Global Closed-form Approximation of Free Boundary for Optimal Investment Stopping Problems
aa r X i v : . [ q -f i n . M F ] O c t Global Closed-form Approximation of Free Boundary for OptimalInvestment Stopping Problems
Jingtang Ma ∗ , Jie Xing † and Harry Zheng ‡ Abstract
In this paper we study a utility maximization problem with both optimal control and opti-mal stopping in a finite time horizon. The value function can be characterized by a variationalequation that involves a free boundary problem of a fully nonlinear partial differential equation.Using the dual control method, we derive the asymptotic properties of the dual value functionand the associated dual free boundary for a class of utility functions, including power and non-HARA utilities. We construct a global closed-form approximation to the dual free boundary,which greatly reduces the computational cost. Using the duality relation, we find the approx-imate formulas for the optimal value function, trading strategy, and exercise boundary for theoptimal investment stopping problem. Numerical examples show the approximation is robust,accurate and fast.
Keywords:
Optimal investment stopping problem, dual control method, free boundary, globalclosed-form approximation.
There has been extensive research in utility maximization. Two main approaches are stochasticcontrol (dynamic programming, HJB equation) and convex duality (static optimization, martin-gale representation). For excellent expositions of these two methods in utility maximization, seeFleming and Soner (1993), Karatzas and Shreve (1998), Pham (2009), and the references therein.A variant of utility maximization of terminal wealth is that investors may stop the investmentbefore or at the maturity to achieve the overall maximum of the expected utility, which naturallyleads to a mixed optimal control and stopping problem. The early work on this line includesKaratzas and Wang (2000) and Dayanik and Karatzas (2003) for properties of the value functionat the initial time, Ceci and Bassan (2004) for existence of viscosity solution of the variationalequation, Henderson and Hobson (2008) for equivalence of the value function in the presence of ∗ School of Economic Mathematics, Southwestern University of Finance and Economics, Chengdu, 611130, P.R.China (Email: [email protected]). The work was supported by National Natural Science Foundation of China (GrantNo. 11671323), Program for New Century Excellent Talents in University (China Grant No. NCET-12-0922) andthe Fundamental Research Funds for the Central Universities, P.R. China (JBK1805001). † School of Economic Mathematics, Southwestern University of Finance and Economics, Chengdu, 611130, P.R.China (Email: [email protected]). ‡ Corresponding author. Department of Mathematics, Imperial College, London SW7 2BZ, UK (Email:[email protected]).
We consider a complete market equipped with a probability space (Ω , F , P ) together with anatural filtration ( F t ) generated by a standard Brownian motion W , satisfying the usual conditions.It consists of one riskless savings account with interest rate r > dS t = µS t dt + σS t dW t , where µ > σ > X t ) ≤ t ≤ T denote the wealth process and π t the amount of wealth an investor holds in riskyasset at time t . With continuous self-financing strategy, the wealth process ( X t ) ≤ t ≤ T evolves as dX t = rX t dt + σπ t ( θdt + dW t ) , where θ = µ − rσ is the market price of risk and ( π t ) ≤ t ≤ T the portfolio process that is F t -progressivelymeasurable and satisfies E [ R T | π t | dt ] < ∞ .The optimal investment stopping problem is given bysup π,τ E h e − βτ U ( X ,x,πτ − K ) i , where U is a utility function, τ ∈ [0 , T ] is an F t -adapted stopping time, β > K > K = 0 then the problem is a standardutility maximization with investment and stopping. It turns out that K > K = 0 wouldlead to completely different optimal trading strategies for non-HARA utility, which indicates thatone cannot simply change a portfolio insurance problem into a standard utility maximization bysetting K = 0 to get a seemingly simplified and equivalent problem, see Examples 3.4 and 3.6 fordetailed discussions. Assumption 2.1. U ∈ C is an increasing and strictly concave function on [0 , ∞ ) , satisfying U (0) = 0 , U ( ∞ ) = ∞ , U ′ (0) = ∞ , U ′ ( ∞ ) = 0 , U ( x ) < C (1+ x p ) for x ≥ , where C > , < p < are constants, and U ( x ) = −∞ for x < . Define the value function as V ( t, x ) = sup τ,π E h e − β ( τ − t ) U ( X t,x,πτ − K ) | X t = x i for ( t, x ) ∈ (0 , T ) × ( K, ∞ ). Then V satisfies the following HJB variational equation (see Guan et al.(2017)): min (cid:26) − ∂V∂t − sup π L π [ V ] , V − U ( x − K ) (cid:27) = 0 (2.1)3or ( t, x ) ∈ (0 , T ) × ( K, ∞ ), where L π [ V ] = rxV x − βV + π ( µ − r ) V x + 12 π σ V xx ,V x denotes ∂∂x V ( t, x ), V and V xx , are defined similarly. The boundary and terminal conditions aregiven by V ( t, K ) = 0 , t ∈ (0 , T ) , V ( T, x ) = U ( x − K ) , x ∈ ( K, ∞ ) . (2.2)Suppose that V ( t, · ) is strictly concave, then the maximum of L π [ V ] is achieved at π ∗ t = − θσ V x V xx , (2.3)and (2.1) is equivalent tomin (cid:26) − ∂V∂t + θ V x V xx − rxV x + βV, V − U ( x − K ) (cid:27) = 0 (2.4)for ( t, x ) ∈ (0 , T ) × ( K, ∞ ).We use the dual method to solve the variational equation (2.4). The dual function of U ( · − K )is defined by ˜ U K ( y ) := sup x>K [ U ( x − K ) − xy ] = ˜ U ( y ) − Ky, y > , where ˜ U is the dual function of U . It is easy to check that ˜ U K is continuously differentiable,decreasing, strictly convex, ˜ U K (0) = ∞ and − Ky ≤ ˜ U K ( y ) ≤ ˜ C + ˜ Cy pp − − Ky, (2.5)where ˜ C = max n C, ( Cp ) p − [ p − − o .Define the dual value function as˜ V ( t, y ) = sup t ≤ τ ≤ T E h e − β ( τ − t ) ˜ U K ( Y τ ) | Y t = y i , where ( Y t ) ≤ t ≤ T is a dual process satisfying the SDE dY t = ( β − r ) Y t dt − θY t dW t . (2.6)Then the dual value function satisfies the following variational equation (see Guan et al. (2017)):min ( − ∂ ˜ V∂t − θ y ˜ V yy − ( β − r ) y ˜ V y + β ˜ V , ˜ V − ˜ U K ) = 0 (2.7)for ( t, y ) ∈ (0 , T ) × (0 , ∞ ), with the terminal condition given by˜ V ( T, y ) = ˜ U K ( y ) , y ∈ (0 , ∞ ) . Define z = log y, τ = θ T − t ) , v ( τ, z ) = ˜ V ( t, y ) . v satisfies the following variational equation:min { L [ v ] , v − g } = 0 (2.8)for ( τ, z ) ∈ Ω T := (0 , θ T / × R , with the initial condition given by v (0 , z ) = g ( z ) for z ∈ R ,where L [ v ] = v τ − v zz + κv z + ρv, g ( z ) = ˜ U K ( e z ) , (2.9)and constants ν, ρ, κ are defined by ν = 2 rθ , ρ = 2 βθ , κ = ν − ρ + 1 . The next result shows the existence of a unique solution of the variational equation (2.8) withmonotonicity properties for each variable. Denote by W , p (Ω T ) the Sobolev space and W , p,loc (Ω T )the local Sobolev space defined by W , p,loc (Ω T ) := { v ∈ W , p ( Q ) , ∀ Q ⊂⊂ Ω T } . Theorem 2.2.
Problem (2.8) has a unique solution v ∈ C ( ¯Ω T ) ∩ W , p,loc (Ω T ) for < p < + ∞ ,satisfying g ( z ) ≤ v ( τ, z ) ≤ ˜ C ( e Bτ + pp − z + 1) , ( τ, z ) ∈ Ω T , (2.10) where B = | ( pp − ) − κ pp − − ρ | + 1 and ˜ C = max n C, ( Cp ) p − [1 /p − o . Furthermore, v satisfies v z ≤ , − v z + v zz > , v τ ≥ , ( τ, z ) ∈ Ω T . (2.11) Proof.
See Appendix.Since ˜ V ( t, y ) = v ( τ, z ), using Theorem 2.2, we can easily derive the corresponding results for ˜ V . Corollary 2.3.
