − r f{3} (z)    u3 − d3    u2 − r f{2} (z) ,  u2 − d2     u1 − r  f{1} (z) u1 − d1

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(ii) If αij ≤ 0, αjk ≥ 0 and αik ≥ 0, where {i, j, k} is an arbitrary permutation of the set {1, 2, 3}, then     (−αijk ) f{i,j} (z) + αik f{j} (z) + αjk f{i} (z)             − r u k     f (z) + {k}     u − d   k k       − r u k     α f (z) + (−α )f (z) + f (z) jk {i} ij {i,j}   {i,k}   uk− dk 1 , (Bf )(z) = max di − r   r   f{j} (z) −     ui− di       − r u   k    αik f{j} (z) + (−αij )f{i,j} (z) + f{j,k} (z)      u d k− k         − r d j     f{i} (z) −   uj− dj (iii) If αij ≥ 0, αjk ≤ 0 and αik ≤ 0, where {i, j, k} is an arbitrary permutation of the set {1, 2, 3}, then   ui − r    αij f{k} (z) + (−αjk )f{j,k} (z) + f{i,k} (z)      ui− di         − r d   k   f (z) −   {j}     u d k− k       − r u   j     α f (z) + (−α )f (z) + f (z) ij {k} ik {i,k} {j,k} 1 u d j− j . (Bf )(z) = max   r dk − r     f{i} (z) −     uk− dk           (α + 1)f (z) − α f (z) − α f (z)   123 jk ik {k} {j,k} {i,k}           dk − r       f{i,j} (z) − uk − dk Addition and scalar multiplication preserves the set of submodular functions and hence Theorem 1 can be applied recursively which allows to get an explicit expression for (B n f )(z) generalizing the CRR formula. Denote by Cnijk the coefficient in the polynomial expansion n

(1 + 2 + 3 + 4 ) =




Cnijk n−i−j−k i2 j3 k4 . 1

i+j+k≤n

The following Corollaries give us the expressions of hedge price Ch when interest rate r is close to upper bounds ui (α123 ≥ 0), or when it is close to lower bounds di for all i = 1, 2, 3.

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Corollary 3. If α123 ≥ 0, the hedge price is equal to:  i  j  k u1 − r u2 − r u3 − r 1
Ch = n Cnijk (α123 )n−i−j−k ρ u1 − d1 u2 − d2 u3 − d3 i,j,k∈Pn   (19) × f di1 un−i S01 , dj2 un−j S02 , dk3 un−k S03 , 2 1 3 and if α123 ≤ −1, hedge price is equal to: Ch =






r − d1 u1− d1 i,j,k∈Pn   × f dn−i ui1 S01 , dn−j uj2 S02 , dn−k uk3 S03 , 2 1 3 1 rn

Cnijk (−α123 − 1)n−j−i−k

i 

r − d2 u2− d2

j 

r − d3 u3− d3

k

(20)

where Pn = {i, j, k ≥ 0 : i + j + k ≤ n}. Proof. Following from (17) and (11) by induction. This formula can be easily specified for concrete f of form (1), (2) and (3). For example: Corollary 4. Let function f has form (3) with J = 3. If α123 ≥ 0, the hedge price is  i  j  k u1 − r u2 − r u3 − r 1
Cnijk (α123 )n−i−j−k Ch = n ρ u1 − d1 u2 − d2 u3 − d3 i,j,k∈Pn˜   j n−j 2 1 k n−k 3 , × max di1 un−i S − K , d u S − K , d u S − K 1 2 3 0 0 3 3 0 2 2 1 where Pn˜ = {0 ≤ i ≤ µ, 0 ≤ j ≤ ν, 0 ≤ k ≤ λ : i + j + k ≤ n} and µ is the S01 ≥ K1 , ν is the maximal integer j such that maximal integer i such that di1 un−i 1 j n−j 2 S03 ≥ K3 . d2 u2 S0 ≥ K2 and λ is the maximal integer k such that dk3 un−k 3 If α123 ≤ −1, the hedge price is  i  j  k 1
r − d1 r − d2 r − d3 ijk n−j−i−k Ch = n Cn (−α123 − 1) ρ u1− d1 u2− d2 u3− d3 i,j,k∈Pn˜   × max dn−i ui1 S01 − K1 , dn−j uj2 S02 − K2 , dn−k uk3 S03 − K3 2 1 3 ui1 S01 ≥ where Pn˜ is the same as above but µ is the minimal integer i such that dn−i 1 n−j j 2 K1 , ν is the minimal integer j such that d2 u2 S0 ≥ K2 and λ is the minimal integer k such that dn−k uk3 S03 ≥ K3 . 3 Proof. Following from (19) and (3) by induction. Theorem 2 give us a result of one step game. Its application to the n-step game is not obvious, because it is not clear under what conditions the set of submodular

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function is preserved by the Bellman operator B from the Theorem 2. That means that investor needs to recalculate and find the best strategy after every step. The proof of the Theorems 1 and 2 is very long. The sketch of the proof of Theorem 1 will be given in the last section. The full proof of both Theorems can be found in our preprint (Hucki and Kolokoltsov (2003)). 3. Discussion 3.1. Probabilistic interpretation d by the following rule: Let us define a Markov process Z t , t = 0, 1, 2, . . . , on R+ t d for the t and Z = z ∈ R+ there are only four possibilities for the position of the process at the next time t + 1, namely (u1 z1 , u2 z2 , u3 z3 ), (d1 z1 , u2 z2 , u3 z3 ), (u1 z1 , d2 z2 , u3 z3 ), (u1 z1 , u2 z2 , d3 z3 ) and they can occur with probabilities u1 − r Pzu1 z1 ,u2 z2 ,u3 z3 = α123 , Pzd1 z1 ,u2 z2 ,u3 z3 = , u1 − d1 u2 − r u3 − r Pzu1 z1 ,d2 z2 ,u3 z3 = , Pzu1 z1 ,u2 z2 ,d3 z3 = , u2 − d2 u3 − d3 respectively. Since there are only finite number of possible jumps this Markov process is in fact a Markov chain.

Theorem 5. If α123 ≥ 0 then (B n f )(z) =Ez f (zn ),

(21)

where Ez denotes the expectation of the process starting at the point z. Proof. For n = 1 this follows from Theorem 1 and for n > 1 it follows from the above made observation that addition and scalar multiplication preserves the set of submodular functions and hence Theorem 1 can be applied recursively. −r Proof. Probabilities ( uu11−d , u2 −r , u3 −r , α123 ) are risk-neutral probabilities. 1 u2 −d2 u3 −d3

In the case when 0 ≥ α123 ≥ −1, observe that the Bellman operator (10) can be written in the form of the Bellman operator of a controlled Markov process, namely (Bf )(z) = max

i=1,2,3

4


I i (z)

Pz j

f (Iji (z)).

