Concave envelopes getdocecde. 254KB Jun 04 2011 12:05:09 AM

If f x y, b = 0, then α ϕ x = α ϕ y, so we are led to the contradictory statement d F α ϕ x 0 and − d F α ϕ x 0. If f x y, b 6= 0, rearranging terms and adding the resulting equations as before, we also obtain a contradiction. If y = a and z b, assume again that the claim of the lemma is false, i.e. η x a = ρx = d G α ψ x d G β ψ x α ψ x − α ψ z β ψ x − β ψ z = f x a, z, η x z = d G α ψ x − α ψ z−α ψ x Gz −Gx d G β ψ x − β ψ z−β ψ x Gz −Gx α ψ x − α ψ z β ψ x − β ψ z = f x a, z. Here we have used the representation of η x involving G and ψ stated in 8. Since β ψ = 1ψ is G-concave and G is decreasing, this leads to a contradiction as before. Finally, if y = a, z = b, since ψ ′ 0 and ϕ ′ 0, clearly min {η x a, η x b} = min ux − u ′ x ψx ψ ′ x , ux − u ′ x ϕx ϕ ′ x ≤ ux. Proof. Theorem 13. By Lemma 10 and Lemma 14, it is immediate that for all x ∈ a, b γx = inf a ≤ yxz≤b f x y, z ≥ inf y 6=x η x y. To see the reverse inequality, note that for z x, η x z = lim u ↑x f x u, z ≥ inf a ≤ux f x u, z ≥ γx. Similarly, for y x, η x y = lim u ↓x f x y, u ≥ inf x u≤b f x y, u ≥ γx. Thus, inf y 6=x η x y ≥ γx.

3.2 Concave envelopes

According to Theorem 13, the computation of the universal stopping signal γ amounts to minimizing the function η x given by 12 for every x ∈ J . In the sequel, we show that under certain concavity conditions on u, the infimum of η x is attained at the boundaries a or b of J which yields a closed form solution for γx since η x a = κx see 10 and η x b = ρx see 11. To this end, if ς: J → R is a strictly monotone function, denote by bu ς the ς-concave envelope of u, i.e. bu ς x = inf{ f x : f ≥ u and f ς-concave}. 1981 Provided that bu ς is finite, bu ς is itself ς-concave. We remark that bu ς x = Ù u ◦ ς −1 ςx where bu = bu id refers to the usual concave envelope. With this concept at hand, we may now state the main result of this section. Theorem 15. Let x ∈ J . Under Assumptions 1 - 5, we have {γ = ρ} = {u = bu ψ } ∩ {u ′ ≥ 0}, and {γ = κ} = {u = bu ϕ } ∩ {u ′ ≤ 0}. In other words, γx = ρx resp. γx = κx iff the ψ-concave resp. ϕ-concave envelope of u coincides with the function u itself and u ′ x ≥ 0 resp. u ′ x ≤ 0. Note that ρx = κx = ux if u ′ x = 0. In order to prove Theorem 15, we need the following lemma. Lemma 16. Let ς: J → R be differentiable and strictly monotonic. 1. Then u is ς-concave iff for all x, y ∈ J ux + d ς ux ς y − ςx ≥ u y. 2. Let x ∈ J . Then x ∈ {u = bu ς } if and only if u y ≤ ux + d ς ux ς y − ςx for all y ∈ J . 13 Proof. 1. Note that u is ς-concave on J if and only if u ◦ ς −1 is concave on ςJ . This is equivalent to u ς −1 x + d d x u ς −1 x · y − x ≥ uς −1 y for all x, y ∈ ςJ . The claim follows upon noting that for all x ∈ ςJ d d x u ς −1 x = u ′ ς −1 x ς ′ ς −1 x = € d ς u Š ς −1 x. 2. Assume that x ∈ {u = bu ς }. By definition of the ς-concave envelope, the differentiability of u and ς and the first part of the lemma, it must hold for all y that u y ≤ bu ς y ≤ bu ς x + d ς bu ς xς y − ςx = ux + d ς ux ς y − ςx. On the other hand, the ς-concave function f y ¬ ux + d ς ux ς y − ςx satisfies f x = ux and f ≥ u if 13 holds. This implies ux = bu ς x. Proof. Theorem 15 Let x, y ∈ J . For y x, η x y ≥ κx is equivalent to α ϕ x − α ϕ y F x − F y − d F α ϕ x ≥ κx · ‚ β ϕ x − β ϕ y F x − F y − d F β ϕ x Œ . 