Problem (2.7) has a unique solution ˜ V ∈ C ([0 , T ] × (0 , ∞ )) ∩ W , p,loc ([0 , T ) × (0 , ∞ )) for < p < + ∞ , satisfying ˜ U K ( y ) ≤ ˜ V ( t, y ) ≤ ˜ C ( e B ( T − t ) y pp − + 1) , ( t, y ) ∈ [0 , T ) × (0 , ∞ ) , where B = Bθ / and B, ˜ C are given in Theorem 2.2. Furthermore, ˜ V is decreasing in t anddecreasing and strictly convex in y , satisfying lim y → − ˜ V y ( t, y ) = + ∞ , lim y →∞ − ˜ V y ( t, y ) := a ≤ K, t ∈ (0 , T ) . (2.12) Proof.
See Section 5.
Remark 2.4.
We can easily find a strong solution V to the variational HJB equation (2.4) withconditions (2.2) by defining V ( t, x ) = inf y> [ ˜ V ( t, y ) + xy ] (2.13) for t ∈ (0 , T ) and x ∈ ( K, ∞ ) , and V is strictly increasing and strictly concave in x , see Jian et al.(2014) for details. Main Results
In this section, we consider the dual utility function of the form˜ U K ( y ) = J X j =1 − q j y q j − Ky, (3.1)where q < q < · · · < q J < Example 3.1. If J = 1 and q = γγ − with < γ < , then ˜ U ( y ) is the dual function of thepower utility U ( x ) = γ x γ . If J = 2 and q = − , q = − , then ˜ U ( y ) is the dual function of thenon-HARA utility U ( x ) = 13 H − ( x ) + H − ( x ) + xH ( x ) , where H ( x ) = ( − √ x ) / , see Bian and Zheng (2015). Define φ := L [ g ], where L is defined in (2.9). Direct computation gives φ ( z ) = L [ g ]( z ) = J X j =1 A j e q j z − νKe z , (3.2)where A j = q j − κ − ρ/q j . Note that A < A < · · · < A J .Define the continuation region in z -coordinate to be C z := { ( τ, z ); v ( τ, z ) > g ( z ) , < τ ≤ θ T / } and the exercise region to be S z := { ( τ, z ); v ( τ, z ) = g ( z ) , < τ ≤ θ T / } . We need thefollowing assumption for our main results. Assumption 3.2.
The parameters of the model satisfy
K > and A > . Now we can prove the existence of the free boundary.
Theorem 3.3.
Let Assumption 3.2 hold. Then there exists a unique free boundary z ( τ ) defined by z ( τ ) := inf { z ; v ( τ, z ) > g ( z ) } , < τ ≤ θ T / . (3.3) such that the continuation region C z and the exercise region S z can be written respectively as C z = (cid:8) ( τ, z ); z > z ( τ ) , < τ ≤ θ T / (cid:9) (3.4) and S z = (cid:8) ( τ, z ); z ≤ z ( τ ) , < τ ≤ θ T / (cid:9) . (3.5) Proof.
See Section 5.
Example 3.4.
In this example, we consider non-HARA utility ( J = 2 , q = − , q = − in (3.1))for K > . Since A < A , we discuss the following three cases.Case 1: A ≥ . There exists a unique free boundary z ( τ ) defined by (3.3). ase 2: A < < A and A + 4 A νK > . There exist two free boundaries z ( τ ) and z ( τ ) defined by z ( τ ) := inf { z ; v ( τ, z ) = g ( z ) } , < τ ≤ θ T / , (3.6) and z ( τ ) := sup { z ; v ( τ, z ) = g ( z ) } , < τ ≤ θ T / , (3.7) such that the continuation region and the exercise region are given by C z = (cid:8) ( τ, z ); z < z ( τ ) or z > z ( τ ) , < τ ≤ θ T / (cid:9) (3.8) and S z = (cid:8) ( τ, z ); z ( τ ) ≤ z ≤ z ( τ ) , < τ ≤ θ T / (cid:9) . (3.9) Moreover, z ( τ ) is increasing and z ( τ ) decreasing with limits lim τ → z ( τ ) = −
12 log − A − p A + 4 A Kν A , (3.10) and lim τ → z ( τ ) = −
12 log − A + p A + 4 A Kν A . (3.11) Case 3: A ≤ or A < < A and A + 4 A νK ≤ . There is no free boundary and it isnot optimal to stop before the maturity.Since the proof is slightly technical, we leave it in Section 5. Figure 1 (a) - (c) illustrates the threecases discussed above with C z the continuation region and S z the exercise region. Remark 3.5.
For Example 3.4, simple algebra shows that A i = a i β + b i , i = 1 , , where a =8 / (3 θ ) , a = 4 /θ , b = − r/θ ) , b = − r/θ ) . Denote by β := − b a = 32 θ + 34 r, β := − b a = 12 θ + 12 r. Then A ≥ is equivalent to β ≥ β and A ≤ is equivalent to β ≤ β . For the case A < < A ,or β < β < β , we need to check the sign of A + 4 A νK , which requires a more detailed but stillsimple analysis. Denote by β := β + s rK (cid:18) θ + 13 r (cid:19) + 49 r K − rK,β := β − s rK (cid:18) θ + 13 r (cid:19) + 49 r K − rK. It is easy to check that β < β < β < β . It turns out that A + 4 A νK ≤ is equivalent to β ≤ β ≤ β . Combining the discussions above, we conclude that the parameter condition of Case1 in Example 3.4 is equivalent to β ≥ β , that of Case 2 to β < β < β , and that of Case 3 to < β ≤ β . Recall that β is the utility discount factor. We see that when β is small ( β ≤ β ),there is no free boundary; when β is in the middle ( β ∈ ( β , β ) ), there are two free boundaries;when β is large ( β ≥ β ), there is one free boundary. The threshold values β and β are criticalin deciding different optimal trading strategies. (a) z (b) z (c) z (d) Figure 1: (a)
K > A ≥
0; (b)
K > A < < A , A + 4 A νK >
0; (c)
K > A ≤ A < < A , A + 4 A νK ≤
0; (d) K = 0, A < < A .The next example is to characterize the optimal exercise and continuation regions for the non-HARA utility discussed in Example 3.4 when the portfolio insurance value K is set to be 0. Example 3.6.
We assume K = 0 and the same non-HARA utility as in Example 3.4. In this casewe have β = β .Case 1: A ≥ (equivalently β ≥ β ). There is no free boundary and it is optimal to stopimmediately.Case 2: A < < A (equivalently β < β < β ). There exists a unique free boundary definedby z ( τ ) := inf { z ; v ( τ, z ) = g ( z ) } , < τ ≤ θ T / . (3.12) Moreover, z ( τ ) is increasing with limits lim τ → z ( τ ) = 12 log (cid:16) − A A (cid:17) , (3.13)lim τ →∞ z ( τ ) = ∞ . (3.14) Case 3: A ≤ (equivalently β ≤ β ). There is no free boundary and it is not optimal to stopbefore the maturity.We leave the proof in Section 5. Figure 1 (d) illustrates the Case 2 discussed above with C z thecontinuation region and S z the exercise region. K = 0 and K >
0. For example, when A < < A , there exists a unique free boundary for K = 0 whereas there exist either two free boundaries or no free boundary for K >
0, which impliesthat one has to use different optimal trading strategies in the presence of portfolio insurance
K > K = 0 to reduce the problem into a standard utility maximization problem.With Assumption 3.2, we can directly verify that φ ( z ) defined by (3.2) is strictly decreasingand there exists a unique z ∈ R such that φ ( z ) = 0 . (3.15) Theorem 3.7.
Let Assumption 3.2 hold. Then the free boundary z ( τ ) defined by (3.3) is strictlydecreasing with lim τ → z ( τ ) = z , where z is defined by (3.15) , and z ( τ ) ∈ C [0 , θ T / ∩ C ∞ (0 , θ T / .Furthermore, z ( τ ) satisfies the following integral equation − Z ∞ z G ( τ, z ( τ ) − y ) φ ( y ) dy + Z τ G ( τ − s, z ( τ ) − z ( s )) φ ( z ( s )) z ′ ( s ) ds = 0 , (3.16) where G is the Green function defined by G ( τ, z ) = 1 √ πτ exp (cid:18) − ( z − κτ ) τ − ρτ (cid:19) . (3.17) Proof.