(22)

j=1

For example, for i = 1, Ij1 (z), j = 1, 2, 3, 4, could be the points I11 (z) = (d1 z1 , d2 z2 , u3 z3 ), I21 (z) =(d1 z1 , u2 z2 , u3 z3 ), I31 (z) = (u1 z1 , d2 z2 , u3 z3 ), I41 (z) =(u1 z1 , u2 z2 , d3 z3 ) and the corresponding probabilities of transitions from z to Ij1 (z) are given by I 1 (z)

Pz 1

I 1 (z)

= −α123 , Pz 2

I 1 (z)

= α23 , Pz 3

I 1 (z)

= α13 , Pz 4

=

u3 − r . u3 − d3

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Representation (22) shows in particular that in this case the solution can not be written in form (21) and hence the obtained formula differs from what one can expect from the usual stochastic analyses approach to option pricing.

3.2. Unpredictable surplus The estimation of unpredictable surplus is an important part of the modern analysis of the financial market (see, e.g. Lions (1995)). Let us give the formula for this surplus in the present game theoretic framework. Copying the previous argument one can see that the maximal income of the investor is given by the formula n f )(S01 , . . . , S0J )), ρn (X0 − (Bmin

where (Bmin f )(z) =

J
1 min [f (ξz) − γ i z i (ξ i − ρ)]|, ρ ξ∈M i=1

and γ is the chosen hedge strategy. Consequently the upper bound for the unpredictable surplus to the income of the investor is n f )(S01 , . . . , S0J ). (B n f )(S01 , . . . , S0J ) − (Bmin

3.3. Generalities A surprisingly simple linear form (17 ) and (18 ) of min-max Bellman operator (10) arises the question whether it can be generalized to other options, see Kulatilaka and Trigeorgis (1994). For example, at the moment we even do not have a proof of the analogous result for the general J > 3. Another point to notice is an unexpectedly long and technical proof of Theorem 1 and 2 resulting from a number of strange coincidence and cancellations. This leads to the following question for the theory of multistep dynamic games. What is the general justif ication for these cancellations and/or what is the class of game theoretic Bellman operators that can be reduced to a simpler Bellman operator of a controlled Markov chain.

3.4. American options In case of American call option buyer is an acting character in the market. Contract allows him to choose moment of exercising the options on his own (any moment before maturity date n < N ). In selecting the corresponding hedging portfolio the investor must keep in mind the buyer’s freedom to exercise the option at any time. It is not difficult to modify our basic formula (8) for this case, namely instead of (8)

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one should write

 

BG(X, S 1 , . . . , S J ) = max min min G ρX + β γ

−ρ

ξ

J


J


γ iξiS i

i=1





i

i

1

1

J

γ S ,ξ S ,...,ξ S

J

225

1

J

, G(X, S , . . . , S )

i=1

Formula (9) remains unchanged. However, the further factorization with the reduced Bellman operator is not possible and one should calculate directly using formula (9). 4. Property of Submodular Functions From now on we shall put β = 1 for simplicity. Hence, r = ρ. In this section we discuss some basic properties of submodular functions and the Bellman operator (10) which are valid for any dimension. J , the inequality Lemma 6. If f (z) ∈ N S on set R+

fI1 ∪I2 ∪I3 (z) + fI1 (z) ≤ fI1 ∪I3 (z) + fI1 ∪I2 (z)

(23)

holds for every disjoint subsets I1 , I2 , I3 of {1, 2, . . . , J}. Proof. Proof is done by trivial induction. For the basis of mathematical induction we will take |I2 | = 1, |I3 | = 1, then inequality (23) holds for an arbitrary I1 because of the submodular property (14). Suppose now that (23) is true for any |I2 | ≤ n and |I3 | = 1 and arbitrary I1 such that I1 , I2 , I3 are disjoint. Let us prove this inequality for I˜2 = I2 ∪{i},where the set I2 contains n elements and i is an arbitrary index not contained in I1 ∪ I2 ∪ I3 . We should prove that fI1 ∪I˜2 ∪I3 (z) + fI1 (z) ≤ fI1 ∪I3 (z) + fI1 ∪I˜2 (z).

(24)

From our assumption with I1 ∪ {i} playing the role of I1 we get fI1 ∪{i}∪I2 ∪I3 (z) + fI1 ∪{i} (z) ≤ fI1 ∪{i}∪I3 (z) + fI1 ∪{i}∪I2 (z).

(25)

By the assumption with {i} playing the role of I2 we get fI1 ∪{i}∪I3 (z) + fI1 (z) ≤ fI1 ∪{i} (z) + fI1 ∪I3 (z).

(26)

Now if we sum (25) and (26) together we get inequality (24). So we have proved (24) and consequently (23) for any I2 . Now, let us suppose that inequality (23) is true for any disjoint sets I1 , I2 and |I3 | ≤ n. Let us prove this inequality for I˜3 = I3 ∪{i}, where the set I3 contains n elements and i is any index not contained in I1 ∪ I2 ∪ I3 .

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By the assumption with {i} playing the role of I3 we have that inequality fI1 ∪{i}∪I2 (z) + fI1 (z) ≤ fI1 ∪{i} (z) + fI1 ∪I2 (z)

(27)

is true. Now if we sum (25) and (27) together we get the inequality fI1 ∪I2 ∪I˜3 (z) + fI1 (z) ≤ fI1 ∪I˜3 (z) + fI1 ∪I2 (z). This completes the proof of (23) for every disjoint subsets I1 , I2 , I3 of {1, 2, . . . , J}. Proposition 7. Let f (z) ∈ N S. If BI1 ∪I2 ∪I3 (z, γ) ≥ BI1 ∪I2 (z, γ) then BI1 ∪I3 (z, γ) ≥ BI1 (z, γ) holds for every disjoint I1 , I2 , I3 subset of {1, 2, . . . , J}. Proof. The inequality BI1 ∪I2 ∪I3 (z, γ) ≥ BI1 ∪I2 (z, γ) is equivalent to

fI1 ∪I2 ∪I3 (z) − γi zi (di − ρ) − γi zi (ui − ρ) i∈I1 ∪I2 ∪I3

≥ fI1 ∪I2 (z) −




i∈I / 1 ∪I2 ∪I3

i∈I1 ∪I2

i.e. fI1 ∪I2 ∪I3 (z) −







γi zi (di − ρ) −

γi zi (ui − ρ)

i∈I / 1 ∪I2

γi zi (di − ρ) ≥ fI1 ∪I2 (z) −

i∈I3




γi zi (ui − ρ).

i∈I3

And this is equivalent to fI1 ∪I2 ∪I3 (z) − fI1 ∪I2 (z) ≥




γi zi (di − ui ).