1982 We have used the fact that β ϕ = 1ϕ is F -concave Remark 5, F is increasing and part 1 of Lemma 16. Since κx = d F α ϕ xd F β ϕ x, we may write η x y ≥ κx ⇐⇒ α ϕ x − α ϕ y F x − F y ≥ κx · β ϕ x − β ϕ y F x − F y . Some algebra shows that this is further equivalent to ux + d ϕ ux ϕ y − ϕx ≥ u y. If y x, a similar calculation leads to the same result showing that η x y ≥ κx = η x b for all y ∈ J if and only if ux + d ϕ ux ϕ y − ϕx ≥ u y for all y ∈ J . 14 In view of part 2 of Lemma 16, we deduce from 14 that κx ≤ η x y for all y ∈ J if and only if x ∈ {u = bu ϕ }. Now κx ≤ η x a = ρx if and only if ux − u ′ x ϕx ϕ ′ x ≤ ux − u ′ x ψx ψ ′ x which is equivalent to u ′ x ≤ 0. This proves the first assertion of the theorem. The proof of the second claim is along the same lines. For a y x, η x y ≥ ρx is equivalent to α ψ x − α ψ y Gx − G y − d G α ψ x ≥ ρx · ‚ β ψ x − β ψ y Gx − G y − d G β ψ x Œ . Since ρ = d G α ψ d G β ψ , η x y ≥ ρx for y x is further equivalent to α ψ x − α ψ y ≥ ρxβ ψ x − β ψ y. One can show that η x y ≥ ρx for all y ∈ J if and only if ux + d ψ ux ψ y − ψx ≥ u y for all y ∈ J . 15 Since ρx ≤ κx = η x a iff u ′ x ≥ 0, the proof is complete in view of 15 and Lemma 16. We can now prove that a monotone stopping signal γ is equal either to κ or to ρ. Thus, by Theorem 15, a nondecreasing resp. nonincreasing signal corresponds to a nondecreasing resp. nonincreas- ing reward function u that is ψ-concave resp. ϕ-concave. In order to prove this claim, we first show that γ is upper semi-continuous. Lemma 17. The function γ is upper semi-continuous on J . Proof. For fixed k, V k · is continuous on J . Moreover, for fixed x ∈ J , k 7→ V k x is Lipschitz con- tinuous with Lipschitz constant 1 which in turn yields that k, x 7→ V k x is continuous. Therefore, the hypograph {x, k : γx ≥ k} = {x, k : V k x = ux − k} Lemma 9 is closed or equivalently, γ is upper semi-continuous. 1983 Corollary 18. Under Assumptions 1 - 5, γ is nonincreasing on J if and only if γ = κ and γ is nondecreasing on J if and only if γ = ρ. Proof. Assume that γ is nonincreasing. By Lemma 17, γ must be left-continuous. Let x ∈ J and set k = γx. If {γ k} = ;, we have that γ y = k for all y ∈ a, x] and V l y = u y − l, l ≤ k = γ y, 0, l γ y. Since k 7→ V k y is continuous for any y ∈ J , we deduce that u y − k = 0 for all y ∈ a, x]. In particular, u ′ x = 0 and therefore, κx = ux = k = γx. Assume next that {γ k} 6= ;. If {γ = k} = {x}, then for y x, we have due to Lemma 1 that V k y = E y ” e −R Tx uX T x − k — = ϕ y ϕx ux − k = sup v y E y ” e −R Tv uX T v − k — = sup v y ϕ y ϕv uv − k. Set gv ¬ ϕ yϕvuv − k for v ≤ y. Note that g y = u y − k V k y since y ∈ Γ k . Thus, g attains its maximum at the interior point x and x must satisfy g ′ x = 0 which is equivalent to κx = k. By definition, also γx = k. Next, consider the case {γ = k} = x 1 , x 2 ] where a x 1 x 2 b and x ∈ x 1 , x 2 ]. For all y ∈ x 1 , x 2 ], we have that V k y = u y − k since γ y = k amounts to stopping immediately. Moreover, we claim that ˆ τ k ¬ inf {t : γX t k} = T x 1 is also an optimal stopping time for the parameter k if the diffusion