See Section 5.In the following, we conduct the asymptotic analysis of the free boundary and construct theglobal approximation for the dual problem. We investigate the asymptotic behaviour of the freeboundary near the expiry by using the integral equation (3.16).
Theorem 3.8.
Let Assumption 3.2 hold. Then the free boundary z ( τ ) defined by (3.3) , for <τ << , satisfies approximately z ( τ ) ≈ z − A √ τ , where A is a positive solution of the following equation e − A − √ π A + A Z e − A η η + η (1 + η ) dη = 0 . (3.18) The numerical solution of equation (3.18) is A ≈ . .Proof. See Section 5.The next result gives the asymptotic property of the free boundary z ( τ ) defined by (3.3) astime to maturity τ tends to infinite. Theorem 3.9.
Let Assumption 3.2 hold and z ∗ be the unique solution of the equation J X j =1 − q j ( q j − λ ) e ( q j − z − K (1 − λ ) = 0 , where λ = ( κ − p κ + 4 ρ ) . Then the free boundary defined by (3.3) satisfies lim τ →∞ z ( τ ) = z ∗ . roof. See Section 5.
Example 3.10.
Simple calculation shows that, for power utility ( J = 1 in (3.1)), we have z = 1 q − KνA , z ∗ = 1 q − Kq (1 − λ ) λ − q , and, for non-HARA utility ( J = 2 , q = − , q = − in (3.1)), we have z = −
12 log − A + p A + 4 A νK A ,z ∗ = −
12 log − ( q − λ ) + q ( q − λ ) + ( q − λ )(1 − λ ) ( q − λ ) . By Assumption 3.2 and q j − λ > , j = 1 , (see (5.28) ), we can verify that the expressions insidethe above logarithmic functions are positive. Now we seek a simple approximation formula for z ( τ ) defined by (3.3) such that (i) it hasasymptotic expansion z − A √ τ for small τ and (ii) it approaches z ∗ for large τ . For this, we seekan approximation of the form z ∗ ( τ ) := z − A r − e − bτ b , where b >
0. To make it match with the large τ behaviour, we need b = A ( z − z ∗ ) . Hence, the globalclosed-form approximation of the free boundary z ( τ ) defined by (3.3) is given by z ∗ ( τ ) := z − ( z − z ∗ ) p − e − b ∗ τ , b ∗ = 4 A ( z − z ∗ ) . (3.19)The next result presents a global closed-form approximation of the free boundary for problem(2.4) with conditions (2.2). Theorem 3.11.
Let the dual utility function be given by (3.1) and Assumption 3.2 hold. Let z ∗ ( τ ) in (3.19) be the global closed form approximation (GCA) to the free boundary z ( τ ) definedby (3.3) . Then the unique free boundary of problem (2.4) with condition (2.2) is strictly decreasingand approximately determined by x ( t ) = − ˜ U ′ K (cid:0) exp (cid:0) z ∗ ( θ ( T − t ) / (cid:1)(cid:1) , ≤ t ≤ T. Furthermore, the primal value function is given by V ( t, x ) = ˜ V ( t, I ( t, x )) + xI ( t, x ) , and the optimal feedback control is given by π ∗ t = θσ I ( t, x ) ˜ V yy ( t, I ( t, x )) , (3.20) where ˜ V is the dual value function, approximately given by ˜ V ( t, y ) = ˜ U K ( y ) − Z τ Z ∞ z ∗ ( s ) G ( τ − s, ln y − w ) φ ( w ) dwds, (3.21) τ = θ ( T − t ) / , and y = I ( t, x ) is the unique solution of the equation ˜ V y ( t, y ) + x = 0 for x > K . roof. See Section 5.
Remark 3.12.
In the continuation region, the optimal feedback control π ∗ can be computed eitherwith (2.3) using the primal value function or with (3.20) using the dual value function. The twomethods would produce the same optimal trading strategy due to the strong duality relation. It isin general more difficult to find the primal value function than to find the dual value function asthe former satisfies a nonlinear PDE in the continuation region whereas the latter a linear PDE.The dual value function has an integral representation which makes possible computing the optimalcontrol, provided the dual free boundary is known. This is where the GCA plays a pivotal role. Itwould be virtually impossible without the GCA to determine the optimal control in the continuationregion as both the primal and dual value functions depend on unknown free boundaries. In this section, we compare the numerical results derived using the global closed-form approxi-mation (GCA) and the binomial tree method (BTM). We now briefly explain to use BTM to solveour problem. BTM cannot be directly applied to solve the original investment stopping problem,however, it can be used to solve the dual optimal stopping problem which is essentially an Americanoptions pricing problem with one additional difficulty, that is, one has to find the initial value y ofthe dual process from the equation ˜ V y ( t, y ) + x = 0 while ˜ V is to be determined. To circumventthe problem, we use the following procedure.First, we fix an arbitrary y > Y up to time T and then use the dynamic programming method to solve the dual optimal stopping problem andfind the value ˜ V y (0 , y ) at time 0. We then check the sign of ˜ V y (0 , y ) + x : if positive, we decreasethe value of ˜ V y (0 , y ) by setting y = y /
10; if negative, we increase the value of ˜ V y (0 , y ) by setting y = 10 y . Repeat the process and get ˜ V y (0 , y ). If ˜ V y (0 , y ) + x and ˜ V y (0 , y ) + x have the samesign, we set y = y and repeat the process above; if they have different signs, we have found aninterval, bounded by y and y , that contains a solution to the equation V y ( t, y ) + x = 0. Wethen use the bisection method to find y with linear convergence. Once the initial value y for thedual process is determined, we can get easily the value ˜ V (0 , y ) and the free boundary for the dualproblem. Finally, using the dual relation, we can find the optimal value and the free boundary forthe primal problem. Example 4.1.
We discuss the free boundary and the optimal strategy of the optimal investmentstopping problem (2.4) with conditions (2.2) for power utility and non-HARA utility defined inExample 3.1.The parameters used are µ = 0 . , β = 0 . , r = 0 . , σ = 0 . , K = 1 , γ = 0 . , T = 1 . Thenumber of time steps for binomial tree method is N = 700 , which gives decimal point accuracy.These parameters satisfy Assumption 3.2.In Figure 2 we plot the optimal exercise boundaries for power and non-HARA utilities usingboth the global closed-form approximation (GCA) and the binomial tree method (BTM). It is clearthat the GCA and the BTM produce the free boundaries with the same shape and very small gaps.In Figure 3 we depict the sample paths of the optimal wealth and the corresponding optimal tradingstrategy using the GCA for power and non-HARA utilities. We can see that the optimal tradingstrategy becomes zero after time τ , the first time the optimal wealth process hits the free boundarybefore the terminal time T , and τ is the optimal stopping time of investing in risky assets and theoptimal wealth becomes X t = X τ e r ( t − τ ) for τ ≤ t ≤ T . t x ( t ) Continuation RegionExercise Region
GCABTM (a) t x ( t ) Continuation RegionExercise Region
GCABTM (b)
Figure 2: (a) The optimal exercise boundary compared with BTM for power utility; (b) Theoptimal exercise boundary compared with BTM for non-HARA utility. t X t ( ,X ) Path 1Path 2Free boundary
Exercise RegionContinuation Region t t * ( X t ) ( , ) Optimal strategy for path1Optimal strategy for path 2 (a) t X t ( ,X ) Path 1Path 2Free boundary
Exercise RegionContinuation Region t t * ( X t ) ( , ) Optimal strategy for path 1Optimal strategy for path 2 (b)
Figure 3: (a) Two different sample paths of wealth with initial wealth x = 1 . x = 1 . xample 4.2. In this example, we compare the optimal values and the optimal strategies obtainedby the closed-form approximation and the binomial tree method at the initial time.(i) For power and non-HARA utility, we compare the numerical results between GCA and BTM.The parameters used are µ = 0 . , β = 0 . , r = 0 . , σ = 0 . , K = 1 , γ = 0 . , T = 1 , t = 0 , initial wealth x = 1 . , number of time steps for binomial tree approach N = 700 . Thenumerical result is shown in Table 1.(ii) In Table 2, we give the mean and standard deviation of the absolute and relative differencebetween BTM and GCA for power and non-HARA utility. We fix K = 1 , T = 1 , t = 0 , initialwealth x = 1 . , and number of time steps for binomial tree approach N = 700 . The restparameters are selected randomly: samples of µ from the uniform distribution on interval [0 . , . , r on [0 . , . , β on [0 . , . , σ on [0 . , . , γ on [0 . , . . We also requirethe parameters satisfy Assumption 3.2.From the numerics in Table 2, we observe that the difference between the GCA and BTM optimalvalues is very small, whereas the computational time for GCA is much less than that for BTM. TheGCA is shown to be correct and fast. Compared to the optimal values, the error for computing theoptimal strategies using both the BTM and GCA is bigger. This is not surprising, as the optimalstrategies are computed with the derivatives of the value functions. Table 1: Comparison between GCA and BTM for Example 4.2 (i).Power utility Non-HARA utilityOptimal value Optimal strategy Optimal value Optimal strategyGCA value 1.4128 0.6558 1.5094 0.6776BTM value 1.4031 0.7454 1.5116 0.6846Difference 0.0096 0.0899 0.0022 0.0069Relative difference 0.0069 0.1206 0.0015 0.0101Time for GCA 22.7s 10.9s 11.4s 5.6sTime for BTM 1683.2s 1664.2s 2777.6s 2744.6sTable 2: Comparison between GCA and BTM for Example 4.2 (ii).Power utility Non-HARA utilityOptimal value Optimal strategy Optimal value Optimal strategyAvg difference 0.0074 0.0969 0.0034 0.0281Std difference 0.0050 0.1495 0.0062 0.0462Avg relative difference 0.0050 0.0745 0.0022 0.0630Std relative difference 0.0033 0.1145 0.0045 0.0919Avg time for GCA 23.2s 8.2s 23.0s 9.3sAvg time for BTM 2640.9s 2609.6s 2878.6s 2844.2s
In this section we give detailed proofs of the results of the paper.13 .1 Proof of Corollary 2.3
Proof.