(28)

i∈I3

Similar, the inequality BI1 ∪I3 (z, γ) ≥ BI1 (z, γ) is equivalent to
fI1 ∪I3 (z) − fI1 (z) ≥ γi zi (di − ui ).

(29)

i∈I3

But clearly (28) implies (29), because fI1 ∪I3 (z) − fI1 (z) ≥ fI1 ∪I2 ∪I3 (z) − fI1 ∪I2 (z). The latter inequality being a consequence of Lemma 6. Lemma 8. (on inclusions) If D, E, F are subsets of {1, 2, . . . , J} such that F ⊂ E ⊂ D and BF (z, γ) = BD (z, γ) = (Bf )(z, γ) then BF (z, γ) = BE (z, γ) = BD (z, γ) = (Bf )(z, γ). Proof. If BD (z, γ) = (Bf )(z, γ)

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then BD (z, γ) ≥ BD\E∪F (z, γ). Consequently, by Proposition 7 BE (z, γ) ≥ BF (z, γ). But BF (z, γ) is the maximum in (12) so BF (z, γ) ≥ BE (z, γ). Consequently, BF (z, γ) = BE (z, γ). This completes the proof. Proposition 9. There exists γ0 such that (Bf )(z, γ0 ) = (Bf )(z)

(30)

Moreover, (Bf )(z, γ0) ≥ 0. Any γ0 satisfying (30) will be called optimal. Proof. Clearly, for any ρ ∈ (di , ui ), i ∈ {1, 2, . . . , J} and any γi we can choose ξi such that γi zi (ξi − ρ) < 0 . Hence, for any γ we can find ξ such that f (ξ 1 z1 , . . . , ξ J zJ ) −

J


γi zi (ξi − ρ) > 0.

i=1

This means that maximum (12) is positive for every fixed γ, which implies (Bf )(z) ≥ 0. From the positivity of f it follows that (Bf )(z, γ 0 ) ≥ max − ξ

J


γi zi (ξi − ρ) ≥ max |γi |zi min(ui − ρ, ρ − di ) i

i=1

and hence (Bf )(z, γ0 ) → +∞ as γ → ∞. Consequently, as (Bf )(z, γ) is continuous, we conclude that there exist γ0 such that (Bf )(z, γ0 ) = min(Bf )(z, γ) γ

and (30) holds. 5. Auxiliary Results Now we shall stick to the case J = 3. For any collection of four different subsets I1 , I2 , I3 , I4 of {1, 2, 3} let γ {I1 ,I2 ,I3 ,I4 } denote any γ for which BI1 (z, γ0 ) = BI2 (z, γ 0 ) = BI3 (z, γ 0 ) = BI4 (z, γ 0 )

(31)

Since (31) is a system of three linear equations on three variables γ, it is clear that γ {I1 ,I2 ,I3 ,I4 } is uniquely defined in generic situation (but not always, of course). Let us denote the right hand side of (31) by B{I1 ,I2 ,I3 ,I4 } .

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Proposition 10. If γ0 is optimal then there exist four different subsets I1 , I2 , I3 , I4 of {1, 2, 3} such that γ0 = γ {I1 ,I2 ,I3 ,I4 } Proof. If we suppose that γ0 is optimal and there exist only one I ⊂ {1, 2, 3} such that (Bf )(z, γ 0 ) = BI1 (z, γ 0 ),

(32)

then we could take γ such that γ i = γ0i for some i ∈ {1, 2, 3}. then we have that BI1 (z, γ 0 ) > BI1 (z, γ) and that is a contradiction with (32). If there exist subsetsI1 and I2 such that (Bf )(z, γ 0 ) = BI1 (z, γ 0 ) = BI2 (z, γ 0 ), then, in the best case, we are getting that one of the components of vector γ is fixed (obtained from BI1 (z, γ 0 ) = BI2 (z, γ 0 )), but still we could change others components to get greater value. Similar situation we have for the three subsets I1 , I2 , I3 . So, the optimal γ0 need to be intersection of four hyper-planes, which have intersection for every two of them, defined by BI1 (z, γ 0 ), BI2 (z, γ 0 ), BI3 (z, γ 0 ) and BI4 (z, γ 0 ). Let us denote γjI =

fI (z)−fI∪{j} (z) . zj (uj − dj )

(33)

As any optimal γ0 has the form γ {I1 ,I2 ,I3 ,I4 } for some sets {I1 , I2 , I3 , I4 } it follows that the number of candidates for optimal γ is finite, not exceeding 70 (the numbers of families {I1 , I2 , I3 , I4 } of four different subsets of {1, 2, 3}). Next results are meant to reduce the number of possibilities. Proposition 11. Suppose that γ0 = γ {I1 ,I2 ,I3 ,I4 } . (i) If ∅ ∈ {I1 , I2 , I3 , I4 } then the only optimal γ is γ {∅,{1},{2},{3}} . (ii) If {1, 2, 3} ∈ {I1 , I2 , I3 , I4 } then the only optimal γ is γ {{1,2},{1,3},{2,3},{1,2,3}} . (iii) Set {I1 , I2 , I3 , I4 } can not be equal to {{i}, {j}, {i, k}, {j, k}} for any permutation i, j, k of {1, 2, 3}. Proof. (i) If we have family of sets {∅, {1}, {2}, {3}}, then by definition (31) we have that B∅ (z, γ) = B{1} (z, γ) = B{2} (z, γ) = B{3} (z, γ) = (Bf )(z, γ).

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From B∅ (z, γ) = B{1} (z, γ) we get γ 1 = we get γ 2 =

f∅ (z)−f{2} (z) z3 (u3 −d3 ) ,

f∅ (z)−f{1} (z) z1 (u1 −d1 ) ,

229

from B∅ (z, γ) = B{2} (z, γ)

and from B∅ (z, γ) = B{3} (z, γ) we get γ 3 = {∅,{1},{2},{3}}

f∅ (z)−f{3} (z) z3 (u3 −d3 ) .

