Optimal liquidation for a risk averse investor in a one-sided limit order book driven by a Levy process
aa r X i v : . [ q -f i n . T R ] O c t Optimal liquidation for a risk averse investor in a one-sidedlimit order book driven by a L´evy process
Arne Løkka ∗ Junwei Xu † November 2, 2020
Abstract
In a one-sided limit order book, satisfying some realistic assumptions, where the un-affected price process follows a L´evy process, we consider a market agent that wants toliquidate a large position of shares. We assume that the agent has constant absolute riskaversion and aims at maximising the expected utility of the cash position at the end oftime. The agent is then faced with the problem of balancing the market risk and the costof a rapid execution. In particular we are interested in how the agent should go aboutoptimally submitting orders. Since liquidation normally takes place within a short periodof time, modelling the risk as a L´evy process should provide a realistic model with goodstatistical fit to observed market data, and thus the model should provide a realistic re-flection of the agent’s market risk. We reduce the optimisation problem to a deterministictwo-dimensional singular problem, to which we are able to derive an explicit solution interms of the model data. In particular we find an expression for the optimal interventionboundary, which completely characterise the optimal liquidation strategy.
This paper is concerned with how a market agent should go about selling (or purchasing) alarge position of shares. This kind of problem has attracted considerable interest over the pastfew years following the introduction of electronic trading platforms. In the model we consider, ∗ Department of Mathematics Columbia House London School of Economics Houghton Street, LondonWC2A 2AE United Kingdom ([email protected]) † Department of Mathematics Columbia House London School of Economics Houghton Street, LondonWC2A 2AE United Kingdom ([email protected])
1e specify the limit order book and how this recovers over time. Thus the optimal liquidationstrategy will explicitly specify the orders the agent submits to the market, as opposed to justspecifying the optimal speed at which to trade that is the case for the popular impact models.We refer the reader to Lehalle and Laruelle (2013), Cartea et al. (2015) and Gu´eant (2016)for an introduction to optimal execution and common models.More precisely, we consider a market agent with constant absolute risk aversion (CARA)that wants to maximise the expected utility of the cash position at the end of time. Thusthe agent does not face any restrictions on the duration of the liquidation, but the rapid-ness is determined by the market risk and the agent’s preference towards risk. Working withan infinite time-horizon also avoids the time dimension of the problem, and hence make theproblem more tractable. We assume that the market risk of the stock price is modelledby a L´evy process, which is allowed to have a drift, but which we assume satisfy a certainexponential moment condition. A number of studies demonstrate that L´evy processes areable to capture the essential statistical properties of stock price movements over short time-horizons (Madan and Seneta, 1990), (Eberlein and Keller, 1995), (Barndorff-Nielsen, 1997)and (Cont and Tankov, 2004). Since the main bulk of the liquidation tends to finish withina short period of time, this model should provide a reasonable reflection of the market riskfaced by the agent. For the same reason, a L´evy process model is a reasonable approx-imation of an exponential L´evy process model. Such a Bachelier-L´evy type model mayseem simplistic, but this kind of modelling of the unaffected price process is widely used inthe optimal liquidation literature (Almgren and Chriss, 2000), (Kissell and Malamut, 2005),(Schied and Sch¨oneborn, 2009) and (Gatheral, 2010). In particular, Forsyth et al. (2012)demonstrated that the linear model provides an excellent approximation to models with pricesmodelled as a geometric Brownian motion and multiplicative impact in the Almgren-Chrissframework.We consider a bid limit order book with general shape and with a general resiliencefunction satisfying some reasonable conditions. In particular, some of these assumptions arecrucial in order to solve the problem since the assumptions imply that a certain function isconcave, which is needed for optimality of our strategy. So with an infinite time horizon, wesolve the problem of maximising the expected utility of the agent’s final cash position. We dothis by showing that the problem can be reduced to a two-dimensional deterministic singularoptimisation problem to which we can obtain an explicit solution in terms of the characteristicsof the limit order book and the agent’s risk aversion. With reference to Løkka (2014) and thenature of the problem, we guess that the optimal strategy consists initially of either a block2ale or a period of waiting, and that the agent thereafter continuously submit sales ordersin such a way that the state process remains on the optimal intervention boundary. Thestate process here is the two-dimensional process consisting of the number of shares the agentcurrently hold and the current state of the order book. The optimal intervention boundaryis associated with the Hamilton-Jacobi-Bellman (HJB) variational inequalities correspondingto the optimisation problem. This intervention boundary might have discontinuities as wellas constant parts. The discontinuities corresponds to periods of waiting while the order bookrecovers, while the constant part corresponds to submitting sales orders at the same rate asthe resilience rate. Following the idea in Løkka (2014), the value function in our problemcan be expressed in an explicit way in terms of the problem’s data, and we characterise theintervention boundary via the HJB variational inequalities. The strategy associated with thisintervention boundary is shown to be optimal by a verification argument.We also provide an example in order to illustrate the optimal strategies for various param-eters of risk aversion for the case when the risk is modelled by a Brownian motion and for thecase when the risk is modelled by a L´evy process with jumps. For the case of a L´evy processwith jumps we choose the L´evy process approximation of the exponential variance-gammaprocess. We find that the two models produce similar optimal strategies if the agent is notvery risk averse, but as the agent’s stock position and the agent’s risk aversion increase, thedifferences become more pronounced.The model we use is a version of the model introduced in Obizhaeva and Wang (2013),which was later generalised in Alfonsi et al. (2010), and then further in Predoiu et al. (2011).However, these papers did not consider risk-aversion, but assume that the agent wants tomaximise the expected value of the cash position. The problem we consider in this paper isan extension of Løkka (2014) in the sense that unaffected price process follows a general L´evyprocess which could have a drift, and not simply a Brownian motion. Moreover, in this paperwe consider an order book with a general resilience function, not just exponential resilience.This paper is structured as follows. In Section 2 we introduce the limit order book modeland the agent’s optimisation problem. We simplify the problem and show that it can bereduced to a deterministic optimisation problem in Section 3. The simplified deterministicoptimisation problem is solved in Section 4. The proofs omitted in the previous sections arecontained in Section 5. 3
Problem formulation
Let (cid:0) Ω , F , ( F t ) t ≥ , P (cid:1) be a complete filtered probability space satisfying the usual conditionsand supporting a one-dimensional non-trivial L´evy process L . Assumption 2.1.
We assume that there exists some δ > E (cid:2) e θL (cid:3) < ∞ , for | θ | < δ .Let κ denote the cumulant generating function of L , i.e. κ ( θ ) = ln (cid:0) E (cid:2) e θL (cid:3)(cid:1) , θ ∈ R . Assumption 2.1 guarantees that the cumulant generating function κ is continuously differen-tiable in a neighbourhood of 0 and that the L´evy process L is square integrable. Hence L admits the representation L t = µt + σW t + Z R \{ } z (cid:0) N ( t, dz ) − tν ( dz ) (cid:1) , t ≥ , where µ ∈ R , W is a standard Brownian motion, N is a Poisson random measure whichis independent of W with compensator π ( t, dz ) = tν ( dz ), where ν denotes the L´evy mea-sure associated with L (Kyprianou, 2006). The cumulant generating function κ can then beexpressed as κ ( θ ) = µθ + 12 σ θ + Z R \{ } (cid:16) e θz − − θz (cid:17) ν ( dz ) , | θ | < δ. (2.1)In particular, κ (0) = 0 , κ ′ (0) = µ and κ ′′ (0) = σ + Z R \{ } z ν ( dz ) . Moreover, κ is strictly convex and continuously differentiable on its effective domain. Assumption 2.2.
We assume that µ ≤
0, i.e. the L´evy process L is a supermartingale.The main reason behind this assumption is that the optimisation problem we consider inthis paper does not have a solution when µ is positive. This is because when µ is positive, itis be optimal for the agent to hold on to some amount of shares for as long as possible, and asa consequence our formulation of the liquidation problem does not have a solution when µ is4ositive (see Remark 4.4). For the remainder of the paper, we therefore restrict ourselves tothe case µ ≤ R + = [0 , ∞ ) and R − = ( −∞ , . We consider a market agent that aims to sell a large amount of shares of a single stockwith no time restrictions. Let Y t denote the number of shares held by the agent at time t .We refer to a process Y as a liquidation strategy if Y t tends to 0 as t tends to infinity. Weconsider the following set of admissible liquidation strategies. Definition 2.3.
For y ∈ R + , let A ( y ) denote the set of all ( F t ) t ≥ -adapted, predictable,decreasing, c`adl`ag processes Y , satisfying Y − = y and Z ∞ κ A (cid:0) k Y t k L ∞ ( P ) (cid:1) dt < ∞ . (2.2)Moreover, let A D ( y ) denote the set of all deterministic strategies in A ( y ). Remark 2.4.
Condition (2.2) amounts to the following Z ∞ k Y t k L ∞ ( P ) dt < ∞ , if µ < , Z ∞ k Y t k L ∞ ( P ) dt < ∞ , if µ = 0 . When µ > µ < L is a Brownianmotion, then the time it takes for the best bid price to reach a given level p is finite almostsurely, for any level p . We skip the mathematical details, but intuitively a higher price level p results in a higher expected utility of the cash position. Consequently the value function willbe infinite if such strategies are admissible.To describe the agent’s execution price, we explicitly model a bid limit order book. Weassume that the unaffected bid price process B , which is the process describing the best bid5rices in the market if the agent does not trade, is given by B t = b + L t , t ≥ , where b > m defined on theBorel σ -algebra on R − , denoted by B ( R − ). If S ∈ B ( R − ), then m ( S ) represents the numberof bid orders with prices in the set B t + S = { B t + s | s ∈ S} , provided that the agent did notmake any trades before time t . Notice that the undisturbed bid order book described by m is relative to the unaffected bid prices in the sense that it shifts together with the movementof the unaffected price. We impose the following assumptions on m . Assumption 2.5.
We assume that(i) there exists some ¯ x ∈ ( −∞ ,
0) such that m ((¯ x, m ( R − ) < ∞ ,(ii) m is absolutely continuous with respect to Lebesgue measure, and is non-zero on anyinterval properly containing the origin,(iii) the function x m (( x, x , for x ∈ R − .The concavity of x m (( x, z = − m ( R − ),which represents the total amount of bid limit orders in the undisturbed limit order book,and introduce the functions φ : [ −∞ , → R − and ψ : R − → [ −∞ ,
0] by φ ( x ) = − m (cid:0) ( x, (cid:1) and ψ ( z ) = φ − ( z ) , where φ ( ψ ( z )) = z , for all z ∈ [¯ z, ψ ( z ) = −∞ , for all z < ¯ z . A consequence ofAssumption 2.5 is that φ is convex, ψ is concave, and φ and ψ are both continuous andstrictly increasing on their effective domain. They also satisfy φ (0) = ψ (0) = 0 , (2.3)6s well as Z ¯ z ψ ( u ) du < ∞ and ψ (¯ z ) > −∞ . (2.4)In order to model the dynamic of the bid order book during trading, we need to introduceone more process that captures the state of the order book. For a given strategy Y , let Z Y bean R − -valued process such that − Z Yt represents the volume spread at time t . That is − Z Yt isequal to the total number of bid orders which have already been executed subtracted by thetotal amount of limit orders which have arrived to refill the book up to time t . We call Z Y thestate process of the bid limit order book associated with a trading strategy Y . Let Z Y − = z ,where z ≥ ¯ z is the initial state of our bid order book. Therefore, we have ψ ( Z Yt ) = B Yt − B t ,where B Yt is the best bid price at time t corresponding to Y , and ψ ( Z Yt ) can be understood asthe extra price spread at time t , caused by the investor who implements a strategy Y . Notethat we have defined ψ ( z ) = −∞ , for all z < ¯ z . This implies that the best bid price dropsdown to −∞ , if one sell more shares than available bids in the book. The rate at which bidorders are refilling the order book is described by a resilience function h : R − → R − whichsatisfies the following. Assumption 2.6.