Everything is a straightforward translation of Theorem 2.2. We only need to show (2.12)holds. For some fixed y > y < y , using ˜ U K (0) = ∞ and the convexity of ˜ V in y , we have˜ V y ( t, y ) ≤ ˜ V ( t, y ) − ˜ V ( t, y ) y − y ≤ ˜ V ( t, y ) − ˜ U K ( y ) y − y , which gives lim y → (cid:16) − ˜ V y ( t, y ) (cid:17) = + ∞ . Similarly, for some fixed y > y < y , using (2.5), weobtain 0 ≤ − ˜ V y ( t, y ) ≤ − ˜ V ( t, y ) − ˜ V ( t, y ) y − y ≤ − ˜ U K ( y ) − ˜ V ( t, y ) y − y ≤ − − Ky − ˜ V ( t, y ) y − y , which gives lim y →∞ (cid:16) − ˜ V y ( t, y ) (cid:17) := a ≤ K . Proof.
In the exercise region, we immediately have v z = g z . In the continuation region, since v z (0 , z ) = g z and v z ( τ, z ) = g z for ( τ, z ) ∈ ∂ C z , and by Assumption 3.2 and q j <
0, we have L [ g z ] = φ ′ ( z ) = J X j =1 A j q j e q j z − Kνe z ≤ . On the other hand, in the continuation region it holds that L [ v z ] = 0. So we have L [ v z − g z ] ≥ v z − g z ≥ . As a consequence, if ( τ, z ) ∈ C z , i.e., v ( τ, z ) > g ( z ), then for any z > z , v ( τ, z ) − g ( z ) ≥ v ( τ, z ) − g ( z ) > , from which we infer that ( τ, z ) ∈ C z . This indicates each τ -section of C z is connected. The existenceof the free boundary z ( τ ) now follows. We obtain (3.3) and (3.4). Moreover, (3.5) follows from(3.4). Proof.
Case 1: If A >
0, from Theorem 3.3, we know there exists a unique free boundary z ( τ )defined by (3.3). If A = 0 , then A > A = 0 , Theorem 3.3 implies that there exists a unique freeboundary z ( τ ) defined by (3.3).Case 2: We now prove (3.6) - (3.9). Denote that Λ := { ( τ, z ); z ( τ ) ≤ z ≤ z ( τ ) , < τ ≤ θ T / } . Since A < < A , A + 4 A νK > φ ( z ) = e z ( A e − z + A e − z − νK ), then thereexists two roots for equation φ ( z ) = 0. We denote the two roots by z I , z II with z I < z II . By adirect computation, we have z I = −
12 log − A − p A + 4 A Kν A , II = −
12 log − A + p A + 4 A Kν A . Then from the definition of the exercise region S z = { ( τ, z ); v ( τ, z ) = g ( z ) , < τ ≤ θ T / } andthe variational equation (2.8) , we have L [ v ] = L [ g ] = φ ( z ) ≥ τ, z ) ∈ S z . This implies that S z ⊆ (cid:8) ( τ, z ); φ ( z ) ≥ , < τ ≤ θ T / (cid:9) = (cid:8) < τ ≤ θ T / , z I ≤ z ≤ z II (cid:9) . This shows that the τ -section { z ; v ( τ, z ) = g ( z ) , < τ ≤ θ T / } of the exercise region S z isbounded. Therefore, z ( τ ) and z ( τ ) in (3.6) - (3.7) are well defined. By the definitions of z ( τ )and z ( τ ), we obtain that S z ⊆ Λ. Now, we prove thatΛ ⊆ S z . (5.1)Since (cid:8) ( τ, z ); z = z ( τ ) or z = z ( τ ) , < τ ≤ θ T / (cid:9) ⊆ S z ⊆ (cid:8) ( τ, z ); φ ( z ) ≥ , < τ ≤ θ T / (cid:9) , we have Λ ⊆ (cid:8) ( τ, z ); φ ( z ) ≥ , < τ ≤ θ T / (cid:9) . Assume that (5.1) is false. Then there exists anon-empty subset N = C z ∩ Λ and the parabolic boundary ∂ p N ⊆ ¯Ω T − C z . Here ¯Ω T denotes theclosure of Ω T . Thus L [ v ] = 0 , ( τ, z ) ∈ N ,L [ g ] = φ ( z ) ≥ , ( τ, z ) ∈ N ,v = g, ( τ, z ) ∈ ∂ p N . By the comparison principle, v ≤ g in N , which implies that N = ∅ . Hence the contradictionarises. Therefore, (5.1) holds. So Λ = S z , i.e., (3.9) holds true. (3.8) follows from (3.9).Next, we prove the monotonicity of the two free boundaries. If z ( τ ) is not increasing, thereexist τ < τ such that z ( τ ) > z ( τ ). Since v τ ≥ g ( z ( τ )) = v ( τ , z ( τ )) ≥ v ( τ , z ( τ )) > g ( z ( τ )) , which is a contradiction. Hence, z ( τ ) is increasing. Similarly, z ( τ ) is decreasing.Finally, we prove (3.10) and (3.11). If lim τ → z ( τ ) > z I , then for any z satisfying lim τ → z ( τ ) > z > z I , and τ = 0, we have 0 = v τ − v zz + κv z + ρv = v τ + φ ( z ) > , where the last inequality follows from the fact that v τ ≥ φ ( z ) > z I < z < z II . This isa contradiction. Hence, (3.10) holds. By a similar argument, we can obtain (3.11).Case 3: In fact, if A ≤ A < < A , A + 4 A νK ≤
0, then L [ g ] = φ ( z ) = e z ( A e − z + A e − z − νK ) ≤
0. Hence, g is a subsolution of problem L [ v ] = 0 , ( τ, z ) ∈ Ω T , (5.2) v (0 , z ) = g ( z ) , z ∈ R . (5.3)Denote the solution of the problem (5.2) - (5.3) by ˜ v . Then by comparison, we obtain that˜ v − g ≥ T . Therefore, ˜ v is also the solution of problem (2.8).15 .4 Proof of Example 3.6 Proof.