Therefore, we have that γ0 = (γ1∅ , γ2∅ , γ3∅ ) (using the notation (33)). Now, suppose that we have ∅, {i, j} ∈ {I1 , I2 , I3 , I4 } for some i, j ∈ {1, 2, 3}. This means that B{i,j} (z, γ) = B∅ (z, γ) = (Bf )(z, γ), then by Lemma 8 we have that B∅ (z, γ) = B{{i,j} (z, γ) = B{i} (z, γ) = B{j} (z, γ) = (Bf )(z, γ) So, we get combination {∅, {i}, {j}, {i, j}}. From B{i} (z, γ) = B∅ (z, γ) we are getting γi = B{i,j} (z, γ) yields γi =

f{j} (z)−f{i,j} (z) . zi (ui −di )

f∅ (z)−f{i} (z) zi (ui −di )

Similar, from B{j} (z, γ) = B∅ (z, γ) we are getting γj = B{i} (z, γ) = B{i,j} (z, γ) yields γj = Now, if

f{i} (z)−f{i,j} (z) . zj (uj −dj )

and B{j} (z, γ) = f∅ (z)−f{j} (z) zj (uj −dj )

and

f∅ (z) − f{i} (z) = f{j} (z) − f{i,j} (z) the situation is impossible. If f∅ (z) − f{i} (z) = f{j} (z) − f{i,j} (z) then (Bf )(z, γ) = f∅ (z)−

f∅ (z) − f{i} (z) f∅ (z) − f{j} (z) (ui − ρ)− (uj − ρ)− γk zk (uk − ρ). (ui − di ) (uj − dj )

However, this is a contradiction with the assumption that γ is optimal, because we still have a free choice of γk . If ∅, {i, j, k} ∈ {I1 , I2 , I3 , I4 } then by Lemma 8 we are getting that B∅ (z, γ) = B{i} (z, γ) = B{i,j} (z, γ) = B{1,2,3} (z, γ) for all i, j ∈ {1, 2, 3}. This means that for all I ⊂ {1, 2, 3} the values BI (z, γ) have to be equal. In this {∅,{1},{2},{3}} . case we are still getting optimal γ0 (ii) The proof is similar to (i). (iii) From B{i} (z, γ) = B{i,k} (z, γ) we get γk = and B{j} (z, γ) = B{j,k} (z, γ) yields γk =

f{i} (z)−f{i,k} (z) zk (uk −dk ) f{j} (z)−f{j,k} (z) . zk (uk −dk )

If f{i} (z) − f{i,k} (z) = f{j} (z) − f{j,k} (z) situation is impossible.

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If f{i} (z) − f{i,k} (z) = f{j} (z) − f{j,k} (z) then (Bf )(γ, z) = f{i} (z) − γi zi (di − ρ) −

f{i} (z) − f{j} (z)−γi (di − ui ) (uj − ρ) (uj − dj )

f{j} (z) − f{j,k} (z) (uk − ρ). (uk − dk ) which is a contradiction with the assumption that γ is optimal, because we still have a free choice of γi . −

As a direct consequence we have: Proposition 12. If γ0 = γ {I1 ,I2 ,I3 ,I4 } , then there are the following possibilities for the collection {I1 , I2 , I3, I4 }: (i) {∅, {1}, {2}, {3}} and γ0 = (γ1∅ , γ2∅ , γ3∅ ) where we used notation (33). {j} {i} {1},{2},{3},{i,j} ) where (ii) {{1}, {2}, {3}, {i, j}} and γ0 = (γi , γj , γk {1},{2},{3},{i,j}

γk

=

f{k} (z) + f{i,j} (z) − f{i} (z) − f{j} (z) , zk (uk − dk )

(34)

where i is arbitrary index from {1, 2, 3}. {it} {i} {i},{1,2},{1,3},{2,3} ) where (iii) {{i}, {1, 2}, {1, 3}, {2, 3}} and γ0 = (γj , γk , γi {i},{1,2},{1,3},{2,3}

γi

=

f{i} (z) + f{j,k} (z) − f{i,k} (z) − f{i,j} (z) , zi (ui − di )

where i, j, k is arbitrary permutation of {1, 2, 3}. {j} {i} {i} (iv) {{i}, {j}, {i, j}, {i, k}} and γ0 = (γi , γj , γk ). {2,3}

(v) {{1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} and γ0 = (γ1

{1,3}

, γ2

{1,2}

, γ3

(35)

).

Let us denote by Γ the set of all γ {I1 ,I2 ,I3, I4 } from Proposition 12. That means that the set Γ contains γ {∅,{1},{2},{3}} , γ {{1},{2},{3},{i,j}} for every i = j, γ {{i},{1,2},{1,3},{2,3}} for every i ∈ {1, 2, 3} and γ {{1,2},{1,3},{2,3},{1,2,3}}. In particular, |Γ| = 14. Direct consequence of Proposition 12 is that (10) can be written as 1 (Bf )(z) = min B(z, γ) ρ γ∈Γ Hence proposition reduces the number of candidates for the optimum from 70 to 14. Let Γ(z) be a subset of Γ defined by the following rule: γ {I1 ,I2 ,I3, I4 } ∈ Γ(z) iff (Bf )(z, γ {I1 ,I2 ,I3, I4 } ) = BI1 (z, γ {I1 ,I2 ,I3, I4 } ) = BI2 (z, γ {I1 ,I2 ,I3, I4 } ) = BI3 (z, γ {I1 ,I2 ,I3, I4 } ) = BI4 (z, γ {I1 ,I2 ,I3, I4 } ). Now we have the following form for the Bellman operator. 1 min (Bf )(z, γ) (Bf )(z) = ρ γ∈Γ(z)

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The following proposition gives criteria for γ ∈ Γ(z). Proposition 13. If f ∈ N S then 3 (a) γ {∅,{1},{2},{3}} , γ {{1,2},{1,3},{2,3},{1,2,3}} ∈ Γ(z) for every z ∈ R+ . {{1},{2},{3},{i,j}} ∈ Γ(z) if and (b) For all i, j ∈ {1, 2, 3} such that i = j one has γ only if the conditions

∆i f{i,k} (z) ≥ ∆i f{i,j} (z) and ∆j f{j,k} (z) ≥ ∆j f{i,j} (z)

(36)

are satisfied, where {k} = {1, 2, 3}\{i, j}. (c) For every i ∈ {1, 2, 3} one has γ {{1,2},{1,3},{2,3},{i}} ∈ Γ(z) if and only if the conditions ∆k f{i,k} (z) ≥ ∆k f{j,k} (z) and ∆j f{i,j} (z) ≥ ∆j f{j,k} (z)

(37)

are satisfied, where {j, k} = {1, 2, 3}\{i}. (d) For every different i, j, k ∈ {1, 2, 3} coefficient γ {{i},{j},{i,j},{i,k}} ∈ Γ(z) if and only if conditions ∆k f{j,k} (z) ≥ ∆k f{i,k} (z) and ∆i f{i,j} (z) ≥ ∆i f{i,k} (z) are satisfied. Proof. In all the cases we need to show that B{I1 ,I2 ,I3 ,I4 } = (Bf )(z, γ {I1 ,I2 ,I3, I4 } ). (a) First, let us show that B{∅,{1},{2},{3}} = (Bf )(z, γ {∅,{1},{2},{3}} ). To show previous inequality it is enough to proof that inequalities (a1 ) (a2 )