We assume that the resilience function h : R − → R − is increasing, locallyLipschitz continuous, satisfies h (0) = 0 with h ( x ) < x <
0, and that the function x /h ( x ) is a concave function for x < h ( x ) = λx , for λ >
0, which corresponds to exponential resilience,satisfies Assumption 2.6. We then consider the state process Z Y with dynamic dZ Yt = − h (cid:0) Z Yt − (cid:1) dt + dY t , Z Y − = z ∈ R − . (2.5)For any admissible strategy Y , we refer to Predoiu et al. (2011) Appendix A for the existenceand uniqueness of a negative, c`adl`ag and adapted solution to this dynamic. From Assump-tion 2.6 and equation (2.5) we observe that the further the best bid price is away from theunaffected bid price, the larger the speed of resilience for the best bid price.If the agent does not make any trades from time t to t , then (cid:0) Z Yt (cid:1) t 0, then t − t = H ( Z Yt ) − H (cid:0) Z Yt (cid:1) . (2.9)Suppose that the agent’s initial cash position is c and that the agent implements a strategy Y ∈ A ( y ). Then the agent’s cash position at time T > C T ( Y ) = c − Z T B Yt − dY ct − X ≤ t ≤ T Z △ Y t (cid:8) B t − + ψ (cid:0) Z Yt − + x (cid:1)(cid:9) dx, (2.10)which corresponds to the best bids offered at all times being executed first so as to match theagent’s orders, where the first integral represents the cost from the continuous component ofthe liquidation strategy and the sum of integrals represents the total cost due to all block sales.We also suppose the agent has a constant absolute risk aversion (CARA). With initial cashposition c , an initial share position y and infinite time-horizon, the agent wants to maximisethe expected utility of the cash position at the end of time. Mathematically, the agent’soptimal liquidation problem is sup Y ∈A ( y ) E (cid:2) U (cid:0) C ∞ ( Y ) (cid:1)(cid:3) , (2.11)where the utility function U is given by U ( c ) = − e − Ac , A > . Observe that if Z Yt < ¯ z , then B Yt = B t + ψ ( Z Yt ) = −∞ . Clearly receiving the price −∞ 8s unfavourable to the agent. Indeed, (2.10) shows that this brings the agent an infinite cost.Due to this consideration, we will from now on only focus on admissible strategies Y for which Z Yt ≥ ¯ z , for all t ≥ 0. Define the function κ A : R + → [0 , ∞ ] by κ A ( y ) = κ ( − Ay ) , y ≥ , and set ¯ y A = sup (cid:8) y ≥ | κ A ( y ) < ∞ (cid:9) . Then κ A is strictly increasing, strictly convex and continuously differentiable on [0 , ¯ y A ), with κ A (0) = 0. In this section, we show that the utility maximisation problem in (2.11) can be reduced toa deterministic optimisation problem. This kind of result was first derived in Schied et al.(2010), which proved that with a certain market structure and an agent with constant absoluterisk aversion, the optimal liquidation strategy is deterministic.Let Y ∈ A ( y ). Then it follows from (2.10) that C T ( Y ) = c + by − ( b + L T ) Y T + Z T Y t − dL t + X ≤ t ≤ T △ L t △ Y t − F T ( Y ) , where F T is given by F T ( Y ) = Z T ψ (cid:0) Z Yt − (cid:1) dY ct + X ≤ t ≤ T Z △ Y t ψ (cid:0) Z Yt − + x (cid:1) dx. (3.1)Since t 7→ k Y T k L ∞ ( P ) is decreasing, condition (2.2) implies that any admissible strategy Y ∈A ( y ) satisfies lim t →∞ t κ A (cid:0) k Y t k L ∞ ( P ) (cid:1) = 0 . (3.2)9lso observe that lim x → κ A ( x ) x = − Aµ. Therefore, if µ < 0, there exists an ǫ > C , C > C x ≤ κ A ( x ) ≤ C x, for x ∈ [0 , ǫ ] . With reference to (3.2), It follows that for every Y ∈ A ( y ),lim t →∞ t k Y t k L ∞ ( P ) = 0 , if µ < . (3.3)If µ = 0, then lim x → κ A ( x ) x = K, for some K > . Therefore, if µ = 0, there exists an ǫ > C , C > C x ≤ κ A ( x ) ≤ C x , for x ∈ [0 , ǫ ] . With reference to (3.2) it follows that for every Y ∈ A ( y ),lim t →∞ t k Y t k L ∞ ( P ) = 0 , if µ = 0 . (3.4)Let Y be an admissible strategy in A ( y ). Then with reference to (3.3) and (3.4), wecalculate lim T →∞ E (cid:2) | L T Y T | (cid:3) ≤ lim T →∞ (cid:0) µ T + κ ′′ (0) T (cid:1) k Y T k L ∞ ( P ) = 0 . We conclude that B T Y T tends to 0 in L ( P ) as T → ∞ . Furthermore, E (cid:20)(cid:18) Z ∞ Y t − dL t (cid:19) (cid:21) ≤ E (cid:20)(cid:18) Z ∞ Y t − µ dt (cid:19) (cid:21) + E (cid:20)(cid:18) Z ∞ Y t − d (cid:0) L t − µt (cid:1)(cid:19) (cid:21) ≤ µ Z ∞ k Y t k L ∞ ( P ) dt + (cid:20) κ ′′ (0) Z ∞ k Y t k L ∞ ( P ) dt (cid:21) < ∞ . R ∞ Y t − dL t is well-defined in L ( P ). Due to the predictability of Y , we also have that E (cid:20)(cid:18) X ≤ t ≤ T △ L t △ Y t (cid:19) (cid:21) = E (cid:20)Z T (cid:0) △ Y t (cid:1) dt (cid:21)(cid:18)Z R \{ } z ν ( dz ) (cid:19) = 0 , for all T > 0, which shows that the quadratic covariation of the jumps of L and Y is almostsurely 0. Moreover, note that F T ( Y ) ≥ T . Therefore, F ∞ is a well defined function from the set of c`adl`ag non-increasing functions into the extendedpositive real numbers. The final cash position is hence given by C ∞ ( Y ) = c + by + Z ∞ Y t − dL t − F ∞ ( Y ) , (3.5)where c + by represents the mark-to-market value of the total wealth of the agent’s positionat the start of the liquidation, R ∞ Y t − dL t represents the profit or loss due to the market risk,and F ∞ ( Y ) represents the cost due to the price impact.Let Y ∈ A ( y ) and define the process M Y by M Yt = exp (cid:18) − A Z t Y s − dL s − Z t κ A (cid:0) Y s − (cid:1) ds (cid:19) , t ≥ . Then it follows from Theorem 3.2 in Kallsen and Shiryaev (2002) that M Y is a uniformlyintegrable martingale. We can therefore define a probability measure e P = P Y by d e P d P = M Y ∞ . Following the idea of the proof of Theorem 2.1 in Schied et al. (2010), we set I = inf Y ∈A D ( y ) Z ∞ κ A (cid:0) Y t − (cid:1) dt + AF ∞ ( Y ) , and note that κ A ( · ) and F ∞ ( · ) are deterministic. Let Y ǫ ∈ A D ( y ) be such that Z ∞ κ A (cid:0) Y ǫt − (cid:1) dt + AF ∞ ( Y ǫ ) ≤ I + ǫ. Y ∈ A ( y ), we calculate that E (cid:2) U (cid:0) C ∞ ( Y ) (cid:1)(cid:3) = − e − A ( c + by ) E (cid:20) exp (cid:18) − A Z ∞ Y t − dL t + AF ∞ ( Y ) (cid:19)(cid:21) = − e − A ( c + by ) E (cid:20) M ∞ exp (cid:18)Z ∞ κ A (cid:0) Y t − (cid:1) dt + AF ∞ ( Y ) (cid:19)(cid:21) = − e − A ( c + by ) e E (cid:20) exp (cid:18)Z ∞ κ A (cid:0) Y t − (cid:1) dt + AF ∞ ( Y ) (cid:19)(cid:21) ≤ − e − A ( c + by ) e − ǫ e E (cid:20) exp (cid:18)Z ∞ κ A (cid:0) Y ǫt − (cid:1) dt + AF ∞ ( Y ǫ ) (cid:19)(cid:21) ≤ − e − A ( c + by ) e − ǫ exp (cid:18) inf Y ∈A D ( y ) (cid:26)Z ∞ κ A (cid:0) Y t − (cid:1) dt + AF ∞ ( Y ) (cid:27)(cid:19) , since Z ∞ κ A (cid:0) Y t − (cid:1) dt + AF ∞ ( Y ) ≥ I ≥ Z ∞ κ A (cid:0) Y ǫt − (cid:1) dt + AF ∞ ( Y ǫ ) − ǫ. By letting ǫ tend to 0 and taking the supremum over all admissible strategies on the left-handside of (3.6), we obtainsup Y ∈A ( y ) E (cid:2) U (cid:0) C ∞ ( Y ) (cid:1)(cid:3) = − e − A ( c + by ) exp (cid:18) inf Y ∈A D ( y ) (cid:26)Z ∞ κ A (cid:0) Y t − (cid:1) dt + AF ∞ ( Y ) (cid:27)(cid:19) (3.6) Lemma 3.1. Let F be given by (3.1). Then for every Y ∈ A D ( y ) and z ∈ [¯ z, , F ∞ ( Y ) = Z z ψ ( s ) ds + Z ∞ h (cid:0) Z Yt − (cid:1) ψ (cid:0) Z Yt − (cid:1) dt. (3.7)With reference to Lemma 3.1 and (3.6), the optimal liquidation problem amounts tosolving V ( y, z ) = inf Y ∈A D ( y ) Z ∞ (cid:18) κ A (cid:0) Y t − (cid:1) + Ah ( Z Yt − ) ψ ( Z Yt − ) (cid:19) dt, (3.8)with y = Y − and z = Z Y − . Since h and ψ are both negative-valued and κ A ≥ 0, we have V ≥ 0. Suppose y > ¯ y A , which is the upper bound for which κ A is finite (¯ y A might be + ∞ ).In this case, the market agent will want to make an immediate block sale to bring the numberof shares less than ¯ y A , since otherwise Y does not satisfy (2.2) and V ( y, z ) = ∞ . However, if12he agent sell more than z − ¯ z number of shares, the value function V ( y, z ) will be infinite.We therefore define the solvency region to be D = (cid:8) ( y, z ) ∈ R + × [¯ z, (cid:12)(cid:12) z > y − ¯ y A + ¯ z (cid:9) , and for the remainder of the paper we focus on this region. For technical reasons, we do notconsider z = y − ¯ y A + ¯ z , as the value function may explode also along this line. Our next aim is to derive a solution to the problem (3.8). The derivation will be based onapplying a time-change, and the principle of dynamic programming. With reference to theresults in Løkka (2014) and the general theory of optimal control (Fleming and Soner, 2006),it is natural to guess that there exists a decreasing c`agl`ad function β = β ∗ : R + → [¯ z, y, z ) domain into two different regions, a region where the agent makesan immediate block sale and another where the agent waits for the order book to recover. Let β ∗ denote the c`adl`ag version of β ∗ , and set S β = (cid:8) ( y, z ) ∈ D | z ≥ β ∗ ( y ) (cid:9) , W β = (cid:8) ( y, z ) ∈ D | z ≤ β ∗ ( y ) (cid:9) ∪ (cid:8) ( y, z ) | y = 0 (cid:9) , G β = S β ∩ W β . S