is started at y ∈ x 1 , x 2 ]. Indeed, since {γ k} 6= ;, there is ˆa such that γˆa k. Set ε = γˆa − γx 0. Then τ k+ εn ↓ ˆ τ k P y -a.s. for any y ∈ J and X τ k+ εn ∈ [ˆa, x P y -a.s. for any y ˆ a. Applying Fatou, we obtain V k y = lim sup n →∞ V k+ εn y = lim sup n →∞ E y • e −Rτ k+ εn uX τ k+ εn − k + ε n ˜ ≤ E y ” e −Rˆ τ k uX ˆ τ k − k — ≤ V k y, which proves the claim that ˆ τ k is optimal. Thus, for any y ∈ x 1 , x 2 ], we have that V k y = u y − k = E y h e −RT x1 uX T x1 − k i = ϕ y ϕx 1 ux 1 − k, showing that u is an affine transformation of ϕ on x 1 , x 2 ], i.e. u y = c ·ϕ y+ k for some constant c = cx 1 0. Now κx = ux − u ′ x ϕx ϕ ′ x = k = γx. The same reasoning applies in the remaining case {γ = k} = x 1 , b. 1984 On the other hand, if γ = κ on J , u must be nonincreasing and ϕ-concave on J due to Theorem 15. By Lemma 16, we have for all x, y ∈ J that ux + d ϕ ux ϕ y − ϕx ≥ u y which we may rewrite as κx = ux − d ϕ ux ϕx ≥ u y − d ϕ ux ϕ y. 16 Note that d ϕ ux = u ◦ ϕ −1 ′ ϕx. Since u ◦ ϕ −1 is concave, its derivative is nonincreasing showing that d ϕ u · is nondecreasing. Thus, if x ≤ y, we see from 16 that κx ≥ κ y. The proof of the second assertion is similar. Let us further investigate the structure of the stopping signal γ if the reward function u is monotone. For instance, if u is nonincreasing, we know from Theorem 15 that γx = κx iff ux = bu ϕ x. Since the set {bu ϕ u} is open, it is the countable union of disjoint open intervals which we denote by l n , r n , n ∈ N. In this setting, we have the following result: Proposition 19. If u is nonincreasing on J and [l n , r n ] ⊂ J , then γl n = κl n = κr n = γr n and γx κl n for all x ∈ l n , b n . Proof. By Theorem 15, we have γl n = κl n and γr n = κr n . From the definition of the ϕ- concave envelope, we see that d ϕ ul n = d ϕ ur n , bu ϕ x = ul n + d ϕ ul n ϕx − ϕl n , x ∈ [l n , r n ]. This implies that κl n = κr n . It remains to show that γx κl n for all x ∈ l n , b n . Indeed, if γx ≥ κl n or equivalently, x ∈ Γ κl n for some x ∈ l n , b n , then V κl n x = ux − κl n ≥ E x h e −RT ln uX T ln − κl n i = ϕx ϕl n ul n − κl n = d ϕ ul n ϕx. This is equivalent to ux ≥ ul n + d ϕ ul n ϕx − ϕl n = bu ϕ x, which is impossible since x ∈ {bu ϕ u}. In particular, the function κ : J → R defined by κx = κx, x ∈ {u = bu ϕ }, κl n , x ∈ l n , r n for some n, defines a nonincreasing majorant of γ and coincides with the optimal stopping signal on {u = bu ϕ }. Note that if the diffusion starts at x ∈ {u = bu ϕ }, optimal stopping times can be derived directly from κ. Thus, the computation of γ on the intervals l n , r n as the infimum of η x as suggested by Theorem 13 is not required in that case. Of course, an analogous result holds if u is nondecreasing. Again, {bu ψ u} is the countable union of disjoint open intervals denoted by l n , r n and we have 1985 Proposition 20. If u is nondecreasing on J and [l n , r n ] ⊂ J , then γl n = ρl n = ρr n = γr n and γx ρl n for all x ∈ l n , b n . 4 Applications and illustrations For simplicity, we illustrate our results for a constant discount rate r 0 which implies that R t = r t. Examples with random discounting can be found in [Dayanik2008] and [Beibel and Lerche2000].

4.1 Optimal stopping