B{i,j} (z, γ {∅,{1},{2},{3}} ) ≤ B{∅,{1},{2},{3}} for every {i, j} ⊂ {1, 2, 3} B{1,2,3} (z, γ {∅,{1},{2},{3}} ) ≤ B{∅,{1},{2},{3}}

are true. The inequality (a1 ) written explicitly is f{i,j} (z) − γi∅ zi (di − ρ) − γj∅ zj (dj − ρ) − γk∅ zk (uk − ρ) ≤ f{i} (z) − γi∅ zi (di − ρ) − γj∅ zj (uj − ρ) − γk∅ zk (uk − ρ) . This is equivalent to f{i,j} (z) − f{i} (z) ≤ γj∅ zj (dj − uj ) . Since γj∅ =

f∅ (z)−f{j} (z) , uj −dj

this is equivalent to

f{i,j} (z) − f{i} (z) ≤ f{j} (z) − f∅ (z) or written in terms of increments ∆j f{i,j} (z) ≥ ∆j f{j} (z)

(38)

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and this is always true by submodular property. The inequality (a2 ) written explicitly is f{1,2,3} (z) − γ1∅ z1 (d1 − ρ) − γ2∅ z2 (d2 − ρ) − γ3∅ z3 (d3 − ρ) ≤ f∅ (z) − γ1∅ z1 (u1 − ρ) − γ2∅ z2 (u2 − ρ) − γ3∅ z3 (u3 − ρ) , i.e. γ1∅ z1 (u1 − d1 ) + γ2∅ z2 (u2 − d2 ) + γ3∅ z3 (u3 − d3 ) ≤ f∅ (z) − f{1,2,3} (z) . Due to notation (33) with I = ∅ γ1∅ =

f∅ (z) − f{1} (z) ∅ f∅ (z) − f{2} (z) f∅ (z) − f{3} (z) , γ2 = and γ3∅ = . z1 (u1 − d1 ) z2 (u2 − d2 ) z3 (u3 − d3 )

Therefore, we have f∅ (z) − f{1} (z) + f∅ (z) − f{2} (z) + f∅ (z) − f{3} (z) ≤ f∅ (z) − f{1,2,3} (z) . Clearly, this could be written as 0 ≤ (−f∅ (z) + f{1} (z) + f{2} (z) − f{1,2} (z)) + (−f∅ (z) + f{3} (z) + f{1} (z) − f{1,3} (z)) + (−f{1} (z) + f{1,3} (z) + f{1,2} (z) − f{1,2,3} (z)) or in the terms of the increments 0 ≤ (∆1 f{1,2} (z) − ∆1 f{1} (z)) + (∆3 f{1,3} (z) − ∆3 f{3} (z)) + (∆3 f{1,2,3} (z) − ∆3 f{1,3} (z)) This inequality is true because all brackets are always positive by the submodular property. The proof that B{{1,2},{1,3},{2.3}{1,2,3}} = (Bf )(z; γ {{1,2},{1,3},{2.3}{1,2,3}}) is similar to previous one. (b) We need to show that (b1 ) (b2 ) (b3 ) (b4 )

B∅ (z, γ {{1},{2},{3},{i,j}} ) ≤ B{{1},{2},{3},{i,j}} B{1,2,3} (z, γ {{1},{2},{3},{i,j}} ) ≤ B{{1},{2},{3},{i,j}} B{i,k} (z, γ {{1},{2},{3},{i,j}} ) ≤ B{{1},{2},{3},{i,j}} B{j,k} (z, γ {{1},{2},{3},{i,j}} ) ≤ B{{1},{2},{3},{i,j}}

The inequality (b1 ) written explicitly is {{1},{2},{3},{i,j}}

f∅ (z) − γij zi (ui − ρ) − γji zj (uj − ρ) − γk

zk (uk − ρ)

{{1},{2},{3},{i,j}}

≤ f{i,j} (z) − γij zi (di − ρ) − γji zj (dj − ρ) − γk

zk (uk − ρ)

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Since, due to notation (33) and (34) γij =

f{j} (z) − f{i,j} (z) i f{i} (z) − f{i,j} (z) , γj = zi (ui − di ) zj (uj − dj )

{1},{2},{3},{i,j}

γk

=

and

f{k} (z) + f{i,j} (z) − f{i} (z) − f{j} (z) . zk (uk − dk )

we are getting 0 ≥ f∅ (z) − f{i,j} (z) − (f{j} (z) − f{i,j} (z)) − (f{i} (z) − f{i,j} (z)), i.e. f{i} (z) − f{i,j} (z) ≥ f∅ (z) − f{j} (z). Written in terms of increments this is equivalent to ∆j f{i,j} (z) ≥ ∆j f{j} (z). This is true by submodular property. The inequality (b2 ) written explicitly is {1},{2},{3},{i,j}

f{1,2,3} (z) − γij zi (di − ρ) − γji zj (dj − ρ) − γk γij zi (di

≤ f{i,j} (z) −

− ρ) −

γji zj (dj

− ρ) −

zk (dk − ρ)

{1},{2},{3},{i,j} γk zk (uk

− ρ),

i.e. {1},{2},{3},{i,j}

0 ≤ f{i,j} (z) − f{1,2,3} (z) − γk

zk (uk − dk ).

Since,due to (35) we have 0 ≤ f{i,j} (z) − f{1,2,3} (z) − (−f{k} (z) − f{i,j} (z) + f{i} (z) + f{j} (z)), If we add +f{j,k} (z) − f{j,k} (z) this is equivalent to 0 ≤ f{i,j} (z) − f{1,2,3} (z) − f{j} (z) + f{j,k} (z) + f{k} (z) − f{j,k} (z) − f{i} (z) + f{i,j} (z) or written in terms of increments 0 ≤ (∆k f{1,2,3} (z) − ∆k f{j,k} (z)) + (∆j f{j,k} (z) − ∆j f{i,j} (z)). The first bracket is always positive by submodular property and the second is positive by the condition (36). The inequality (b3 ) is equivalent to {1},{2},{3},{i,j}

f{i,k} (z) − γij zi (di − ρ) − γji zj (uj − ρ) − γk ≤ f{i,j} (z) −

γij zi (di

− ρ) −

γji zj (dj

− ρ) −

zk (dk − ρ)

{1},{2},{3},{i,j} γk zk (uk

i.e. {1},{2},{3},{i,j}

f{i,k} (z) − f{i,j} (z) ≤ γji zj (uj − dj ) − γk

zk (uk − dk ).