β represents the immediate sales region, W β the waiting region, and G β is the graph ofthe intervention boundary β and represents the continuous sales region. For y > 0, theHamilton-Jacobi-Bellman equation corresponding to V given by (3.8) takes the form D − y v ( y, z ) + v z ( y, z ) = 0 , for ( y, z ) ∈ S β , (4.1) h ( z ) v z ( y, z ) − κ A ( y ) − Ah ( z ) ψ ( z ) ≤ , for ( y, z ) ∈ S β \ G β , (4.2) When the volume spread is small, but the stock position is large, it seems intuitive to sell rapidly. Onthe other hand, if the volume spread is large, but the stock position is small, it seems intuitive to wait for theorder book to recover. This motivates us to guess that the optimal intervention boundary is decreasing andseparates the ( y, z ) domain. h ( z ) v z ( y, z ) − κ A ( y ) − Ah ( z ) ψ ( z ) = 0 , for ( y, z ) ∈ W β , (4.3) D − y v ( y, z ) + v z ( y, z ) ≤ , for ( y, z ) ∈ W β \ G β , (4.4)with associated boundary condition v (0 , z ) = A R z ψ ( u ) du for all z ∈ [¯ z, D − y v ( y, z ) = lim ǫ → − ǫ (cid:18) v ( y + ǫ, z ) − v ( y, z ) (cid:19) . The equations (4.1)–(4.4) can be motivated as follows. When the market agent is trying tooptimise over deterministic strategies, the agent basically has two options. The agent caneither sell a certain number △ > y, z ), it may or may notbe optimal to sell △ amount of shares, thus v ( y, z ) ≤ v (cid:0) y − △ , z − △ (cid:1) , because the share position is decreased from y to y − △ , due to △ number of shares beingsold, while at the same time the state of the bid order book changes from z to z − △ . Thisinequality should hold for all 0 < △ ≤ y , thereforemax < △≤ y (cid:8) v ( y, z ) − v ( y − △ , z − △ ) (cid:9) ≤ . (4.5)On the other hand, during a period of time △ t > 0, it may or may not be optimal to wait,hence v ( y, z ) ≤ v (cid:0) y, Z △ t (cid:1) + Z △ t (cid:18) κ A ( y ) + Ah ( Z u − ) ψ (cid:0) Z u − (cid:1)(cid:19) du = v ( y, z ) + Z △ t (cid:18) κ A ( y ) + Ah ( Z u − ) (cid:0) Z u − (cid:1) − v z (cid:0) y, Z u − (cid:1) h ( Z u − ) (cid:19) du, The value function turns out to be continuously differentiable in z , but only continuous with a one-sidedderivative in y (see Proposition 4.7). dZ u = − h ( Z u ) du , for 0 ≤ u ≤ △ t . Multiplying the above inequality by ( △ t ) − andsending △ t to 0, we obtain h ( z ) v z ( y, z ) − κ A ( y ) − Ah ( z ) ψ ( z ) ≤ . (4.6)Since one of these strategies should be optimal, equality should hold in either (4.5) or (4.6).We therefore getmax (cid:26) max < △≤ y (cid:8) v ( y, z ) − v ( y − △ , z − △ ) (cid:9) , h ( z ) v z ( y, z ) − κ A ( y ) − Ah ( z ) ψ ( z ) (cid:27) = 0 , from which (4.1)–(4.4) follow.We define the liquidation strategy Y β corresponding to an intervention boundary β asthe c`adl`ag function with the following properties:(i) If ( y, z ) ∈ S β , then the agent makes an immediate block trade of size △ such that( Y β , Z Y β ) = ( y − △ , z − △ ) ∈ G β , and set t w = 0.(ii) If ( y, z ) ∈ W β , then the agent waits until the time t w = inf (cid:8) t ≥ | Z Y β t = β ( y ) (cid:9) , where Z Y β t = z − Z t h (cid:0) Z Y β u (cid:1) du, ≤ t ≤ t w . (iii) For t ≥ t w , the agent continuously sell shares in such a way that ( Y βt , Z Y β t ) ∈ G β , where Z Y β t = Z Y β t w − Z tt w h (cid:0) Z Y β u (cid:1) du + Y βt − Y βt w , t ≥ t w . (iv) The agent takes no further action once Y βt = 0.Figure 1 provides an illustration of such a strategy. We will later characterise an optimalintervention boundary, and prove that the corresponding strategy exists, is admissible andoptimal. The key to characterise the optimal intervention boundary is that we are able toobtain expressions for the performance of the strategy corresponding to a given interventionboundary. But first, let us examine in more detail the strategy corresponding to a givenintervention boundary function β . We will consider any intervention boundary β : R + → [¯ z, β ( y ) < 0, for all y > 0, lim y →∞ β ( y ) = ¯ z and β (0) = 0.Now, given an intervention boundary β , one may ask whether the corresponding liquidation15 ✻ (cid:0)(cid:0) (cid:0)(cid:0) (cid:0)(cid:0) (cid:0)(cid:0) (cid:0)(cid:0) (cid:0)(cid:0) (cid:0)(cid:0) (cid:0)(cid:0) (cid:0)(cid:0) (cid:0)(cid:0)❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆ ❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆❆ . ......................................... ........................................ ........................................ ....................................... ....................................... ......................................... ........................................ ........................................ ....................................... ....................................... ......................................... ........................................ ........................................ ....................................... ...................................... ................................................................................................................................................................................................................................................................ .................... ................... ..................... ....................... .......................... .................... ................... ..................... ....................... .......................... .................... ................... ..................... ....................... ......................... . ..................................................................................... ..................................................................................... .................................................................................... . ..................................... .................................. ............................... ........................... ........................ ..................... ................... ................. .............. ............ ........... ..................................... .................................. ............................... ........................... ........................ ..................... ................... ................. .............. ............ ........... ..................................... .................................. ............................... ........................... ........................ ..................... ................... ................. .............. ............ .......... G β S β W β ¯ z ¯ y A z = y − ¯ y A + ¯ z (cid:0)(cid:0)(cid:0)✠ ( y, z ) ❛ ✻ ( y, z ) ❛ ❦ ❨✻ ✛ z yA Figure 1: An illustratation of the strategy Y β corresponding to an intervention boundary β (the graph of β is G β ). The solvency region D is the region that is not shaded. For aninitial state ( y, z ) ∈ S β , the strategy Y β consists of an initial block sale that brings the stateonto G β . For an initial state ( y, z ) ∈ W β , the strategy Y β consists of waiting while the orderbook recovers until the state reach G β . Once the state is on G β , the strategy Y β consists ofcontinuously submitting sales orders in such a way that the state stays on G β .16trategy Y β exists and is unique. In order to answer this, we need to introduce the followingfunctions related to β . γ β ( y ) = β ( y ) − y, for y ∈ R + , (4.7) ρ β ( z ) = z − β − ( z ) , for z ∈ [¯ z, . (4.8)We also introduce the inverse functions β − ( z ) = inf (cid:8) y ∈ R + (cid:12)(cid:12) β ( y ) ≤ z (cid:9) , for z ∈ [¯ z, γ − β ( x ) = inf (cid:8) y ∈ R + (cid:12)(cid:12) γ β ( y ) ≤ x (cid:9) , for x ∈ R − ; (4.10) ρ − β ( x ) = inf (cid:8) z ∈ [¯ z, (cid:12)(cid:12) ρ β ( z ) ≥ x (cid:9) , for x ∈ R − . (4.11)Note that β and γ β are c`agl`ad , β − and ρ β are c`adl`ag , and γ − β as well as ρ − β are continuous .Moreover, β , β − and γ − β are decreasing, γ β is strictly decreasing, ρ β is strictly increasing,and ρ − β is increasing. Furthermore, it follows directly from the definitions of β − , γ β , γ − β , ρ β and ρ − β that the following three identities hold. ρ − β ( x ) = x + γ − β ( x ) , for all x ∈ R − , (4.12) γ − β (cid:0) ρ β ( z ) (cid:1) = β − ( z ) , for all z ∈ [¯ z, , (4.13) ρ − β (cid:0) γ β ( y ) (cid:1) = β ( y ) , for all y ∈ R + . (4.14)Also, by the definitions of G β , β and β − , we see that the set G β is the union of the graphs ofthe functions β and β − restricted to D .Observe that if z > β ( y ) , then the strategy Y β corresponding to the intervention bound-ary described by β consists of an initial sale of △ number of shares so that ( y − △ , z − △ ) isin G β (see Figure 1). Let Y β − = y and Y β = y − △ . Suppose ( y − △ , z − △ ) is on