Case 1 and Case 3 can be easily proved as follows: Since K = 0, we have L [ g ] = φ ( z ) = A e − z + A e − z ≥ A ≥ φ ( z ) ≤ A ≤ A < A . If φ ( z ) ≤
0, thenby the same argument as in the proof of Example 3.4, we conclude that there is no free boundaryand it is not optimal to stop before the maturity. If φ ( z ) ≥
0, then v = g is the solution to problem(2.8), which implies that there is no free boundary and it is optimal to stop immediately.Next we prove Case 2. We can show (3.12), (3.13) and monotonicity of z ( τ ) following a similarargument as in the proof of Example 3.4. We only need to prove (3.14). If z ( τ ) is bounded, thenwe have lim τ →∞ z ( τ ) < ∞ . Denote lim τ →∞ z ( τ ) := a .We rewrite problem (2.8) as L [ v ] = I { z ≥ z ( τ ) } φ ( z ) , ( τ, z ) ∈ Ω T ,v (0 , z ) = g ( z ) , z ∈ R , where I A is the indicator function of set A . By Green’s identity, we have v ( τ, z ) = Z ∞−∞ G ( τ, z − y ) g ( y ) dy + Z τ Z ∞ z ( τ − s ) G ( s, z − y ) φ ( y ) dyds, where G is the Green function defined by (3.17). We setΛ ( τ ) = 13 e − z ( τ ) − A τ + e − z ( τ ) − A τ , Λ ( τ ) = A √ π e − z ( τ ) Z τ e − A s Z ∞ z ( τ − s ) − z ( τ )+( κ +6) s √ s e − η dηds, Λ ( τ ) = A √ π e − z ( τ ) Z τ e − A s Z ∞ z ( τ − s ) − z ( τ )+( κ +2) s √ s e − η dηds. Since v ( τ, z ( τ )) = g ( z ( τ )), we haveΛ ( τ ) + Λ ( τ ) + Λ ( τ ) = g ( z ( τ )) . By dominated convergence theorem, we havelim τ →∞ Λ ( τ ) = A √ π e − a Z ∞ e − A s Z ∞ ( κ +6) √ s e − η dηds, = − κ + 66 √ π e − a Z ∞ e − ( κ + ρ ) t dt < ∞ . Since A < < A , we have lim τ →∞ Λ ( τ ) ≤ A e − a Z ∞ e − A s ds < ∞ , lim τ →∞ Λ ( τ ) = ∞ . As lim τ →∞ g ( z ( τ )) = g ( a ) < ∞ , this leads to a contradiction. Hence, we obtain that z ( τ ) is increasingand unbounded, i.e., (3.14) holds. 16 .5 Proof of Theorem 3.7 Proof.
From Theorem 3 .
3, the variational problem (2.8) can be written as L [ v ] = 0 for z > z ( τ ) , < τ ≤ θ T / , (5.4) v ( τ, z ) = g ( z ) for z ≤ z ( τ ) , < τ ≤ θ T / , (5.5) v z ( τ, z ( τ )) = g z ( z ( τ )) , < τ ≤ θ T / , (5.6) v (0 , z ) = g ( z ) , z ∈ R , where L and g is defined as in (2.9).Firstly, we claim that z ( τ ) is non-increasing. Otherwise, there exists some 0 < τ < τ suchthat z ( τ ) < z ( τ ). Then since v τ ≥ v ( τ , z ( τ )) − g ( z ( τ )) ≥ v ( τ , z ( τ )) − g ( z ( τ )) > , where the second inequality follows from the definition of the free boundary z ( τ ). This leads tocontradiction. Then we claim that z ( τ ) < z for τ > . (5.7)Let ¯ v = v − g . We rewrite the variational problem (2.8) asmin { L [¯ v ] + φ ( z ) , ¯ v } = 0 , ( τ, z ) ∈ Ω T , (5.8)¯ v (0 , z ) = 0 , z ∈ R , (5.9)where φ ( z ) is defined by (3.2).Let U be the solution to L [ U ] = − φ ( z ) , ( τ, z ) ∈ Ω := { ( τ, z ) ∈ Ω T ; z > z } , (5.10) U ( τ, z ) = 0 , ( τ, z ) ∈ ∂ p Ω , (5.11)where ∂ p Ω is the parabolic boundary of Ω.Since φ ( z ) is strictly decreasing and z > z , by (3.15), we have L [ U ] = − φ ( z ) > U > U z ( τ, z ) > τ >
0. By (5.8) - (5.11), wehave L [¯ v ] ≥ − φ ( z ) = L [ U ] and ¯ v ( τ, z ) ≥ U ( τ, z ) for ( τ, z ) ∈ ∂ p Ω. By the comparison principle(Lieberman, 1996, Corollary 2.5), we see that ¯ v ≥ U > z ( τ ) ≤ z .Otherwise, there exists some z ∈ ( z , z ( τ )) such that ¯ v ( τ, z ) = 0.If there exists some τ > z ( τ ) = z , then we have¯ v ( τ , z ( τ )) = U ( τ , z ( τ )) = 0 . Hopf’s lemma (see (Lieberman, 1996, Lemma 2.8)) implies that¯ v z ( τ , z ( τ )) > U z ( τ , z ( τ )) > . Since ¯ v z ( τ, z ( τ )) = 0, for any τ >
0, this leads to contradiction. So (5.7) is proved.Hence, we have lim τ → z ( τ ) ≤ z . If lim τ → z ( τ ) < z , then for some z ∈ ( lim τ → z ( τ ) , z ), by (5.4), wehave L [ v ] | τ =0 = [ v τ − v zz + κv z + ρv ] | τ =0 = 0 . This leads to v τ | τ =0 = [ v zz − κv z − ρv ] | τ =0 = − φ ( z ) < , φ is strictly decreasing, φ ( z ) = 0 and z < z . This contradictswith the fact that v τ (0 , z ) ≥ z ( τ ) ∈ C [0 , θ T / τ ∈ [0 , θ T / z < z , where z = lim τ → τ +0 z ( τ ) , z = lim τ → τ − z ( τ ) . By (5.7), we have z , z ≤ z . For any z ∈ [ z , z ], by (5.4), we have L [ v ] | τ = τ = [ v τ − v zz + κv z + ρv ] | τ = τ = 0 . For any z ∈ [ z , z ], this leads to v τ | τ = τ = [ v zz − κv z − ρv ] | τ = τ = − φ ( z ) ≤ , where the last inequality follows from φ is decreasing, φ ( z ) = 0 and z , z ≤ z . By (2.11), thismeans φ ( z ) = 0 for any z ∈ [ z , z ], while φ is a strictly decreasing function. The contradictionarises. Therefore z ( τ ) ∈ C [0 , θ T /
2] is true. Furthermore we can use the bootstrap argumentdeveloped by Friedman (1975) to conclude that z ( τ ) ∈ C ∞ (0 , θ T / v ( τ, z ( τ )) = g ( z ( τ )) , τ > . (5.12)Differentiating (5.12) in τ , by (5.6), we obtain v τ ( τ, z ( τ )) = 0 . (5.13)Furthermore, (5.4) implies L [ v ]( τ, z ( τ )) = v τ ( τ, z ( τ )) − v zz ( τ, z ( τ )) + κv z ( τ, z ( τ )) + ρv ( τ, z ( τ )) = 0 , which leads to v zz ( τ, z ( τ )) = κg z ( z ( τ )) + ρg ( z ( τ )) , τ > . (5.14)By (5.13) and Theorem 2.2, we derive L [ v τ ] = 0 in continuation region ,v τ ( τ, z ( τ )) = 0 , < τ < θ T / ,v τ (0 , z ) ≥ , z ∈ R . Hopf’s lemma and the maximum principle imply that v τ > v τz > τ, z ( τ )). Differentiating (5.6) in τ , we have v τz ( τ, z ( τ )) + v zz ( τ, z ( τ )) z ′ ( τ ) = g zz ( z ( τ )) z ′ ( τ ) . By (5.14), v zτ ( τ, z ( τ )) = − φ ( z ( τ )) z ′ ( τ ) > , (5.15)Since φ is strictly decreasing, by (5.7), we derive that − φ ( z ( τ )) < − φ ( z ) = 0 , τ > . z ′ ( τ ) < G defined by (3.17), which satisfies L [ G ] = G τ − G zz + κG z + ρG = 0 . Denote by u ( τ, z ) = v τ ( τ, z ) and I ( z, τ, s ) = Z ∞ z ( s ) G ( τ − s, z − y ) u ( s, y ) dy. (5.16)Note that lim s → τ G ( τ − s, z − y ) = δ ( z − y ), where δ is a Dirac delta function, therefore for any z > z ( τ ),lim s → τ I ( z, τ, s ) = u ( τ, z ) . With this in mind, we can relate the solution u ( τ, z ) to the initial condition by integrating I s ( z, τ, s )between s = 0 and s = τ .Differentiating (5.16), also noting u ( s, z ( s )) = v s ( s, z ( s )) = 0 (see (5.13)), yields I s ( z, τ, s ) = − Z ∞ z ( s ) G τ ( τ − s, z − y ) u ( s, y ) dy + Z ∞ z ( s ) G ( τ − s, z − y ) u s ( s, y ) dy. Simple computation, using integration by parts, gives Z ∞ z ( s ) G ( τ − s, z − y ) u s ( s, y ) dy = Z ∞ z ( s ) G ( τ − s, z − y ) [ u zz − κu z − ρu ] ( s, y ) dy = − G ( τ − s, z − z ( s )) u z ( s, z ( s )) + Z ∞ z ( s ) [ G zz − κG z − ρG ]( τ − s, z − y ) u ( s, y ) dy = − G ( τ − s, z − z ( s )) u z ( s, z ( s )) + Z ∞ z ( s ) G τ ( τ − s, z − y ) u ( s, y ) dy. Hence, I s ( z, τ, s ) = − G ( τ − s, z − z ( s )) u z ( s, z ( s )) . Integrating I s ( z, τ, s ) from s = 0 to s = τ , we obtain u ( τ, z ) − Z ∞ z G ( τ, z − y ) u (0 , y ) dy = − Z τ G ( τ − s, z − z ( s )) u z ( s, z ( s )) ds. (5.17)Now we calculate u (0 , y ) for y ≥ z . By (5.4), we have u (0 , y ) = v τ (0 , y ) = v zz (0 , y ) − κv z (0 , y ) − ρv (0 , y ) = − L [ g ] = − φ ( y ) . Also u z ( s, z ( s )) = v τz ( s, z ( s )) is given by (5.15). Since u ( τ, z ( τ )) = v τ ( τ, z ( τ )) = 0 (see (5.13)),letting z = z ( τ ) in (5.17), we have (3.16). 19 .6 Proof of Theorem 3.8 Proof.