− ρ)

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Since, γj

=

f{i} −f{i,j} zj (uj −dj )

and (34) this inequality becomes

f{i,k} (z) − f{i,j} (z) ≤ f{i} (z) − f{i,j} (z) + f{k} (z) + f{i,j} (z) − f{i} (z) − f{j} (z) i.e. f{j} (z) − f{i,j} (z) ≤ f{k} (z) − f{i,k} (z) This is equivalent to the condition ∆i f{i,j} (z) ≤ ∆i f{i,k} (z), from the (36) Proof for the case (b4 ) is similar to the case (b3 ). (c) To prove this we need to show that (c1 ) B∅ (z, γ {{1,2},{1,3},{2,3},{i}} ) ≤ B{{1,2},{1,3},{2,3},{i}} (c2 ) B{j} (z, γ {{1,2},{1,3},{2,3},{i}} ) ≤ B{{1,2},{1,3},{2,3},{i}} (c3 ) B{k} (z, γ {{1,2},{1,3},{2,3},{i}} ) ≤ B{{1,2},{1,3},{2,3},{i}} . (c4 ) B{1,2,3} (z, γ {{1,2},{1,3},{2,3},{i}} ) ≤ B{{1,2},{1,3},{2,3},{i}} are true. The inequality (c1 ) is equivalent to {{1,2},{1,3},{2,3},{i}}

f∅ (z) − γi

{i}

{i}

zi (ui − ρ) − γj zj (uj − ρ) − γk zk (uk − ρ)

{{1,2},{1,3},{2,3},{i}}

≤ f{i} (z) − γi

{i}

{i}

zi (di − ρ) − γj zj (uj − ρ) − γk zk (uk − ρ)

i.e. {{1,2},{1,3},{2,3},{i}}

0 ≤ f{i} (z) − f∅ (z) + γi

zi (ui − di ) .

Due to (35) we get 0 ≤ f{i} (z) − f∅ (z) + f{i} (z) + f{j,k} (z) − f{i,k} (z) − f{i,j} (z) If we add +f{k} (z) − f{k} (z) to the right hand side and write in terms of increments 0 ≤ (∆i f{i,k} (z) − ∆i f{i} (z)) + (∆j f{i,j} (z) − ∆j f{j,k} (z)). First bracket is positive by submodular property and the second one is positive by condition ∆j f{i,j} (z) ≥ ∆j f{j,k} (z) from (37). The inequality (c2 ) is equivalent to {{1,2},{1,3},{2,3},{i}}

f{j} (z) − γi

{{1,2},{1,3},{2,3},{i}}

≤ f{i,j} (z) − γi

{i}

{i}

zi (ui − ρ) − γj zj (dj − ρ) − γk zk (uk − ρ) {i}

{i}

zi (di − ρ) − γj zj (dj − ρ) − γk zk (uk − ρ)

This is equivalent to {{1,2},{1,3},{2,3},{i}}

0 ≤ f{i,j} (z) − f{j} (z) + γi

zi (ui − di )

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Due to (35) we get 0 ≤ f{i,j} (z) − f{j} (z) + f{i} (z) + f{j,k} (z) − f{i,k} (z) − f{i,j} (z) Written in terms of increments we get the condition ∆k f{i,k} (z) ≥ ∆k f{j,k} (z). The proof of the case (c3 ) is similar to the previous one. The inequality (c4 ) is equivalent to {{1,2},{1,3},{2,3},{i}}

f{i,j,k} (z) − γi

{i}

{i}

zi (di − ρ) − γj zj (dj − ρ) − γk zk (dk − ρ)

{{1,2},{1,3},{2,3},{i}}

≤ f{j,k} (z) − γi

{i}

{i}

zi (ui − ρ) − γj zj (dj − ρ) − γk zk (dk − ρ),

i.e. {{1,2},{1,3},{2,3},{i}}

0 ≤ f{j,k} (z) − f{i,j,k} (z) − γi ({i},{1,2},{1,3},{2,3})

Due to γi

=

zi (ui − di )

f{i} (z)+f{j,k} (z)−f{i,k} (z)−f{i,j} (z) zi (ui −di )

we get

0 ≤ f{j,k} (z) − f{i,j,k} (z) − f{i} (z) − f{j,k} (z) + f{i,k} (z) + f{i,j} (z) Written in terms of increments this is (∆k f{i,k} (z) − ∆k f{1,2,3} (z)) ≥ 0. This is true by submodular property. (d) To prove this we need to show that (d1 ) B∅ (z, γ {{i},{j},{i,j},{i,k}} ) ≤ B{{i},{j},{i,j},{i,k}} (d2 ) B{k} (z, γ {{i},{j},{i,j},{i,k}} ) ≤ B{{i},{j},{i,j},{i,k}} (d3 ) B{j,k} (z, γ {{i},{j},{i,j},{i,k}} ) ≤ B{{i},{j},{i,j},{i,k}} (d4 ) B{1,2,3} (z, γ {{i},{j},{i,j},{i,k}} ) ≤ B{{i},{j},{i,j},{i,k}} Inequality (d1 ) written explicitly is {j}

f∅ (z) − γi

{i}

{i}

zi (ui − ρ) − γj zj (uj − ρ) − γk zk (uk − ρ) {j}

≤ f{i} (z) − γi

{i}

{i}

zi (di − ρ) − γj zj (uj − ρ) − γk zk (uk − ρ) ,

i.e. {j}

0 ≤ f{i} (z) − f∅ (z) + γi

zi (ui − di )

Due to (33) we have 0 ≤ f{i} (z) − f∅ (z) + f{j} (z) − f{i,j} (z) . Written in terms of the increments we get the condition 0 ≤ ∆i f{i,j} (z) − ∆i f{i} (z) and this is one of submodular property inequalities.

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Inequality (d2 ) written explicitly is {j}

f{k} (z) − γi

{i}

{i}

zi (ui − ρ) − γj zj (uj − ρ) − γk zk (dk − ρ) {j}

≤ f{i,k} (z) − γi

{i}

{i}

zi (di − ρ) − γj zj (uj − ρ) − γk zk (dk − ρ) ,

i.e. {j}

0 ≤ f{i,k} (z) − f{k} (z) + γi

zi (ui − di ) .