the graphof β . Then ( y − △ , z − △ ) = ( y − △ , β ( y − △ )) and this equality is equivalent to γ β (cid:0) Y β (cid:1) = β ( Y β ) − Y β = z − y, from which it follows that Y β = γ − β ( z − y ) and △ = y − γ − β ( z − y ). Now suppose ( y −△ , z −△ )is on the graph of β − , and let Z Y β − = z and Z Y β = z − △ . Then ( y − △ , z − △ ) = It can be checked that for x ∈ R − , γ − β ( x ) and ρ − β ( x ) is respectively the y -coordinate and the z -coordinateof the intersection of the line z = y + x and G β . β − ( z − △ ) , z − △ ), which is equivalent to ρ β (cid:0) Z Y β (cid:1) = Z Y β − β − ( Z Y β ) = z − y, and it follows that Z Y β = ρ − β ( z − y ) and △ = z − ρ − β ( z − y ). According (4.12), the number △ of shares in both of the aforementioned two cases can be expressed by △ = y − γ − β ( z − y ) = z − ρ − β ( z − y ) . On the other hand, if z ≤ β ( y ), then the strategy Y β consists of an initial waiting perioduntil (cid:0) Y βt , Z Y β t (cid:1) is on the graph of β (see Figure 1). As long as no action is taken, we have Y βt = y , and with reference to (2.8) and (2.9), we obtain Z Y β t = H − (cid:0) H ( z ) − t (cid:1) . The firsttime t w that the state process is on the graph of β is therefore given by t w = H ( z ) − H (cid:0) β ( y ) (cid:1) . (4.15)Once the state process ( Y β , Z Y β ) is in the set G β , the strategy Y β consists of selling sharesin such a way that the state process remains in G β (see Figure 1). Therefore, (cid:0) Y βt , Z Y β t (cid:1) = (cid:0) Y βt , β ( Y βt ) (cid:1) whenever β (cid:0) Y βt + (cid:1) = β (cid:0) Y βt (cid:1) . With reference to (2.5), this implies that Y βt shouldsolve dβ (cid:0) Y βt (cid:1) = − h (cid:0) β (cid:0) Y βt (cid:1)(cid:1) dt + dY βt , which is equivalent to dγ β (cid:0) Y βt (cid:1) = − h (cid:0) β (cid:0) Y βt (cid:1)(cid:1) dt. If β − (cid:0) Z Y β t (cid:1) = β − (cid:0) Z Y β t − (cid:1) , then (cid:0) Y βt , Z Y β t (cid:1) = (cid:0) β − ( Z Y β t ) , Z Y β t (cid:1) . According to (2.5) and thedefinition of β − , Z Y β should solve dZ Y β t = − h (cid:0) Z Y β t − (cid:1) dt. Set t w = , if z > β ( y ) ,H ( z ) − H (cid:0) β ( y ) (cid:1) , if z ≤ β ( y ) , (4.16)18nd ¯ t = inf { t ≥ | Y βt = 0 } . (4.17)Denote by { y n } n ∈ I the set of discontinuity points of β . Then I is countable since β is c`agl`ad.Define { t n } n ∈ I by t n = inf (cid:8) t ≥ t w | Y βt = y n (cid:9) , (4.18)and { s n } n ∈ I by s n = inf (cid:8) t ≥ t w | Y βt < y n (cid:9) . (4.19)If { t ≥ t w | Y βt = y n (cid:9) = ∅ , set t n = ∞ ; and set s n = ∞ , if { t ≥ t w | Y βt < y n (cid:9) = ∅ . Thefollowing result establish existence and uniqueness of such a strategy Y β corresponding to agiven intervention boundary β . Lemma 4.1. Let ( y, z ) ∈ D where y > , and let β be an intervention boundary function. Let H , β − , γ β , γ − β , t w , ¯ t , y n , t n and s n be given by (2.7), (4.9), (4.7), (4.10) and (4.16)–(4.19),respectively. Set (cid:0) Y βt (cid:1) t ≥ = (cid:0) Y βt ∧ ¯ t (cid:1) t ≥ , with Y β − = y , which denotes the decreasing c`adl`agliquidation strategy corresponding to β , and let (cid:0) Z Y β t (cid:1) t ≥ , with Z Y β − = z , be the state processof the bid order book associated with Y β . Suppose Y β satisfies the following description:(i) If z > β ( y ) ,(a) when y ∈ ∪ n ∈ I (cid:0) z − β ( y n ) + y n , z − β ( y n +) + y n (cid:3) , immediately sell y − γ − β ( z − y ) number of shares. This block trade ensures Y β = β − (cid:0) Z Y β (cid:1) .(b) when y ∈ (cid:0) z, ∞ (cid:1) \ ∪ n ∈ I (cid:0) z − β ( y n ) + y n , z − β ( y n +) + y n (cid:3) , immediately sell y − γ − β ( z − y ) number of shares. This block trade ensures Z Y β = β (cid:0) Y β (cid:1) .Then continuously sell shares so that (cid:0) Y βt , Z Y β t (cid:1) ∈ G β for all t ∈ [ t w , ¯ t ] .(ii) If z ≤ β ( y ) , then wait until time t w . The time t w has the property that Z Y β t w = β ( y ) .Continuously sell shares so that (cid:0) Y βt , Z Y β t (cid:1) ∈ G β for all t ∈ [ t w , ¯ t ] .Such a strategy Y β exists and is unique, and it is continuous for all t > . In particular, Y βt = y n for t ∈ [ t w , ¯ t ] ∩ ∪ n ∈ I [ t n , s n ) , (4.20)19 ith corresponding Z Y β t being the unique solution to dZ Y β t = − h (cid:0) Z Y β t (cid:1) dt, (4.21) where Z Y β t w = ρ − β ( z − y ) if z > β ( y ) , and Z Y β t n = β (cid:0) Y βt n − (cid:1) for t n > t w . (4.22) Moreover, Z Y β t = β (cid:0) Y βt (cid:1) , for t ∈ [ t w , ¯ t ] \ ∪ n ∈ I [ t n , s n ) , (4.23) where Y β is the unique solution to dγ β (cid:0) Y βt (cid:1) = − h (cid:0) β (cid:0) Y βt (cid:1)(cid:1) dt, (4.24) with Y βt w = y if z ≤ β ( y ) , Y βt w = γ − β ( z − y ) if z > β ( y ) , and Y βs n = y n for s n > t w . (4.25) If t w > , then Y βt = y and Z Y β t = H − (cid:0) H ( z ) − t (cid:1) , for ≤ t ≤ t w . We can also describe Z Y β t for t ∈ [¯ t, ∞ ) since it satisfies (4.21) with initial condition Z Y β ¯ t = Z Y β t w , if ¯ t = t w ,z, if ¯ t < t w ,β (0+) , if ¯ t > t w . (4.26)The value β (0+) can then be used to determine whether the liquidation period is finite ornot. More specifically, we have that β (0+) < t < ∞ . To see this, it is enough toconsider γ β (cid:0) Y β ¯ t (cid:1) − γ β (cid:0) Y βt (cid:1) = Z ¯ tt − h (cid:0) β (cid:0) Y βu (cid:1)(cid:1) du which follows from (4.24) when there is no waiting period between the times t and ¯ t . To geta contradiction, suppose ¯ t = ∞ . Then it is clear that R ¯ tt − h (cid:0) β (cid:0) Y βu (cid:1)(cid:1) du = ∞ , as β (cid:0) Y βu (cid:1) is20ounded away from 0 on the interval ( t, ¯ t ). However, γ β (cid:0) Y β ¯ t (cid:1) − γ β (cid:0) Y βt (cid:1) is finite, so we get acontradiction.It follows from the dynamics of Z Y β t that Z Y β is c`adl`ag and increasing to 0. Moreover,the continuity of Y βt for t > Z Y β is also continuous for all t > Y β described by Lemma 4.1 for an arbitrary intervention boundary β . Thisexpression can then later be used to derive an explicit expression for the value function of ourproblem. For the strategy Y β with associated state process Z Y β , given an initial state ( y, z ),and with reference to (3.8), we define the performance function J β by J β ( y, z ) = Z ∞ (cid:18) κ A (cid:0) Y βt (cid:1) + Ah ( Z Y β t ) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt, (4.27)where Y β − = y , Z Y β − = z and ( y, z ) ∈ D . Since κ A (0) = 0, it follows that Z ∞ ¯ t (cid:18) κ A (cid:0) Y βt (cid:1) + Ah ( Z Y β t ) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt = A Z Z Y β ¯ t ψ ( u ) du. (4.28)Therefore, J β (0 , z ) = A Z z ψ ( u ) du. (4.29) Lemma 4.2. Let β , Y β , Z Y β , t w and ¯ t be defined as in Lemma 4.1. If t w < ¯ t , then Z ∞ t w (cid:18) κ A (cid:0) Y βt (cid:1) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt = Z Z Y βtw − Y βtw β (0+) (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du + A Z β (0+)0 ψ ( u ) du, where γ − β and ρ − β are defined by (4.10) and (4.11), respectively. In case (i) (a) of Lemma 4.1, the strategy Y β consists of an initial sale of y − γ − β ( z − y ) = z − ρ − β ( z − y ) number of shares. The state after the block sale is (cid:0) Y β , Z Y β (cid:1) = (cid:0) β − (cid:0) ρ − β ( z − ) (cid:1) , ρ − β ( z − y ) (cid:1) , Hence, according to (4.27) and Lemma 4.2, J β ( y, z ) = J β (cid:16) β − (cid:0) ρ − β ( z − y ) (cid:1) , ρ − β ( z − y ) (cid:17) = Z ρ β ( ρ − β ( z − y )) β (0+) (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du + A Z β (0+)0 ψ ( u ) du = Z z − yβ (0+) (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du + A Z β (0+)0 ψ ( u ) du. In case (i) (b), we immediately sell y − γ − β ( z − y ) number of shares at the beginning. Thestate after the block sale is (cid:0) Y β , Z Y β (cid:1) = (cid:0) γ − β ( z − y ) , β (cid:0) γ − β ( z − y ) (cid:1)(cid:1) . Hence, similar to theabove calculation, we have J β ( y, z ) = J β (cid:16) γ − β ( z − y ) , β (cid:0) γ − β ( z − y ) (cid:1)(cid:17) = Z z − yβ (0+) (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du + A Z β (0+)0 ψ ( u ) du. We therefore conclude that for case (i) of Lemma 4.1, J β ( y, z ) = Z z − yβ (0+) (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du + A Z β (0+)0 ψ ( u ) du. (4.30)Moreover, in case (ii), z ≤ β ( y ). So we need to wait until time t w > Z Y β t w = β ( y ). With reference to (2.7) and (4.15), we have t w = H ( z ) − H (cid:0) β ( y ) (cid:1) = Z zβ ( y ) h ( u ) du. Also, observe that Z t w h (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1) dt = − Z t w ψ (cid:0) Z Y β t (cid:1) dZ Y β t = − Z β ( y ) z ψ ( u ) du. J β ( y, z ) = Z t w (cid:18) κ A ( y ) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt + J β (cid:0) y, β ( y ) (cid:1) = κ A ( y ) Z zβ ( y ) h ( u ) du − A Z β ( y ) z ψ ( u ) du + Z γ β ( y ) β (0+) (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du + A Z β (0+)0 ψ ( u ) du. (4.31)Although this provides an explicit expression for J β ( y, z ), it is not entirely straightforward toconclude about properties of continuity and differentiability for J β ( y, z ) in y since β is only ac`agl`ad function. However, we can further calculate that Z γ β ( y ) β (0+) (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du = Z y (cid:18) κ A ( u ) h (cid:0) β ( u ) (cid:1) + Aψ (cid:0) β ( u ) (cid:1)(cid:19) dγ cβ ( u )+ X