We postulate that z ( τ ) = z − A √ τ + o ( √ τ ) , τ → . (5.18)A direct computation shows that the first term in (3.16) is given by − Z ∞ z G ( τ, z ( τ ) − y ) φ ( y ) dy = J X j =1 − A j e q j A j τ + q j z ( τ ) erfc (cid:18) z − z ( τ ) + ( κ − q j ) τ √ τ (cid:19) + 12 νKe − ντ + z ( τ ) erfc (cid:18) z − z ( τ ) + ( ν − ρ − τ √ τ (cid:19) , (5.19)where erfc( z ) is the complementary error function defined byerfc( z ) = 2 √ π Z ∞ z e − η dη. By Taylor’s expansion and (5.18), we have e q j A j τ + q j z ( τ ) = e o ( √ τ )+ q j ( z − A √ τ + o ( √ τ )) = e q j z (1 − q j A √ τ + o ( √ τ )) (5.20) e − ντ + z ( τ ) = e z (1 − A √ τ + o ( √ τ )) . Similarly, Taylor’s expansion giveserfc (cid:18) A (1 + o (1)) + κ − q j √ τ (cid:19) = erfc ( A (1 + o (1))) − κ − q j √ π e − A √ τ + o ( √ τ ) (5.21)and erfc( A (1 + o (1)) + ν − ρ − √ τ ) = erfc( A (1 + o (1))) − ν − ρ − √ π e − A √ τ + o ( √ τ ) . (5.22)Since φ ( z ) = 0, by (5.19) - (5.22), we derive that − Z ∞ z G ( τ, z ( τ ) − y ) φ ( y ) dy = J X j =1 − A j e q j z (cid:2) − q j A √ τ + o ( √ τ ) (cid:3) (cid:20) erfc( A (1 + o (1))) − κ − q j √ π e − A √ τ + o ( √ τ ) (cid:21) + 12 νKe z (cid:2) − A √ τ + o ( √ τ ) (cid:3) (cid:20) erfc( A (1 + o (1))) − ν − ρ − √ π e − A √ τ + o ( √ τ ) (cid:21) = − φ ( z )erfc( A (1 + o (1))) − J X j =1 A j e q j z (cid:20) ( − q j A √ τ )erfc( A (1 + o (1))) − κ − q j √ π e − A √ τ (cid:21) + 12 νKe z (cid:20) ( − A √ τ )erfc( A (1 + o (1))) − ν − ρ − √ π e − A √ τ (cid:21) + o ( √ τ )= (cid:18) A erfc( A (1 + o (1))) − √ π e − A (cid:19) √ τ J X j =1 q j A j e q j z − νKe z + o ( √ τ ) . (5.23)20e use the transformation t = ζτ and denote ¯ ζ = 1 − ζ . Then the second term in (3.16) can becalculated by Z τ G ( τ − s, z ( τ ) − z ( s )) φ ( z ( s )) z ′ ( s ) ds = J X j =1 A j Z τ √ πt e − [ z ( τ ) − z ( τ − t ) − κt ]24 t − ρt + q j z ( τ − t ) z ′ ( τ − t ) dt − νK Z τ √ πt e − [ z ( τ ) − z ( τ − t ) − κt ]24 t − ρt + z ( τ − t ) z ′ ( τ − t ) dt = J X j =1 A j √ τ Z √ πζ e − [ z ( τ ) − z (¯ ζτ ) − κζτ ]24 ζτ − ρζτ + q j z (¯ ζτ ) z ′ (¯ ζτ ) dζ − νK √ τ Z √ πζ e − [ z ( τ ) − z (¯ ζτ ) − κζτ ]24 ζτ − ρζτ + z (¯ ζτ ) z ′ (¯ ζτ ) dζ := (Term 1) + (Term 2) . Using the expansions z ′ (¯ ζτ ) = − A (¯ ζτ ) − / (1 + o (1)) ,e q j z (¯ ζτ ) − ρζτ = e q j z (1 − Aq j q ¯ ζτ + o ( √ τ )) , we derive that (Term 1) = J X j =1 A j e q j z τ Z √ πζ e − [ z ( τ ) − z (¯ ζτ ) − κζτ ]24 ζτ z ′ (¯ ζτ ) dζ − J X j =1 AA j e q j z Z √ πζ e − [2 A ( √ ¯ ζτ −√ τ )+ o ( √ τ )]24 ζτ · (1 + o (1)) (cid:0) − Aq j q ¯ ζτ + o ( √ τ ) (cid:1) (¯ ζ ) − / dζ. Similarly, one can obtain that(Term 2) = − νKe z τ Z √ πζ e − [ z ( τ ) − z (¯ ζτ ) − κζτ ]24 ζτ z ′ (¯ ζτ ) dζ + AνKe z Z √ πζ e − [2 A ( √ ¯ ζτ −√ τ )+ o ( √ τ )]24 ζτ · (1 + o (1)) (cid:18) − A q ¯ ζτ + o ( √ τ ) (cid:19) (¯ ζ ) − / dζ. Since φ ( z ) = 0, this leads to(Term 1) + (Term 2) = 2 A J X j =1 q j A j e q j z − νKe z · (1 + o (1)) √ τ Z √ πζ e − [2 A ( √ ¯ ζτ −√ τ )+ o ( √ τ )]24 ζτ dζ + o ( √ τ ) . (5.24)21y (3.16), (5.23) and (5.24), we derive that1 √ π e − A − A erfc( A ) = 2 A Z √ πζ e − [2 A ( √ ¯ ζτ −√ τ )]24 ζτ dζ. By the transformation η = − √ ¯ ζ √ ζ A , we derive1 √ π e − A − A √ π Z ∞ A e − η dη = 2 √ π Z A e − η A ( A − η )( A + η ) dη. (5.25)Let F ( A ) = √ πA e − A − erfc( A ) − √ π Z A e − η A ( A − η )( A + η ) dη . By a direct computation, we have F ′ ( A ) = − A √ π e − A + 2 √ π Z A e − η ( − A + 2 Aη ) η ( A + η ) dη < , with F (0) = + ∞ and F (+ ∞ ) = −
1. This implies there exists a unique solution to the equation(5.25). Finally, (3.18) follows from Z ∞ A e − η dη = √ π − Z A e − η dη and (5.25). To prove Theorem 3.9, we need the following lemma.
Lemma 5.1.
There exists some z ∗ ∈ R such that lim τ →∞ z ( τ ) = z ∗ . Proof.