Due to (33) we have 0 ≤ f{i,k} (z) − f{k} (z) + f{j} (z) − f{i,j} (z) Written in terms of the increments we get the condition 0 ≤ ∆i f{i,j} (z) − ∆i f{i,k} (z). Inequality (d3 ) written explicitly is {j}

f{j,k} (z) − γi

{i}

{i}

zi (ui − ρ) − γj zj (dj − ρ) − γk zk (dk − ρ) {j}

≤ f{j} (z) − γi

{i}

{i}

zi (ui − ρ) − γj zj (dj − ρ) − γk zk (uk − ρ) ,

i.e. {i}

0 ≤ f{j} (z) − f{j,k} (z) − γk zk (uk − dk ) Due to (33) we have 0 ≤ f{j} (z) − f{j,k} (z) − f{i} (z) + f{i,k} (z) Written in terms of the increments we get the condition 0 ≤ ∆k f{j,k} (z) − ∆k f{i,k} (z). And inequality (d4 ) written explicitly is {j}

f{i,j,k} (z) − γi

{i}

{i}

zi (di − ρ) − γj zj (dj − ρ) − γk zk (dk − ρ) {j}

≤ f{i,k} (z) − γi

{i}

{i}

zi (di − ρ) − γj zj (uj − ρ) − γk zk (dk − ρ) ,

i.e. {i}

0 ≤ f{i,k} (z) − f{i,j,k} (z) − γj zj (uj − dj ) . Due to ( 33) we have 0 ≤ f{i,k} (z) − f{i,j,k} (z) − f{i} (z) + f{i,j} (z) . Written in terms of the increments we get 0 ≤ ∆j f{1,2,3} (z) − ∆j f{i,j} (z). This is true by submodular property. Lemma 14. If we have conditions ∆i f{i,k} (z) ≥ ∆i f{i,j} (z)

and

∆j f{i,j} (z) ≥ ∆j f{j,k} (z)

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then ∆k f{i,k} (z) ≥ ∆k f{j,k} (z) is true for all different i, j, k ∈ {1, 2, 3}. Proof. Adding together the first and the second inequality we get third one. 6. Proof of Theorem 1 We will prove only the (17) (formula (18) can be shown similarly. It is not difficult to see that one of the following three cases is always realized: (i) ∆1 f{1,3} (z) ≥ ∆1 f{1,2} (z)

(39)

∆2 f{1,2} (z) ≥ ∆2 f{2,3} (z),

(40)

∆1 f{1,3} (z) ≥ ∆1 f{1,2} (z)

(41)

∆2 f{2,3} (z) ≥ ∆2 f{1,2} (z),

(42)

∆1 f{1,2} (z) ≥ ∆1 f{1,3} (z)

(43)

∆2 f{2,3} (z) ≥ ∆2 f{1,2} (z).

(44)

and

(ii)

and

(iii)

and

We shall consider only the case (i) (other being similar) By Lemma 14 we have ∆3 f{1,3} (z) ≥ ∆3 f{2,3} (z).

(45)

B{∅,{1},{2},{3}} = min B(z, γ).

(46)

So, we need to show that γ∈Γ(z)

From Proposition 13 we have that the set Γ(z) is equal to Γ(z) = {γ {∅,{1},{2},{3}} , γ {{1},{2},{3},{2,3}} , γ {{1,2},{1,3},{2,3},{1}} , γ {{1},{2},{1,2},{2,3}} , γ {{1},{3},{1,3},{2,3}} , γ {{1,2},{1,3},{2,3},{1,2,3}}}.

(47)

According to notation (31), we need to show that B{∅,{1},{2},{3}} = min{B{∅,{1},{2},{3}}, B{{1},{2},{3},{2,3}} , B{{1,2},{1,3},{2,3},{1}} , B{{1},{2},{1,2},{2,3}} , B{{1},{3},{1,3},{2,3}} , B{{1,2},{1,3},{2,3},{1,2,3}}}.

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We will show the following inequalities: (a1 )

B{∅,{1},{2},{3}} ≤ B{{1},{2},{3},{2,3}}

(a2 )

B{∅,{1},{2},{3}} ≤ B{{1,2},{1,3},{2,3},{1}}

(a3 )

B{∅,{1},{2},{3}} ≤ B{{1},{2},{1,2},{2,3}}

(a4 )

B{∅,{1},{2},{3}} ≤ B{{1},{3},{1,3},{2,3}}

(a5 )

B{∅,{1},{2},{3}} ≤ B{{1,2},{1,3},{2,3},{1,2,3}}

(a1 ) We need to show that u1 − ρ u2 − ρ u3 − ρ f{1} (z) + f{2} (z) + f{3} (z) u1 − d1 u2 − d2 u3 − d3 u1 − ρ ≤ (−α123 ) f{2,3} (z) + α12 f{3} (z) + α13 f{2} + f{1} (z). u1 − d1

α123 f∅ (z) +

This is equivalent to

  u2 − ρ 0 ≤ −α123 f∅ (z) − α123 f{2,3} (z) + α13 − f{2} (z) u2 − d2   u3 − ρ + α12 − f{3} (z), u3 − d3

i.e. 0 ≤ α123 (f{3} (z) − f{2,3} (z) + f{2} (z) − f∅ (z)). Written in terms of the increments we get 0 ≤ α123 (∆2 f{2,3} (z) − ∆2 f{2} (z)). The term in bracket is positive by the submodular property. (a2 ) Secondly, we need to prove the inequality α123 f∅ +

u1 − ρ u2 − ρ u3 − ρ f{1} (z) + f{2} + f{3} (z) u1 − d1 u2 − d2 u3 − d3

≤ (α123 + 1) f{1} (z) + (−α12 ) f{1,2} (z) + (−α13 ) f{1,3} (z) −

d1 − ρ f{2,3} (z). u1 − d1

This is equivalent to u2 − ρ u3 − ρ f{2} (z) − f{3} (z) u2 − d2 u3 − d3     u3 − ρ u2 − ρ − α123 + f{1,2} (z) − α123 + f{1,3} (z) u3 − d3 u2 − d2   u2 − ρ u3 − ρ − α123 + + f{2,3} (z) u2 − d2 u3 − d3   u2 − ρ u3 − ρ + + 2α123 + f{1} (z), u2 − d2 u3 − d3

0 ≤ −α123 f∅ (z) −

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which after tidying up is 0 ≤ −α123 (f∅ (z) − 2f{1} (z) + f{1,2} (z) + f{1,3} (z) − f{2,3} (z)) u2 − ρ (f{2} (z) − f{2,3} (z) − f{1} (z) + f{1,3} (z)) u2 − d2 u3 − ρ (f{3} (z) − f{2,3} (z) − f{1} (z) + f{1,2} (z)). − u3 − d3

−

The first bracket is equal to (f∅ (z) − f{1} (z)−f{2} (z) + f{1,2} (z)+f{2} (z) − f{2,3} (z) − f{1} (z) + f{1,3} (z)) = (∆1 f{1} (z)−∆1 f{1,2} (z))+(∆3 f{2,3} (z) − ∆3 f{1,3} (z)). So, written in terms of the increments we get the inequality 0 ≤ −α123 ((∆1 f{1} (z)−∆1 f{1,2} (z))+(∆3 f{2,3} (z) − ∆3 f{1,3} (z))) u2 − ρ (∆3 f{2,3} (z) − ∆3 f{1,3} (z)) u2 − d2 u3 − ρ − (∆2 f{2,3} − ∆2 f{1,2} (z)). u3 − d3 −