For y ∈ (0 , ¯ y A ) , define the function Γ( · ; y ) : [¯ z, → R by Γ( x ; y ) = Aψ ( x ) + κ A ( y ) h ( x ) + κ ′ A ( y ) H ( x ) , for x ∈ (¯ z, , (4.33) and Γ(0; y ) = lim x → Γ( x ; y ) , Γ(¯ z ; y ) = lim x → ¯ z Γ( x ; y ) . Let β ∗ = β ∗ ( y ) and β ∗ = β ∗ ( y ) denote the functions defined as the largest and smallest ∈ [¯ z, satisfying max x ∈ [¯ z, Γ( x ; y ) = Γ (cid:0) β ; y (cid:1) , (4.34) respectively. Then for all y ∈ (0 , ¯ y A ) , we have ¯ z ≤ β ∗ ( y ) ≤ β ∗ ( y ) < . Furthermore, if ¯ y A < ∞ , write β ∗ ( y ) = β ∗ ( y ) = ¯ z , for all y > ¯ y A . Set β ∗ (0) = 0 , β ∗ (0) = lim y → + β ∗ ( y ) , and β ∗ (¯ y A ) = lim y → ¯ y A − β ∗ ( y ) , β ∗ (¯ y A ) = lim y → ¯ y A + β ∗ ( y ) . This uniquely defines two decreasing functions β ∗ , β ∗ : R + → [¯ z, that are c`agl`ad and c`adl`ag,respectively, and they are the left and the right-continuous version of each other. Remark 4.4. We will see later that the optimal intervention boundary will be given by β ∗ ,and we have previously assumed that µ ≤ 0. But lets examine some of the properties that β ∗ would have if µ > 0. If µ > 0, then κ A is no longer strictly increasing, but will be strictlydecreasing on [0 , y ) and then strictly increasing on ( y, ¯ y A ), where y denotes the unique pointwhere κ A attains its minimum. In particular κ A ( y ) < κ ′ A ( y ) = 0. ThereforeΓ( x ; y ) = Aψ ( x ) + κ A ( y ) h ( x ) , which is maximised for x = 0. Thus β ∗ ( y ) = β ∗ ( y ) = 0 if µ > 0. This means that if theoptimal intervention boundary is given by β ∗ , then the optimal strategy would involve holdingon to at least y number of shares until eternity if the initial stock position is larger than y .But such a strategy is not admissible as we want to restrict the set of admissible strategies tostrategies which tend to zero sufficiently fast (we would also have to rephrase the admissibilitycondition if µ > κ A is not strictly increasing). We conclude that if µ > Lemma 4.5. Let β = β ∗ where β ∗ is as in Lemma 4.3. For ¯ z − ¯ y A < s < and ρ − β ( s ) < z < t holds that Γ (cid:0) ρ − β ( s ); γ − β ( s ) (cid:1) − Γ (cid:0) z ; γ − β ( s ) (cid:1) ≥ . Lemma 4.5 is needed for the proof of Proposition 4.7 (see in particular equation (5.20)).The result relies on the assumptions that x ν ([ x, x /h ( x ) are concave functions.If we do not impose these conditions then one can show that the strategy Y β with β = β ∗ given by (4.34) may not be optimal (compare Theorem 4.8). Thus while these concavityconditions may seem realistic, they are also needed in order to solve the problem. If theseconditions does not hold, we simply do not know what the solution to the liquidation problemlooks like. Lemma 4.6. Let β ∗ be given by Lemma 4.3, it follows that if lim x → y − κ A ( x ) = ∞ or lim x → y − κ ′ A ( x ) = ∞ , then lim x → y − β ∗ ( x ) = ¯ z . Furthermore, we have lim y → κ A ( y ) h (cid:0) β ∗ ( y ) (cid:1) = 0 and lim y → κ ′ A ( y ) H (cid:0) β ∗ ( y ) (cid:1) = 0 . (4.35)Clearly, the function β ∗ given in Lemma 4.3 satisfies the properties we require of anintervention boundary. With this intervention boundary, the proposition below provides anexplicit expression for the value function that solves (4.1)–(4.4) with associated boundarycondition v (0 , z ) = A R z ψ ( u ) du , for all z ∈ [¯ z, Proposition 4.7. Let β = β ∗ denote the largest solution to (4.34), and let γ − β and ρ − β bethe corresponding functions defined by (4.10) and (4.11). Then the function v : D → R givenby v ( y, z ) = Z z − yβ (0+) (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du + A Z β (0+)0 ψ ( u ) du (4.36) for z > β ( y ) , and v ( y, z ) = κ A ( y ) H ( z ) + A Z z ψ ( u ) du − Z y (cid:18) κ A ( u ) h (cid:0) β ( u ) (cid:1) + Aψ (cid:0) β ( u ) (cid:1) + κ ′ A ( u ) H (cid:0) β ( u ) (cid:1)(cid:19) du, (4.37)26 or z ≤ β ( y ) , is a C , ( D ) solution to (4.1)–(4.4) with the boundary condition v (0 , z ) = A Z z ψ ( u ) du, for all z ∈ [¯ z, . Moreover, D − y v ( y, z ) is c`agl`ad in y and continuous in z . The following theorem verifies that the function v given by (4.36)-(4.37) is equal to thevalue function V given by (3.8), and that the strategy Y β corresponding to β = β ∗ in Lemma4.3 is an optimal liquidation strategy. Hence, such a Y β provides a solution to the utilitymaximization problem in (2.11). Theorem 4.8. Let β denote the largest solution to (4.34), let v be given by (4.36) and (4.37),and let V be given by (3.8). Then v = V on D and sup Y ∈A ( y ) E (cid:2) U (cid:0) C ∞ ( Y ) (cid:1)(cid:3) = − exp (cid:18) − A ( c + by ) + A Z z − yz ψ ( s ) ds (cid:19) exp (cid:0) v ( y, z ) (cid:1) , where A denotes the agent’s risk aversion, b is the initial unaffected price, c is the agent’sinitial cash position, z = Z Y − is the initial state of the bid order book, and y is the agent’sinitial share position. The optimal strategy Y ∗ is equal to Y β ∈ A D ( y ) , where Y β is thestrategy described in Lemma 4.1 corresponding to β with Y β − = y . Example 1. Suppose that the the bid order book has an equal number n of orders atevery price point to a level ¯ x below the unaffected best bid price. This corresponds to x m ([ x, − n ( x ∧ ¯ x ). Therefore ψ ( z ) = n z , for ¯ z = n ¯ x < z ≤ 0, and −∞ , for z ≤ ¯ z . We calculate thatΓ ′ ( x ; y ) = An − κ A ( y ) λx + κ ′ A ( y ) λx , for n ¯ x < x < . In order to find the maximiser, we want to solve Γ ′ ( x ; y ) = 0. This amounts to solving λAn x + κ ′ A ( y ) x − κ A ( y ) = 0 . (4.38)The unique solution is x = n λA (cid:26) − κ ′ A ( y ) − r(cid:0) κ ′ A ( y ) (cid:1) + 4 λAn κ A ( y ) (cid:27) . x ; y ) is concave in x for every 0 ≤ y < ¯ y A and equation (4.38) has a unique solutionfor x ≤ 0, it follows that β ( y ) = n λA (cid:26) − κ ′ A ( y ) − r(cid:0) κ ′ A ( y ) (cid:1) + 4 λAn κ A ( y ) (cid:27) , for 0 ≤ y < ¯ y A , (4.39)provided that β ( y ) > n ¯ x .(i) ( BM case ) If L is a Brownian motion with drift µ ≤ 0, then κ A ( y ) = − Aµy + 12 A σ y , y ≥ , and hence, β ( y ) = n λA (cid:26) µA − σ A y − s µ A − (cid:18) λµA n − µσ A (cid:19) y + (cid:18) σ A + 2 λσ A n (cid:19) y (cid:27) . If µ = 0, then it is straightforward to check that the equation for the optimal interventionboundary coincides with the result obtained in the example in Løkka (2014).(ii) ( LVG case ) Consider a L´evy process L with drift µ ≤ ν ( dx ) = − η ln( x +1) ( x + 1) C + D − dx, x ∈ ( − , , η ln( x +1) ( x + 1) C − D − dx, x ∈ (0 , ∞ ) . (4.40)where C = θρ and D = q θ + ρ η ρ . (4.41)If the initial stock price is 1 (if the initial price is s = 1, one can deal with this case byreplacing the risk aversion A with ˜ A = sA ) and µ = − η ln (cid:0) − ρ η − θη (cid:1) , (4.42)then this choice of L´evy process corresponds to the (linear) L´evy process approximationof the exponential variance-gamma process with parameters ( ρ, η, θ ). We refer the reader28o Løkka and Xu (2020) for more details. The common L´evy processes considered asmodels for financial price data assume models of the exponential type, and taking L tosimply be one of these L´evy processes will in general not do a particularly good job asa model for financial price data. We then have κ A ( y ) = − Aµy + Z ( − , ∞ ) \{ } (cid:18) e − Ayz − Ayz (cid:19) ν ( dz ) ,κ ′ A ( y ) = − Aµ + A Z ( − , ∞ ) \{ } (cid:18) − e − Ayz (cid:19) z ν ( dz ) , where ν is given by (4.40). With reference to (4.39) we can calculate the optimalintervention boundary β .From Figure 2 we see that the difference between the BM case and the LVG case increase withthe agent’s share position. Moreover the difference becomes more pronounced the more riskaverse the agent is. If the agent is not particularly risk averse, then the more heavy-tailed riskin the LVG model does not make much difference to the agent, but if the agent is risk aversethen it does. However, from Figure 2 we see that while for A = 10 − the optimal interventionboundary for the LVG case and the BM case are quite different, in terms of the correspondingstrategy Y β , the initial block sale in the LVG case is about 200 shares and in the BM caseabout 150 shares, if the initial stock position is 10 . An initial block sale of 200 shares wouldeat into the order book and reduce the best bid price by , so the best bid immediatelyafter the block sale would be 0 . − = 0 . -3 A=10 -2 A=10 -2 A=10 -2.3 A=10 -2.3 Figure 2: The optimal intervention boundary β for the two cases of L being a Brownianmotion (BM case) and L being the L´evy process approximating the exponential variance-gamma (LVG case). We take n = 1000 and λ = 5. For the LVG case, we take θ = − . ρ = 0 . 