Firstly, we consider the following problem − Ψ ′′ ( z ) + κ Ψ ′ ( z ) + ρ Ψ( z ) = 0 for z > a, Ψ( z ) = g ( z ) for z ≤ a, Ψ ′ ( a ) = g ′ ( a ) , (5.26)lim z →∞ Ψ( z ) = 0 . Denote p ( z ) := J X j =1 − q j ( q j − λ ) e ( q j − z − K (1 − λ ) , (5.27)where λ = κ − √ κ +4 ρ . By Assumption 3.2, and A < A < . . . < A J , we have κ + 4 ρ − ( κ − q j ) = − q j A j >
0, which leads to q j − λ = p κ + 4 ρ − κ + 2 q j > . (5.28)This implies that p ′ ( z ) <
0, lim z →∞ p ( z ) = − K (1 − λ ) <
0, and lim z →−∞ p ( z ) = ∞ . Hence, there exists aunique a ∈ R such that p ( a ) = 0 . (5.29)Now the solution to problem (5.26) is given byΨ( z ) = g ( a ) e λ ( z − a ) for z > a, Ψ( z ) = g ( z ) for z ≤ a, (5.30)22here a is defined by (5.29).We shall prove that the function Ψ( z ) defined by (5.30) satisfies the following variational equa-tion min {− Ψ ′′ + κ Ψ ′ + ρ Ψ , Ψ − g } = 0 , z ∈ R . (5.31)Firstly, for any z > a, since Ψ is the solution to problem (5.26), we only need to verify Ψ( z ) > g ( z ).Denote Φ( z, c ) = g ( c ) e λ ( z − c ) . Differentiating Φ( z, c ) in c we have ∂∂c Φ( z, c ) = e λ ( z − c )+ c p ( c ) , where p ( c ) is defined by (5.27). This implies that Φ( z, · ) is strictly increasing in ( −∞ , a ) and strictlydecreasing in ( a, z ). Hence, we have Ψ( z ) = Φ( z, a ) > Φ( z, z ) = g ( z ) for any z > a. Consequently,Ψ satisfies (5.31) for any z > a .Secondly, for any z ≤ a, since φ ( z ) = P Jj =1 A j e q j z − Kνe z = 0, κ = ν − ρ + 1, q j − λ > νe z p ( z ) = νe z p ( z ) − φ ( z )= J X j =1 e q j z [ − νq j ( q j − λ ) − A j ] + λKνe z = J X j =1 e q j z [ − νq j ( q j − λ ) + ( λ − A j ]= J X j =1 − q j e q j z ( q j − λ )[ κ (2 − q j ) + ( − q j ) p κ + 4 ρ + 4 q j − < . Since p ( z ) is strictly decreasing and p ( a ) = 0, we derive that z > a . This leads to − g ′′ + κg ′ + ρg = L [ g ] = φ ( z ) > z ≤ a, where the last inequality follows from the fact that φ is strictly decreasing, φ ( z ) = 0, and a < z .Thus Ψ satisfies (5.31) for any z ≤ a .Now the variational inequality (5.31) implies thatmin { L [Ψ] , Ψ − g } = 0 , ( τ, z ) ∈ Ω T , Ψ( z ) ≥ g ( z ) = v (0 , z ) , z ∈ R . By the comparison principle (see Lemma A.1), we have v ( τ, z ) ≤ Ψ( z ) for ( τ, z ) ∈ Ω T . Then wederive that z ( τ ) ≥ a . Otherwise, by the definition of z ( τ ) (see (3.3)) and (5.30), there exists some z ∈ ( z ( τ ) , a ) such that v ( τ, z ) > g ( z ) = Ψ( z ) . The contradiction arises. Since z ( τ ) is decreasing (See Theorem 3.7 ) and has a lower bound, thereexists some z ∗ ∈ R such that lim τ →∞ z ( τ ) = z ∗ .We can now prove Theorem 3.9. 23 roof. We only need to show that z ∗ = a , where a is defined in (5.29).We rewrite problem (2.8) as L [ v ] = I { z ≤ z ( τ ) } φ ( z ) , ( τ, z ) ∈ Ω T ,v (0 , z ) = g ( z ) , z ∈ R , where I A is the indicator function of set A . By Green’s identity, we have v ( τ, z ) = Z ∞−∞ G ( τ, z − y ) g ( y ) dy + Z τ Z z ( τ − s ) −∞ G ( s, z − y ) φ ( y ) dyds, where G is the Green function defined by (3.17). Since v ( τ, z ( τ )) = g ( z ( τ )) on the free boundary( τ, z ( τ )), a direct computation shows that g ( z ( τ )) = Z ∞−∞ G ( τ, z ( τ ) − y ) g ( y ) dy + Z τ Z z ( τ − s ) −∞ G ( s, z ( τ ) − y ) φ ( y ) dyds = J X j =1 − q j √ πτ e q j z ( τ )+ q j A j τ Z ∞−∞ exp h − ( y − z ( τ ) + κτ − q j τ ) τ i dy − K √ πτ e z ( τ ) − ντ Z ∞−∞ exp h − ( y − z ( τ ) + κτ − τ ) τ i dy + J X j =1 A j Z τ √ πs e q j z ( τ )+ q j A j s Z z ( τ − s ) −∞ exp h − ( y − z ( τ ) + κs − q j s ) s i dyds − νK Z τ √ πs e z ( τ ) − νs Z z ( τ − s ) −∞ exp h − ( y − z ( τ ) + κs − s ) s i dyds = J X j =1 − q j e q j z ( τ )+ q j A j τ − Ke z ( τ ) − ντ + J X j =1 A j Z τ e q j z ( τ )+ q j A j s N (cid:18) z ( τ − s ) − z ( τ ) + κs − q j s √ s (cid:19) ds − νK Z τ e z ( τ ) − νs N (cid:18) z ( τ − s ) − z ( τ ) + κs − s √ s (cid:19) ds, where N ( · ) is the cumulative distribution function of a standard normal variable. Letting τ → ∞ ,by the dominated convergence theorem and the integration by parts, we have g ( z ∗ ) = J X j =1 A j Z ∞ e q j z ∗ + q j A j s N (cid:18) κs − q j s √ s (cid:19) ds − νK Z ∞ e z ∗ − νs N (cid:18) κs − s √ s (cid:19) ds = J X j =1 − q j e q j z ∗ (cid:16) κ − q j p κ + 4 ρ (cid:17) − Ke z ∗ (cid:16) κ − p κ + 4 ρ (cid:17) = 12 g ( z ∗ ) − J X j =1 κ − q j q j p κ + 4 ρ e q j z ∗ − κ − p κ + 4 ρ Ke z ∗ . Simple algebraic computation gives p ( z ∗ ) = 0 = p ( a ), where p is defined in (5.27). Hence,lim τ →∞ z ( τ ) = z ∗ = a . 24 .8 Proof of Theorem 3.11 Proof.
Define the continuation region in y -coordinate to be C y = { ( t, y ); ˜ V ( t, y ) > ˜ U K ( y ) , ≤ t
This paper provides a rigorous analysis of the optimal investment stopping problem using thedual control method. The analysis covers a class of utility functions, including power and non-HARA utilities. The approximate formulas for the optimal value functions and optimal strategiesare derived by developing the approximate formulas for the dual problems. For non-HARA utility,if Assumption 3.2 does not hold, then there may exist two free boundaries or no free boundaryfor the dual problem and we cannot use the method developed in this paper to characterize thelimiting behaviour of the free boundary as time to maturity tends to zero or infinity, which makesimpossible to find a global closed-form approximation to the free boundary. We leave this for thefuture work.
Acknowledgments . The authors are very grateful to two anonymous reviewers whose constructivecomments and suggestions have helped to improve the paper of the previous version.
References
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A Appendix: Proof of Theorem 2.2
Theorem 2.2 is considered a known result in the PDE theory, but for the convenience of thereader, we give a proof. Note that the payoff function for vanilla American option is Lipschitzcontinuous, but the function g in (2.9) is not Lipschitz continuous in the infinite region. So theanalysis is different from that of Liang et al. (2007).Firstly, we prove the following comparison principle: Lemma A.1.
Let v , v ∈ W , p,loc (Ω T ) ∩ C ( ¯Ω T ) be functions satisfying | v i | ≤ C ( e αz + e − γz ) forsome positive constants C, α, γ , i = 1 , , and F [ v ] ≥ F [ v ] , ( τ, z ) ∈ Ω T ,v (0 , z ) ≥ v (0 , z ) , z ∈ R , where F [ v ] := min { L [ v ] , v − g } . Then v ( τ, z ) ≥ v ( τ, z ) , ( τ, z ) ∈ ¯Ω T . Proof.