First bracket is negative by the submodular property. The rest is negative by conditions (45) and (40). (a3 ) Third, we need to show that u1 − ρ u2 − ρ u3 − ρ f{1} (z) + f{2} (z) + f{3} (z) u1 − d1 u2 − d2 u3 − d3 u3 − ρ d2 − ρ ≤ (−α12 ) f{1,2} (z) + (α13 ) f{2} (z) + f{2,3} (z) − f{1} (z). u3 − d3 u2 − d2

α123 f∅ +

This is equivalent to u3 − ρ u3 − ρ f{3} (z) + f{2,3} (z) + α123 f{2} (z) u3 − d3 u3 − d3   u3 − ρ u3 − ρ )f{1} (z), + −α123 − f{1,2} (z) + (α123 + u3 − d3 u3 − d3

0 ≤ −α123 f∅ −

which after tidying up is 0 ≤ −α123 (f∅ (z) − f{1} (z) − f{2} (z) + f{1,2} (z)) u3 − ρ − (f{3} (z) − f{2,3} (z) − f{1} (z) + f{1,2} (z)). u3 − d3 Written in terms of the increments we get 0 ≤ α123 (∆1 f{1,2} (z) − ∆1 f{1} (z)) +

u3 − ρ (∆2 f{1,2} (z) − ∆2 f{2,3} (z)). u3 − d3

First bracket is positive by the submodular property and the second one by condition (40).

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(a4 ) We need to prove that u1 − ρ u2 − ρ u3 − ρ f{1} (z) + f{2} (z) + f{3} (z) u1 − d1 u2 − d2 u3 − d3 u2 − ρ d3 − ρ ≤ −α13 f{1,3} (z) + (α12 ) f{3} (z) + f{2,3} (z) − f{1} (z). u2 − d2 u3 − d3

α123 f∅ (z) +

This is equivalent to

  u2 − ρ u2 − ρ α123 f∅ (z) + f{2} (z) − α123 + f{1} (z) u2 − d2 u2 − d2   u2 − ρ u2 − ρ ≤ − α123 + f{2,3} (z), f{1,3} (z) + (α123 ) f{3} (z) + u2 − d2 u2 − d2

which after tidying is 0 ≤ α123 (f{1} (z) − f{1,3} (z) − f∅ (z) + f{3} (z)) u2 − ρ + (f{1} (z) + f{1,3} (z) − f{2} (z) + f{2,3} (z)). u2 − d2 Written in terms of the increments we get 0 ≤ α123 (∆3 f{1,3} (z) − ∆3 f{3} (z)) +

u2 − ρ (∆3 f{1,3} (z) − ∆3 f{2,3} (z)). u2 − d2

The first bracket is positive by the submodular property and the second one by condition (45). (a5 ) finally, we need to prove the inequality α123 f∅ (z) +

u1 − ρ u2 − ρ u3 − ρ f{1} (z) + f{2} (z) + f{3} (z) u1 − d1 u2 − d2 u3 − d3

≤ (−α123 − 1)f{1,2,3} (z) − −

d1 − ρ f{2,3} (z) u1 − d1

d3 − ρ d2 − ρ f{1,3} (z) + f{1,2} (z). u2 − d2 u3 − d3

This is equivalent to u1 − ρ u2 − ρ u3 − ρ f{1} (z) + f{2} (z) + f{3} (z) u1 − d1 u2 − d2 u3 − d3   u2 − ρ u3 − ρ + − α123 + f{2,3} (z) u2 − d2 u3 − d3   u1 − ρ u3 − ρ + − α123 + f{1,3} (z) u1 − d1 u3 − d3   u1 − ρ u2 − ρ − α123 + + f{1,2} (z) u1 − d1 u2 − d2   u2 − ρ u3 − ρ u1 − ρ + + − 2α123 + f{1,2,3} (z), u2 − d2 u3 − d3 u1 − d1

0 ≥ α123 f∅ (z) +

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which after tidying up is 0 ≤ α123 (−f∅ (z) − 2f{1,2,3} (z) + f{1,2} (z) + f{1,3} (z) + f{2,3} (z)) u1 − ρ (f{1,2} (z) − f{1,2,3} (z) − f{1} (z) + f{1,3} (z)) u1 − d1 u2 − ρ + (f{1,2} (z) − f{1,2,3} (z) − f{2} (z) + f{2,3} (z)) u2 − d2 u3 − ρ + (f{1,3} (z) − f{1,2,3} (z) − f{3} (z) + f{2,3} (z)). u3 − d3 +

The first term is equal to −f∅ (z) − 2f{1,2,3} (z) + f{1,2} (z) + f{1,3} (z) + f{2,3} (z) = (∆2 f{1,2,3} (z) − ∆2 f{1,2} (z)) + (∆1 f{1,2} (z) − ∆1 f{1} (z)) + (∆1 f{1,2,3} (z) − ∆1 f{1,3} (z)). Shortly written, in terms of the increments, we get 0 ≤ α123 ((∆2 f{1,2,3} (z) − ∆2 f{1,2} (z)) + (∆1 f{1,2} (z) − ∆1 f{1} (z)) + (∆1 f{1,2,3} (z) − ∆1 f{1,3} (z)) u1 − ρ (∆3 f{1,2,3} (z) − ∆3 f{1,3} (z)) u1 − d1 u2 − ρ + (∆3 f{1,2,3} (z) − ∆3 f{2,3} (z)) u2 − d2 u3 − ρ + (∆2 f{1,2,3} (z) − ∆2 f{2,3} (z)). u3 − d3

+

finally, we could see that all brackets are positive by the submodular property. This completes the proof. References Altman, E., Gaujal, B. and Hordijk, A. [1999] Multimodularity, Convexity and Optimazation Properties, preprint, URL: http://www-sop.inria.fr/mistral/personnel/ Eitan.Altman/me.html Altman, E. and Koole, G. [1998] On submodular value funstions and complex dynamic programming, Stochastic Models 14, 1051–1072. Bellman, R. E. [1957] Dynamic Programming (Princeton University Press: Princeton). Bernard, P. [2005] Robust control approach to option pricing, including transaction costs. Advances in dynamic games, Ann. Internal. Soc. Dynam. Games 7, Birkhauser Boston, Boston, MA, 391–416. Cox, J. C., Ross, S. A. and Rubinstein, M. [1979] Option Pricing: A Simplified Approach, J. Financial Economics 7, 229–263. David, D. Y. [1995] S-Modular games with queueing applications, Queueing Systems 21, 445–475. Dixit, A. K. and Pindyck, R. S. [1997] Investment Under Uncertainty (Priceton University Press, Priceton, NJ)

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