02 and η = 0 . µ = − . µ = − . σ = 4 . ∗ − , which match the variance of L in the LVG case.Solid graphs correspond to the LVG case and a dotted graphs to the BM case. When A = 10 − we see that the LVG and the BM graphs are pretty much identical.30 Proofs Proof of Lemma 3.1 . With reference to the dynamic of Z Y , we calculate that for z ≥ ¯ z , Z Z YT ψ ( u ) du = Z z ψ ( u ) du + Z T ψ ( Z Yt − ) dY ct − Z T h ( Z Yt − ) ψ ( Z Yt − ) dt + X ≤ t ≤ T Z Z Yt − + △ Y t Z Yt − ψ ( u ) du = Z z ψ ( u ) du + Z T ψ ( Z Yt − ) dY ct − Z T h ( Z Yt − ) ψ ( Z Yt − ) dt + X ≤ t ≤ T Z △ Y t ψ (cid:0) Z Yt − + u (cid:1) du. Then, F T ( Y ) = Z T ψ (cid:0) Z Yt − (cid:1) dY ct + X ≤ t ≤ T Z △ Y t ψ (cid:0) Z Yt − + x (cid:1) dx = Z Z YT ψ ( u ) du − Z z ψ ( u ) du + Z T h ( Z Yt − ) ψ ( Z Yt − ) dt = Z Z YT z ψ ( u ) du + Z T h ( Z Yt − ) ψ ( Z Yt − ) dt. Notice that for any admissible liquidation strategy Y , we have that either Y and Z Y become0 at the same time or Y becomes 0 at some time s while Z Ys < 0. In the second case, for all t > s , Z Y satisfies dZ Yt = − h ( Z Yt ) dt. According to (2.8), we know that the solution to the above dynamic tends to 0, as t → ∞ .Therefore, Z Yt → 0, as t → ∞ in any case. It then follows from the above expression for F T ( Y ) that F ∞ ( Y ) = Z z ψ ( u ) du + Z ∞ h ( Z Yt − ) ψ ( Z Yt − ) dt. roof of Lemma 4.1 . We first prove that on any time interval I contained in [ t w , ¯ t ] \∪ n ∈ I [ t n , s n ), there exists a unique solution to the dynamic (4.24). On such an interval I , theprocess Y β does not cross any jump of β . Thus, in terms of the function β , we shall onlyfocus on those parts without jumps. Also, it is sufficient to consider Y starting from time 0(rather than starting at any time in [ t w , ¯ t ] \ ∪ n ∈ I [ t n , s n )). Write Y t = Y > Y k +1 t = γ − β (cid:18)n γ β ( Y ) − Z t h (cid:0) β ( Y ku ) (cid:1) du o ∧ β (0+) (cid:19) . (5.1)Let T ∈ [0 , ∞ ). Thensup ≤ t ≤ T (cid:12)(cid:12) β ( Y k +1 t ) − β ( Y kt ) (cid:12)(cid:12) = sup ≤ t ≤ T (cid:12)(cid:12)(cid:12)(cid:12)n γ β ( Y ) − Z t h (cid:0) β ( Y ku ) (cid:1) du o ∧ β (0+) − n γ β ( Y ) − Z t h (cid:0) β ( Y k − u ) (cid:1) du o ∧ β (0+)+ γ − β (cid:18)n γ β ( Y ) − Z t h (cid:0) β ( Y ku ) (cid:1) du o ∧ β (0+) (cid:19) − γ − β (cid:18)n γ β ( Y ) − Z t h (cid:0) β ( Y k − u ) (cid:1) du o ∧ β (0+) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ≤ t ≤ T (cid:12)(cid:12)(cid:12)(cid:12) n γ β ( Y ) − Z t h (cid:0) β ( Y ku ) (cid:1) du o ∧ β (0+) − n γ β ( Y ) − Z t h (cid:0) β ( Y k − u ) (cid:1) du o ∧ β (0+) (cid:12)(cid:12)(cid:12)(cid:12) ≤ ≤ t ≤ T (cid:12)(cid:12)(cid:12)(cid:12) Z t h (cid:0) β ( Y ku ) (cid:1) − h (cid:0) β ( Y k − u ) (cid:1) du (cid:12)(cid:12)(cid:12)(cid:12) ≤ L Z T (cid:12)(cid:12)(cid:12) β ( Y ku ) − β ( Y k − u ) (cid:12)(cid:12)(cid:12) du ≤ L Z T sup ≤ t ≤ u (cid:12)(cid:12)(cid:12) β ( Y kt ) − β ( Y k − t ) (cid:12)(cid:12)(cid:12) du, (5.2)where the first equality holds because when β has no jumps we have β (cid:0) γ − β ( x ) (cid:1) = x + γ − β ( x ),the first inequality is due to the triangle inequality and | γ − β ( x ) − γ − β ( y ) | ≤ | x − y | , and thethird inequality follows from the boundedness of the processes β ( Y k ) and β ( Y k − ) and thelocal Lipschitz continuity of h with a Lipschitz constant L . By induction and with referenceto (5.2), it can be shown thatsup ≤ t ≤ T (cid:12)(cid:12) β ( Y k +1 t ) − β ( Y kt ) (cid:12)(cid:12) ≤ (2 LT ) k k ! 2 (cid:12)(cid:12) β ( Y ) (cid:12)(cid:12) . k to infinity, we have that β ( Y kt ) converges uniformly on [0 , T ]. Define β t =lim k →∞ β ( Y kt ), for t ∈ [0 , T ]. Since T ∈ [0 , ∞ ) is arbitrary, it follows that β t = lim k →∞ β ( Y kt )for all t ∈ [0 , ∞ ). With reference to (5.1) and the dominated convergence theorem it followsthat, for every t ∈ [0 , ∞ ), (cid:0) Y kt (cid:1) ∞ k =0 is convergent. We define Y βt = lim k →∞ Y kt . It can bechecked that Y β decreases to 0. Then since β is continuous, we obtain β t = β ( Y βt ), for all t ∈ [0 , ∞ ). Therefore, by sending k to infinity in (5.1), since we only consider Y βt before time¯ t , we have that Y βt = γ − β (cid:18) γ β ( Y β ) − Z t h (cid:0) β ( Y βu ) (cid:1) du (cid:19) , for t ≤ ¯ t. This proves the existence of a solution to the dynamic (4.24) on any time interval containedin [ t w , ¯ t ] \ ∪ n ∈ I [ t n , s n ). For uniqueness, let’s assume that Y (1) and Y (2) satisfy (4.24), where Y (1) t = Y (2) t for 0 ≤ t ≤ t , and Y (1) t < Y (2) t for t < t < t . Then for t < t < t , Y (1) t = γ − β (cid:18) γ β ( Y (1)0 ) − Z t h (cid:0) β ( Y (1) u ) (cid:1) du (cid:19) ≥ γ − β (cid:18) γ β ( Y (2)0 ) − Z t h (cid:0) β ( Y (2) u ) (cid:1) du (cid:19) = Y (2) t , which contradicts the assumption that Y (1) t < Y (2) t for t < t < t . So uniqueness holds. Theexistence and uniqueness of a solution to the dynamic in (4.21) on any time interval containedin [ t w , ¯ t ] ∩ ∪ n ∈ I [ t n , s n ) follow from the locally Lipschitz continuity of the function h .Now let Y β and Z Y β be processes satisfying (4.20)–(4.26) with (cid:0) Y β − , Z Y β − (cid:1) = ( y, z ) ∈ D .Note that (cid:0) Y βt , Z Y β t (cid:1) ∈ G β for all t ∈ [ t w , ¯ t ]. We need to show that (2.5) is satisfied. We firstfocus on the case when t ≤ t w . Suppose z > β ( y ), i.e. t w = 0. Then in case (i) (a), Y β − Y β − = β − (cid:0) ρ − β ( z − y ) (cid:1) − y = γ − β ( z − y ) − y = (cid:0) z − y + γ − β ( z − y ) (cid:1) − z = Z Y β − Z Y β − , where we have used the identity β − (cid:0) ρ − β ( z − y ) (cid:1) = γ − β ( z − y ), which follows from (4.13) and33s valid under the condition of (i) (a). In case (i) (b), we obtain Z Y β − Z Y β − = β (cid:0) γ − β ( z − y ) (cid:1) − z = z − y + γ − β ( z − y ) − z = γ − β ( z − y ) − y = Y β − Y β − , where we have used that β (cid:0) γ − β ( z − y ) (cid:1) = ρ − β ( z − y ). Suppose z ≤ β ( y ), i.e. t w > 0. It can bechecked that Z Y β t = H − (cid:0) H ( z ) − t (cid:1) has dynamic (4.21). Because Y βt is now constant, (2.5)is satisfied. In the case when t > t w , Y βt and Z Y β t follow (4.20)–(4.26), which satisfy (2.5).We next like to prove that Y β is c`adl`ag and decreasing. Note that by the definitions of t n , s n , t w and ¯ t and (4.21), (4.24) and the first part of the proof, we have Y βt and Z Y β t arecontinuous when ( Y βt , Z Y β t ) is in each continuous part of the graph of β or β − , for t > Y βt and Z Y β t to be continuous at t n , s n and t w when t w > 0. It can also be seen that Y β and Z Y β are right continuous at t = 0. These together with the well-defined Y β − and Z Y β − imply that Y β and Z Y β are continuous for t > t = 0. That Y β decreases to 0 follows from (4.20), (4.21), (4.24), and the first part of thisproof. Finally, that Z Y β t = H − (cid:0) H ( z ) − t (cid:1) , for 0 ≤ t ≤ t w , follows from (2.8). Proof of Lemma 4.2 . Let { y n } n ∈ I be the set of all points at which the intervention bound-ary β is discontinuous. Consider a time interval [ t, s ] ⊆ [ t n , s n ), for some n ∈ I , where t n and s n are given by (4.18) and (4.19). With reference to (2.5), we note that formally, dt = − dρ β (cid:0) Z Y β t (cid:1) h (cid:0) Z Y β t (cid:1) ∀ t ∈ [ t n , s n ) , Z st (cid:18) κ A (cid:0) Y βr (cid:1) + Ah (cid:0) Z Y β r (cid:1) ψ (cid:0) Z Y β r (cid:1)(cid:19) dr = Z ts (cid:18) κ A (cid:0) β − ( Z Y β r ) (cid:1) h ( Z Y β r ) + Aψ (cid:0) Z Y β r (cid:1)(cid:19) dρ β (cid:0) Z Y β r (cid:1) = Z ρ β ( Z Y βt ) ρ β ( Z Y βs ) (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du = Z Z Y βt − Y βtn Z Y βs − Y βsn (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du, (5.3)where we have used the identity in (4.13). Similarly, since dt = − dγ β (cid:0) Y βt (cid:1) h (cid:0) β ( Y βt ) (cid:1) ∀ t ∈ [ t w , ¯ t ] \ ∪ n ∈ I [ t n , s n ) , applying (4.14), it can be calculated that on some time interval [ s, t ] ⊂ [ t w , ¯ t ] \ ∪ n ∈ I [ t n , s n ),for some n ∈ I , Z ts (cid:18) κ A (cid:0) Y βr (cid:1) + Ah (cid:0) Z Y β r (cid:1) ψ (cid:0) Z Y β r (cid:1)(cid:19) dr = Z Z Y βs − Y βs Z Y βt − Y βt (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du. (5.4)Let t w < ¯ t . Suppose the number of t n and s n in the interval [ t w , ¯ t ] is equal to m < ∞ (possibly m = 0). Consider r ≤ r < ... < r m < r m +1 , where r = t w , r m +1 = ¯ t and for k = 1 , ..., m , r