Note that on the set Ω := { ( τ, z ) ∈ Ω T ; v ( τ, z ) − g ( z ) ≤ L [ v ] } , we automatically have v ( τ, z ) − g ( z ) ≥ F [ v ] ≥ F [ v ] = v ( τ, z ) − g ( z ), so that v ( τ, z ) ≥ v ( τ, z ). On the set Ω := { ( τ, z ) ∈ Ω T ; v ( τ, z ) − g ( z ) > L [ v ] } , we have L [ v ] ≥ F [ v ] ≥ F [ v ] = L [ v ].We are now in a situation where L [ v ] ≥ L [ v ] for ( τ, z ) ∈ Ω and v ( τ, z ) ≥ v ( τ, z ) for( τ, z ) ∈ ¯Ω T − Ω . We can apply the maximum principle (see (Lieberman, 1996, Theorem 2.7)) onΩ to conclude that v ( τ, z ) ≥ v ( τ, z ) on Ω .To prove the existence of the solution of problem (2.8), we construct a penalty function β ǫ ( t ) ∈ C ( R ) satisfying (see Friedman (1982)) β ǫ ( t ) ≤ , β ǫ (0) = − C ( C > ,β ǫ ( t ) = 0 , t ≥ ǫ,β ′ ǫ ( t ) ≥ , β ′′ ǫ ( t ) ≤ ,β ǫ ( t ) → , if t > , ǫ → ,β ǫ ( t ) → −∞ , if t < , ǫ → , where C is a constant to be determined. 27ince system (2.8) lies in an unbounded domain, we apply a bounded domain to approximateit: min (cid:8) L [ v R ] , v R − g (cid:9) = 0 , ( τ, z ) ∈ Ω RT := (cid:0) , θ T / (cid:1) × ( − R, R ) , (A.1) v R ( τ, z ) = g ( z ) , ( τ, z ) ∈ ∂ p Ω RT , (A.2)where ∂ p Ω RT is parabolic boundary, the operator L and g ( z ) is defined in (2.9). Consider the penaltyproblem of (A.1) - (A.2): L [ v ǫ,R ] + β ǫ ( v ǫ,R − g ) = 0 , ( τ, z ) ∈ Ω RT , (A.3) v ǫ,R ( τ, z ) = g ( z ) , ( τ, z ) ∈ ∂ p Ω RT . (A.4)By (Friedman, 1982, Theorem 8.2), For fixed ǫ and R , problem (A.3) - (A.4) has a unique solution v = v ǫ,R ∈ W , p (Ω RT ), 1 < p < + ∞ . Lemma A.2.
For any fixed
R > , there exists a unique solution v R ∈ C ( ¯Ω RT ) ∩ W , p (Ω RT ) ofproblem (A.1) - (A.2), < p < + ∞ . Moreover g ( z ) ≤ v R ( τ, z ) ≤ ˜ C ( e Bτ + pp − z + 1) , ( τ, z ) ∈ Ω RT , (A.5) where ˜ C is defined as in (2.5), B = | ( pp − ) − κ pp − − ρ | + 1 .Proof. By (Friedman, 1982, Theorem 8.2), we immediately obtain that there exists a unique solutiondefined by v R := lim ǫ → v ǫ,R of the problem (A.1) - (A.2) and v R ∈ C ( ¯Ω RT ) ∩ W , p (Ω RT ). Thevariational inequality (A.1) implies the first inequality in (A.5).To obtain the second inequality in (A.5), denote w ( τ, z ) = ˜ C (1 + e Bτ + pp − z ). By (2.5), we notethat w − g = w − ˜ U K ( e z ) ≥ w − ( ˜ C (1 + e pp − z ) − Ke z ) ≥ Ke z ≥ Ke − R ≥ ǫ for small ǫ and ( τ, z ) ∈ Ω RT . By the definition of β ǫ , this implies that β ǫ ( w − g ) = 0 . Hence, by choosing B = | ( pp − ) − κ pp − − ρ | + 1, we have L [ w ] + β ǫ ( w − g ) = ˜ Ce Bτ + pp − z (cid:18) B − (cid:16) pp − (cid:17) + pp − κ + ρ (cid:19) + ˜ Cρ ≥ . The last inequality above follows from the definition of A and B . By the comparison principle, weobtain v ǫ,R ≤ w in Ω RT . Now by letting ǫ →
0, we complete the proof.We can now complete the proof of Theorem 2.2.
Proof.
By setting R = n ( n ∈ Z + ) in (A.1) - (A.2), we rewrite the variational problem (A.1) - (A.2)as L [ v n ] = f ( τ, z ) , ( τ, z ) ∈ Ω nT ,v n ( τ, z ) = g ( z ) , z ∈ ∂ p Ω nT , f ( τ, z ) = I { v = g } L [ g ]( z ) , where I A is the indicator function of set A . Combining (A.5), we deduce that for any fixed ξ > W , p interior estimate holds for n > ξ : k v n k W , p (Ω ξT ) ≤ C ξ , (A.6)where C ξ is a constant depending on ξ but not on n , and k · k W , p (Ω ξT ) is the norm in the Sobolevspace W , p (Ω ξT ).Letting ξ = 1 in (A.6). By the weak compactness and Sobolev embedding, there is a subsequence { v n (1) } of { v n } such that v n (1) → v (1) weakly in W , p (Ω T )and k v n (1) − v (1) k C (Ω T ) → . Letting ξ = 2 in (A.6) with subsequence { v n (1) } instead of { v n } . By the weak compactness andSobolev embedding, there is a subsequence { v n (2) } of { v n (1) } such that v n (2) → v (2) weakly in W , p (Ω T )and k v n (2) − v (2) k C (Ω T ) → . Moreover, we have v (2) = v (1) in Ω T . By induction, we conclude that there exists a subsequence v n ( m ) of v n ( m − on Ω mT such that v n ( m ) → v ( m ) weakly in W , p (Ω mT )and k v n ( m ) − v ( m ) k C (Ω mT ) → . Moreover, v ( m ) = v ( j ) in Ω jT , ≤ j ≤ m − . We define v = v ( m ) if ( τ, z ) ∈ Ω mT for any m >
0. We consider the sequence v m ( m ) in diagram. Forany N >
0, since v m ( m ) is a subsequence of v m ( N ) if m > N , we derive that v m ( m ) → v ( N ) = v weakly in W , p (Ω NT )and k v m ( m ) − v k C (Ω NT ) = k v m ( m ) − v ( N ) k C (Ω NT ) → . Letting m → ∞ in the system min { L [ v m ( m ) ] , v m ( m ) − g } = 0 , ( τ, z ) ∈ Ω mT ,v m ( m ) (0 , z ) = g ( z ) , z ∈ ∂ p Ω mT , we find that v is the solution of problem (2.8). 29he inequality (2.10) follows by letting R → ∞ in the inequality (A.5). Lemma A.1 and (2.10)imply the uniqueness.Finally, we prove (2.11). In the exercise region S z , we have v z ( τ, z ) = g ′ ( z ) = ˜ U ′ K ( e z ) e z ≤ − v z ( τ, z ) + v zz ( τ, z ) = ˜ U ′′ K ( e z ) e z > . Note that the above inequalities also hold at time τ = 0 and at the boundary of C z . Since L [ v ] = 0in C z , we have L [ v z ] = 0 and L [ − v z + v zz ] = 0 for ( τ, z ) ∈ C z . The maximum principle implies that v z ≤ − v z + v zz > τ, z ) ∈ C z .To prove v τ ≥
0, we define w ( τ, z ) = v ( τ + δ, z ) , for small δ > . From (2.8), we know that w ( τ, z ) satisfiesmin { L [ w ] , w − g } = 0 , ( τ, z ) ∈ ˜Ω T := (cid:0) , θ T / − δ (cid:1) × R ,w (0 , z ) = v ( δ, z ) ≥ g ( z ) = v (0 , z ) , z ∈ R . Applying the comparison principle (Lemma A.1), we obtain that w ( τ, z ) = v ( τ + δ, z ) ≥ v ( τ, z ) , τ ∈ (cid:0) , θ T / − δ (cid:1) , z ∈ R . Thus we have v τ ≥≥