k are equal to those t n , s n ∈ [ t w , ¯ t ]. We assume r , ..., r m are in an ascendingorder. Then it follows from (5.3), (5.4) and the continuity of Y βt and Z Y β t when t > Z ¯ tt w (cid:18) κ A (cid:0) Y βt (cid:1) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt = m X k =0 Z r k +1 r k (cid:18) κ A (cid:0) Y βt (cid:1) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt = Z Z Y βtw − Y βtw Z Y β ¯ t (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du. t n and s n in the interval [ t w , ¯ t ]. Let r ∈ [ t w , ¯ t ] be anaccumulation point of the sequence { t n } n ∈ I . Then without loss of generality, consider asubsequence { t n k } ∞ k =1 ⊂ [ t w , ¯ t ] increasing to r . Consider some time interval [ t, s ] in which r is the only accumulation point of { t n } n ∈ I . Then, it follows that Z st (cid:18) κ A (cid:0) Y βt (cid:1) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt = lim n →∞ Z t n t (cid:18) κ A (cid:0) Y βt (cid:1) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt + Z sr (cid:18) κ A (cid:0) Y βt (cid:1) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt = lim n →∞ Z Z Y βt − Y βt Z Y βtn − Y βtn (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du + Z Z Y βr − Y βr Z Y βs − Y βs (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du, = Z Z Y βt − Y βt Z Y βs − Y βs (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du. This implies that Z ¯ tt w (cid:18) κ A (cid:0) Y βt (cid:1) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt = Z Z Y βtw − Y βtw Z Y β ¯ t (cid:18) κ A (cid:0) γ − β ( u ) (cid:1) h (cid:0) ρ − β ( u ) (cid:1) + Aψ (cid:0) ρ − β ( u ) (cid:1)(cid:19) du. Therefore the result follows from the above equality as well as (4.26) and (4.28). Proof of Lemma 4.3 . First notice that, for any y ∈ (0 , ¯ y A ), the function Γ( x, y ) is concavein x , but that this concavity may not be strict. Observe that for y ∈ (0 , ¯ y A ),lim x → − Γ( x ; y ) = −∞ . Also, Γ( x ; y ) ∈ R , for x ∈ [¯ z, z ≤ β ∗ ( y ) ≤ β ∗ ( y ) < 0, for all0 < y < ¯ y A . The largest and smallest solution to (4.34) uniquely define the functions β ∗ and36 ∗ . For 0 < y < y + △ < ¯ y A and x ∈ [¯ z, ddx (cid:20) Γ( x ; y + △ ) − Γ( x ; y ) (cid:21) = − (cid:0) κ A ( y + △ ) − κ A ( y ) (cid:1) h ′ ( x ) h ( x ) + κ ′ A ( y + △ ) − κ ′ A ( y ) h ( x ) < , (5.5)since κ A is convex and κ ′ A ( u ) > 0, for u > 0. We want to show that β ∗ and β ∗ are decreasingfunctions. In order to get a contradiction, suppose that there exists y ∈ (0 , ¯ y A ) and △ > β ∗ ( y + △ ) > β ∗ ( y ). With reference to (5.5), we obtainΓ (cid:0) β ∗ ( y + △ ); y + △ (cid:1) − Γ (cid:0) β ∗ ( y + △ ); y (cid:1) < Γ (cid:0) β ∗ ( y ); y + △ (cid:1) − Γ (cid:0) β ∗ ( y ); y (cid:1) . However, this contradicts the definitions of β ∗ and β ∗ , which imply thatΓ (cid:0) β ∗ ( y + △ ); y + △ (cid:1) ≥ Γ (cid:0) β ∗ ( y ); y + △ (cid:1) and Γ (cid:0) β ∗ ( y ); y (cid:1) ≥ Γ (cid:0) β ∗ ( y + △ ); y (cid:1) . Therefore, for all 0 < y < ¯ y A , β ∗ ( y + △ ) ≤ β ∗ ( y + △ ) ≤ β ∗ ( y ) ≤ β ∗ ( y ) , (5.6)from which it follows that β ∗ and β ∗ are decreasing. By (4.33), we know that for ¯ z ≤ x < x ; y ) is continuous in y . Then for y ∈ (0 , ¯ y A ), we haveΓ (cid:0) β ∗ ( y +); y + (cid:1) = Γ (cid:0) β ∗ ( y +); y (cid:1) ≤ Γ (cid:0) β ∗ ( y ); y (cid:1) = Γ (cid:0) β ∗ ( y ); y + (cid:1) Γ (cid:0) β ∗ ( y − ); y (cid:1) = Γ (cid:0) β ∗ ( y − ); y − (cid:1) ≥ Γ (cid:0) β ∗ ( y ); y − (cid:1) = Γ (cid:0) β ∗ ( y ); y (cid:1) . Since β ∗ and β ∗ are defined as respectively the largest and smallest maximiser to (4.34), and β ∗ and β ∗ are decreasing, it follows that β ∗ ( y +) = β ∗ ( y ) and β ∗ ( y − ) = β ∗ ( y ). By monotonicity,the right limit of β ∗ and the left limit of β ∗ exist. Hence, we have proved that β ∗ is c`agl`adand β ∗ is c`adl`ag . The claim that β ∗ is the c`agl`ad version of β ∗ and that β ∗ is the c`adl`agversion of β ∗ follows from (5.6). Proof of Lemma 4.5 . With reference to (4.14), we have that β (cid:0) γ − β ( s ) (cid:1) = ρ − β (cid:0) γ β (cid:0) γ − β ( s ) (cid:1)(cid:1) = ρ − β ( s ) if γ β (cid:0) γ − β ( s ) (cid:1) = s. Moreover, γ β (cid:0) γ − β ( s ) (cid:1) = s , unless β has a jump at γ − β ( s ). Thus if β does not have a jump at37 − β ( s ) thenΓ (cid:0) ρ − β ( s ); γ − β ( s ) (cid:1) − Γ (cid:0) z ; γ − β ( s ) (cid:1) = Γ (cid:0) β (cid:0) γ − β ( s ) (cid:1) ; γ − β ( s ) (cid:1) − Γ (cid:0) z ; γ − β ( s ) (cid:1) ≥ , (5.7)by the definition of β . If on the other hand β has a jump at γ − β ( s ), then γ − β is flat on theinterval [ s ∗ , s ∗ ], where s ∗ = β (cid:0) γ − β ( s )+ (cid:1) − γ − β ( s ) and s ∗ = β (cid:0) γ − β ( s ) (cid:1) − γ − β ( s ). Also β (cid:0) γ − β ( s ∗ )+ (cid:1) = ρ − β (cid:0) γ β (cid:0) γ − β ( s ∗ +) (cid:1)(cid:1) = ρ − β ( s ∗ )and β (cid:0) γ − β ( s ∗ ) (cid:1) = ρ − β (cid:0) γ β (cid:0) γ − β ( s ∗ ) (cid:1)(cid:1) = ρ − β ( s ∗ ) . In particular, β (cid:0) γ − β ( s ∗ )+ (cid:1) = ρ − β ( s ∗ ) ≤ ρ − β ( s ) ≤ ρ − β ( s ∗ ) = β (cid:0) γ − β ( s ∗ ) (cid:1) (5.8)and so Γ( ρ − β ( s ∗ ) , γ − β ( s ) (cid:1) = Γ( ρ − β ( s ∗ ) , γ − β ( s ) (cid:1) ≥ Γ( z, γ − β ( s ) (cid:1) , by the definition of β . With reference to Assumption 2.5 and Assumption 2.6, it followsthat x Γ( x ; y ) is concave. According to (5.8), there exists λ ∈ [0 , 1] such that ρ − β ( s ) = λρ − β ( s ∗ ) + (1 − λ ) ρ − β ( s ∗ ). Hence,Γ (cid:0) ρ − β ( s ); γ − β ( s ) (cid:1) − Γ (cid:0) z ; γ − β ( s ) (cid:1) ≥ Γ (cid:0) β (cid:0) γ − β ( s ) (cid:1) ; γ − β ( s ) (cid:1) − Γ (cid:0) z ; γ − β ( s ) (cid:1) ≥ . (5.9) Proof of Lemma 4.6 . If y > ¯ y A , then by the definition of β ∗ , it holds that if lim x → y − κ A ( x ) = ∞ or lim x → y − κ ′ A ( x ) = ∞ , then lim x → y − β ∗ ( x ) = ¯ z . The remaining case is when y = ¯ y A .We will prove this case by contradiction. Suppose β ∗ (¯ y A ) > ¯ z . For any x ∈ (cid:0) ¯ z, β ∗ (¯ y A ) (cid:1) and38 ∈ (0 , ¯ y A ) such that β ∗ ( y ) ≥ β ∗ (¯ y A ), we have Aψ ( x ) ≤ A (cid:18) ψ ( x ) − ψ (cid:0) β ∗ ( y ) (cid:1)(cid:19) ≤ κ A ( y ) (cid:18) h (cid:0) β ∗ ( y ) (cid:1) − h ( x ) (cid:19) + κ ′ A ( y ) (cid:18) H (cid:0) β ∗ ( y ) (cid:1) − H ( x ) (cid:19) ≤ κ A ( y ) (cid:18) h (cid:0) β ∗ (¯ y A ) (cid:1) − h ( x ) (cid:19) + κ ′ A ( y ) (cid:18) H (cid:0) β ∗ (¯ y A ) (cid:1) − H ( x ) (cid:19) . Taking y to be arbitrarily close to ¯ y A implies ψ ( x ) = −∞ . This means x < ¯ z , which contradicts x > ¯ z . Hence, we conclude that β ∗ (¯ y A ) = −∞ .Next we prove (4.35). Observe that if β ∗ (0+) < 0, then (4.35) is true. However, if β ∗ (0+) = 0, then κ A ( y ) h (cid:0) β ∗ ( y ) (cid:1) ≥ Γ( x ; y ) − Aψ (cid:0) β ∗ ( y ) (cid:1) − κ ′ A ( y ) H (cid:0) β ∗ ( y ) (cid:1) ≥ Γ( x ; y ) − κ ′ A ( y ) H (cid:0) β ∗ ( y ) (cid:1) , from which it follows that for any x ∈ (¯ z, ≥ lim inf y → κ A ( y ) h (cid:0) β ∗ ( y ) (cid:1) ≥ Aψ ( x ) − lim sup y → κ ′ A ( y ) H (cid:0) β ∗ ( y ) (cid:1) , (5.10)0 ≥ lim sup y → κ A ( y ) h (cid:0) β ∗ ( y ) (cid:1) ≥ Aψ ( x ) − lim inf y → κ ′ A ( y ) H (cid:0) β ∗ ( y ) (cid:1) . (5.11)Therefore, 0 ≥ lim sup y → κ ′ A ( y ) H (cid:0) β ∗ ( y ) (cid:1) ≥ Aψ ( x ) , ≥ lim inf y → κ ′ A ( y ) H (cid:0) β ∗ ( y ) (cid:1) ≥ Aψ ( x ) . By letting x tend to 0, then with reference to (2.3), we get lim y → κ ′ A ( y ) H (cid:0) β ∗ ( y ) (cid:1) = 0. Also,by letting x tend to 0 in (5.10) and (5.11), lim y → κ A ( y ) h ( β ∗ ( y )) = 0 follows. Proof of Proposition 4.7 . To show that v is continuous, we first prove it is finite. Withreference to (4.28)-(4.32), it is sufficent to show that the function J β given by (4.27) is finitefor β defined by Lemma 4.3. By the continuity of Y β and Z Y β after time 0 and condition392.4), we have that there exists some s > Z s (cid:18) κ A (cid:0) Y βt (cid:1) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt < ∞ (5.12)and Y βs < ¯ y A . According to the condition in Lemma 4.6,lim y → + κ A ( y ) h (cid:0) β ( y ) (cid:1) = 0 , so it follows that there exists C > < ǫ < ¯ y A such that κ A ( y ) ≤ − C h (cid:0) β ( y ) (cid:1) , for all y ∈ [0 , ǫ ] . Since ψ (cid:0) Z Y β t (cid:1) is bounded for all t ≥ s (it increases to 0), this and the above inequality implythat Z ∞ s (cid:18) κ A (cid:0) Y βt (cid:1) + Ah (cid:0) Z Y β t (cid:1) ψ (cid:0) Z Y β t (cid:1)(cid:19) dt ≤ Z ∞ s (cid:18) − C h (cid:0) β ( Y βt ) (cid:1) − C h (cid:0) Z Y β t (cid:1)(cid:19) dt ≤ Z ∞ s (cid:18) − C h (cid:0) Z Y β t (cid:1) − C h (cid:0) Z Y β t (cid:1)(cid:19) dt ≤ ( C + C ) (cid:0) Y βs − Z Y β s (cid:1) < ∞ , (5.13)where C > v is finite.Note that both of the expressions (4.36) and (4.37) are continuous in y and z . It istherefore sufficient to prove that v is continuous across G β . Let J u ( y, z ) denote the expressionfor v ( y, z ) given by (4.36), and let J l ( y, z ) denote the expression in (4.37). Suppose ( y, z ) isa point on the graph of β , i.e, z = β ( y ). Consider a sequence of points ( y n , z n ) ∞ n =1 containedin S β \ G β , converging to ( y, z ). With reference to (4.31) and(4.32), we calculate thatlim n →∞ v ( y n , z n ) = J u (cid:0) y, β ( y ) (cid:1) = J l (cid:0) y, β ( y ) (cid:1) = v (cid:0) y, β ( y ) (cid:1) . (5.14)If ( y, z ) lies on the graph of β − , i.e, y = β − ( z ), then using the property that β − ( u ) = β − ( z ), for u ∈ (cid:0) z, β ( β − ( z )) (cid:1) , direct calculation results in (5.14). We therefore conclude40hat v is a continuous function. Differentiating v gives D − y v ( y, z ) = − κ A (cid:0) γ − β ( z − y ) (cid:1) h (cid:0) ρ − β ( z − y ) (cid:1) − Aψ (cid:0) ρ − β ( z − y ) (cid:1) , z > β ( y ) , (5.15) v z ( y, z ) = κ A (cid:0) γ − β ( z − y ) (cid:1) h (cid:0) ρ − β ( z − y ) (cid:1) + Aψ (cid:0) ρ − β ( z − y ) (cid:1) , z > β ( y ) , (5.16) D − y v ( y, z ) = κ ′ A ( y ) H ( z ) − κ A ( y ) h (cid:0) β ( y ) (cid:1) − Aψ (cid:0) β ( y ) (cid:1) − κ ′ A ( y ) H (cid:0) β ( y ) (cid:1) , z ≤ β ( y ) , (5.17) v z ( y, z ) = κ A ( y ) h ( z ) + Aψ ( z ) , z ≤ β ( y ) . (5.18)These expressions are left-continuous with right limit in y and continuous in z (all of theseexpressions are continuous at (0 , y n , z n ) ∞ n =1 ⊆ S β , ( y, z ) ∈ G β and lim n →∞ ( y n , z n ) = ( y, z ), we have v z ( y n , z n ) → v z ( y, z ),as n → ∞ . Further, lim z → β ( y )+ D − y v ( y, z ) = D − y v (cid:0) y, β ( y ) (cid:1) . Therefore, we conclude that v z ( y, z ) is continuous, and D − y v ( y, z ) is c`agl`ad in y and continuous in z .Standard calculations show that v satisfies (4.1) and (4.3). When z = 0, (4.2) is clearlytrue. In order to verify (4.2) for z = 0, we compute that when z > β ( y ), h ( z ) v z ( y, z ) − κ A ( y ) − Ah ( z ) ψ ( z )= h ( z ) (cid:26) κ A (cid:0) γ − β ( s ) (cid:1) h (cid:0) ρ − β ( s ) (cid:1) − κ A ( z − s ) h ( z ) + A (cid:8) ψ (cid:0) ρ − β ( s ) (cid:1) − ψ ( z ) (cid:9)(cid:27) , (5.19)where s = z − y . Observe that h (cid:0) ρ − β ( s ) (cid:1) = 0 implies y = 0, but (4.1)–(4.4) are under thecondition that y > 0. So h (cid:0) ρ − β ( s ) (cid:1) is non-zero. By the definition of γ − β , we must have γ − β ( s ) ∈ (0 , ¯ y A ) if β (¯ y A ) = ¯ z , or γ − β ( s ) ∈ (0 , ¯ y A ] if β (¯ y A ) > ¯ z . Then according to the limitingbehaviour of β in Lemma 4.6, κ A (cid:0) γ − β ( s ) (cid:1) must be finite, and therefore also κ ′ A (cid:0) γ − β ( s ) (cid:1) .However, κ A ( z − s ) may be infinite, but then it follows that (5.19) is negative. Otherwise, if κ A ( y ) < ∞ , write G ( s ; z ) = κ A (cid:0) γ − β ( s ) (cid:1) h (cid:0) ρ − β ( s ) (cid:1) − κ A ( z − s ) h ( z ) + A (cid:8) ψ (cid:0) ρ − β ( s ) (cid:1) − ψ ( z ) (cid:9) . Then in order to verify (4.2), it is sufficient to show that G ( s ; z ) ≥ 0, for all ρ − β ( s ) < z < G ( s ; y ) can be expressed by G ( s ; z ) = h Γ (cid:0) ρ − β ( s ); γ − β ( s ) (cid:1) − Γ (cid:0) z ; γ − β ( s ) (cid:1)i − κ ′ A (cid:0) γ − β ( s ) (cid:1)h H (cid:0) ρ − β ( s ) (cid:1)(cid:1) − H ( z ) i + 1 h ( z ) (cid:20) κ A (cid:0) γ − β ( s ) (cid:1) − κ A ( z − s ) (cid:21) , (5.20)According to Lemma 4.5, Γ (cid:0) ρ − β ( s ); γ − β ( s ) (cid:1) − Γ (cid:0) z ; γ − β ( s ) (cid:1) ≥ . Furthermore, we calculate that1 h ( z ) h κ A (cid:0) γ − β ( s ) (cid:1) − κ A ( z − s ) i − κ ′ A (cid:0) γ − β ( s ) (cid:1)h H (cid:0) ρ − β ( s ) (cid:1) − H ( z ) i = Z zρ − β ( s ) (cid:2) κ A ( u − s ) − κ A (cid:0) ρ − β ( s ) − s (cid:1)(cid:3) h ′ ( u ) h ( u ) + κ ′ A (cid:0) ρ − β ( s ) − s (cid:1) − κ ′ A ( u − s ) h ( u ) ! du ≥ . (5.21)(4.2) then follows from (5.20)–(5.21). Moreover, from the definition of β we get D − y v ( y, z ) + v z ( y, z ) = κ ′ A ( y ) H ( z ) − κ A ( y ) h (cid:0) β ( y ) (cid:1) + Aψ ( z − y ) − Aψ (cid:0) β ( y ) (cid:1) − κ ′ A ( y ) H (cid:0) β ( y ) (cid:1) + κ A ( y ) h ( z ) + Aψ ( z ) − Aψ ( z − y )= Γ( z ; y ) − Γ (cid:0) β ( y ); y (cid:1) ≤ , which verifies (4.4).Finally, the expression in (4.36) satisfies the boundary condition since, for any u ∈ [ β (0+) , z ], we have γ − β ( u ) = 0 and ρ − β ( u ) = u . The expression in (4.37) clearly satisfiesthe boundary condition. Proof of Theorem 4.8 . Let δ be a positive-valued C ∞ ( R ) function with support on [0 , R δ ( x ) dx = 1, and define a sequence of functions { δ n } ∞ n =1 by δ n ( s ) = n δ ( ns ) , s ≥ . 42e mollify v to obtain a sequence of functions { v ( n ) } ∞ n =1 which are given by v ( n ) ( y, z ) = Z v ( y − s, z ) δ n ( s ) ds. (One may extend the lower bound of the domain of v ( · , z ) properly so that v ( n ) is well-definedat y = 0.) Then v ( n ) ∈ C , ( D ), for all n ∈ N , and v ( y, z ) = lim n →∞ v ( n ) ( y, z ) ,v z ( y, z ) = lim n →∞ v ( n ) z ( y, z ) ,D − y v ( y, z ) = lim n →∞ v ( n ) y ( y, z ) , where the last equality is due to D − y v ( y, z ) being c`agl`ad in y . Moreover, for every ( y , z ) ∈ D there exists a K > (cid:8) ( y, z ) ∈ D (cid:12)(cid:12) z ≥ y + z − y (cid:9) , v ( n ) ( y, z ) ≤ K, n ∈ N , (5.22) v ( n ) y ( y, z ) ≤ K, n ∈ N , (5.23) v ( n ) z ( y, z ) ≤ K, n ∈ N . (5.24)(If Y is admissible and ( Y − , Z Y − ) = ( y , z ), then ( Y t , Z Yt ) ∈ (cid:8) ( y, z ) ∈ D (cid:12)(cid:12) z ≥ y + z − y (cid:9) ,for all t ≥ v ( n ) (cid:0) Y T , Z YT (cid:1) + Z T (cid:18) κ A (cid:0) Y t − (cid:1) + Ah (cid:0) Z Yt − (cid:1) ψ (cid:0) Z Yt − (cid:1)(cid:19) dt = v ( n ) ( y, z ) + Z T (cid:18) v ( n ) y ( Y t − , Z Yt − ) + v ( n ) z ( Y t − , Z Yt − ) (cid:19) dY ct + Z T (cid:18) κ A (cid:0) Y t − (cid:1) + Ah (cid:0) Z Yt − (cid:1) ψ (cid:0) Z Yt − (cid:1) − v ( n ) z (cid:0) Y t − , Z Yt − (cid:1) h (cid:0) Z Yt − (cid:1)(cid:19) dt + X ≤ t ≤ T (cid:26) v ( n ) (cid:0) Y t − + △ Y t , Z Yt − + △ Y t (cid:1) − v ( n ) (cid:0) Y t − , Z Yt − (cid:1)(cid:27) , (5.25)for all Y ∈ A D ( y ). Observe that for t ≥ ≤ − Z t h (cid:0) Z Yu (cid:1) du = Z Yt − Y t − Z Y + Y ≤ y − z. Z ∞ sup n ∈ N v ( n ) z (cid:0) Y t − , Z Yt − (cid:1) h (cid:0) Z Yt − (cid:1) dt ≤ K ( y − z ) . Similarly, Z ∞ sup n ∈ N v ( n ) y (cid:0) Y t − , Z Yt − (cid:1) + v ( n ) z (cid:0) Y t − , Z Yt − (cid:1) d ( − Y ct ) ≤ Ky and X ≤ t sup n ∈ N v ( n ) (cid:0) Y t − + △ Y t , Z Yt − + △ Y t (cid:1) − v ( n ) (cid:0) Y t − , Z Yt − (cid:1) ≤ Ky. Hence, by (5.25) and the boundary condition v (0 , z ) = A R z ψ ( u ) du , it follows from thedominated convergence theorem that for any Y ∈ A D ( y ), Z ∞ (cid:18) κ A (cid:0) Y t − (cid:1) + Ah (cid:0) Z Yt − (cid:1) ψ (cid:0) Z Yt − (cid:1)(cid:19) dt = v ( y, z ) + Z ∞ (cid:18) D − y v (cid:0) Y t − , Z Yt − (cid:1) + v z (cid:0) Y t − , Z Yt − (cid:1)(cid:19) dY ct + Z ∞ (cid:18) κ A (cid:0) Y t − (cid:1) + Ah (cid:0) Z Yt − (cid:1) ψ (cid:0) Z Yt − (cid:1) − v z (cid:0) Y t − , Z Yt − (cid:1) h (cid:0) Z Yt − (cid:1)(cid:19) dt + X t ≥ (cid:26) v (cid:0) Y t − + △ Y t , Z Yt − + △ Y t (cid:1) − v (cid:0) Y t − , Z Yt − (cid:1)(cid:27) , (5.26)as n → ∞ and T → ∞ . According to Proposition 4.7, v satisfies (4.1)–(4.4), and therefore, Z ∞ (cid:18) κ A (cid:0) Y t − (cid:1) + Ah (cid:0) Z Yt − (cid:1) ψ (cid:0) Z Yt − (cid:1)(cid:19) dt ≥ v ( y, z ) . (5.27)Hence, V ≥ v .From from (5.12)-(5.13), we know that with β being the largest solution to (4.34) and Y β being the strategy described in Lemma 4.1 corresponding to β , Y β is admissible, in particular(2.2) is satisfied. Therefore, with reference to (5.27), in order to complete the proof, we needto show that (5.27) holds with equality for Y β . Observe that △ Y β < t = 0 and z > β ( y ). But by (4.1) and Proposition 4.7, we have that D − y v ( y, z ) + v z ( y, z ) = 0, for44 > β ( y ). Therefore, X t ≥ (cid:26) v (cid:0) Y βt − + △ Y βt , Z Y β t − + △ Y βt (cid:1) − v (cid:0) Y βt − , Z Y β t − (cid:1)(cid:27) = 0 . For any z ≤ 0, if 0 ≤ t ≤ t w , where t w is defined by (4.16), then d (cid:0) Y βt (cid:1) c = 0, hence Z t w (cid:18) D − y v (cid:0) Y βt − , Z Y βt − (cid:1) + v z (cid:0) Y βt − , Z Y β t − (cid:1)(cid:19) d (cid:0) Y βt (cid:1) c = 0;if t > t w , then (cid:0) Y βt , Z Y β t (cid:1) ∈ G β , which implies Z ∞ t w (cid:18) D − y v (cid:0) Y βt − , Z Y β t − (cid:1) + v z (cid:0) Y βt − , Z Y β t − (cid:1)(cid:19) d ( Y βt ) c = 0 . Finally we have Z ∞ (cid:18) κ A (cid:0) Y βt − (cid:1) + Ah (cid:0) Z Y β t − (cid:1) ψ (cid:0) Z Y β t − (cid:1) − v z (cid:0) Y βt − , Z Y β t − (cid:1) h (cid:0) Z Y β t − (cid:1)(cid:19) dt = 0 , since the integrand is equal to 0, for all (cid:0) Y βt , Z Y β t (cid:1) ∈ W β , and the Lebesgue measure of the setof t ≥ (cid:0) Y βt , Z Y β t (cid:1) ∈ S βs \ G β is 0. 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