Existence of a solution

C. Hipp, M. Plum Insurance: Mathematics and Economics 27 2000 215–228 219 Now ε → 0 yields δ ∗ s ≥ δs. On the other hand, δ ∗ s + ε = Eδ ∗ T s + ε, θ ∗ , t ∧ τ ∗ which, with t tending to infinity, implies δ ∗ s + ε ≤ P {τ ∗ = ∞}.

3. Existence of a solution

If δs is a solution of 1.3, then for a constant α, the function αδs is a solution, too. Hence, we may fix δ0 and replace the resulting function δs by αδs if it is necessary to achieve the property δs → 1 for s → ∞. From A0 = 0, we derive δ ′ = δ0λc 0. We shall use the norming δ ′ = 1. Substituting the maximizing As, we obtain the equation λ Z ∞ [δs − x − δs]Qdx + cδ ′ s = a 2 2b 2 δ ′ s 2 δ ′′ s . Substituting λ and c by λb 2 a 2 and cb 2 a 2 , respectively, and denoting these new constants with the same symbols λ and c, we obtain the standard form λ Z ∞ [δs − x − δs]Qdx + cδ ′ s = 1 2 δ ′ s 2 δ ′′ s . With H t = Qt, ∞ and integration by parts Z ∞ [δs − x − δs]Qdx = −δ0H s − Z s δ ′ s − xH x dx, we transform this equation into δ ′′ s −λ Z s δ ′ s − xH x dx + cδ ′ s − H s = 1 2 δ ′ s 2 . Hence our Bellman equation is equivalent to the following problem for u = δ ′ : u ′ s −λ Z s us − xH x dx + cus − H s = 1 2 us 2 , u0 = 1. 3.1 Theorem 3.1. Let Q have a locally bounded density. Then there exists a solution u ∈ C 1 0, ∞ ∩ C[0, ∞ of problem 3.1 satisfying u 0, u ′ 0 on 0, ∞, and us = 1 − r s c + o √ s as s → 0. If moreover H has a finite integral over [0, ∞, then also u has a finite integral. 220 C. Hipp, M. Plum Insurance: Mathematics and Economics 27 2000 215–228 Using the above transformation δ ′ = u and the condition δ ′ = δ0λc, we immediately obtain from Theorem 3.1, the following corollary. Corollary 3.2. If Q has a locally bounded density, there exists a positive, strictly increasing and strictly concave solution δ ∈ C 2 0, ∞ ∩ C 1 [0, ∞ of the Bellman equation satisfying δs = c λ + s − 2 3 √ c s 32 + os 32 as s → 0. If, moreover, H has a finite integral over [0, ∞, δ is bounded on [0, ∞. Proof of Theorem 3.1. a First we prove the existence of a solution u with the asserted properties on [0, ε 2 ] for some ε 0. Via the transformation, vs = us 2 , we obtain the equivalent problem v ′ sφ[v]s = vs 2 , v0 = 1, 3.2 where φ[v]s = −2λs Z 1 tvstH s 2 1 − t 2 dt + c vs − H s 2 s . Defining Q ε [v] : = sup 0s ≤ε 1 s |v ′ s − v ′ | for ε 0, v ∈ C 1 [0, ε], we find that R ε : = {v ∈ C 1 [0, ε] : Q ε [v] ∞}, endowed with the norm kvk ε : = max {kvk ∞ , |v ′ |, εQ ε [v] }, is a Banach space, and D ε,M : = v ∈ R ε : v0 = 1, v ′ = − 1 √ c , kv − 1k ∞ ≤ 1 3 , Q ε [v] ≤ M is a closed subset. A lengthy but elementary calculation shows that the operator T defined by T vs : = 1 + Z s vx 2 φ[v]x dx, v ∈ D ε,M , s ∈ [0, ε], maps D ε,M into itself and is a contraction on D ε,M with respect to k · k ε , if M 0 is sufficiently large and ε 0 is sufficiently small. Therefore, Banach’s Fixed-Point Theorem provides the existence of a fixed point v ∈ D ε,M of T and thus, of a solution of 3.2 on [0, ε]. For reasons of continuity, we have v 0 and v ′ 0 on [0, ε] after possibly further reduction of ε. Moreover, vs = 1 − s √ c + os as s → 0. Thus, us = v √ s has the corresponding properties on 0, ε 2 ]. b Next, we show that if u ∈ C 1 0, b ∩ C[0, b is a solution of 3.1 on [0, b for some b 0, satisfying us 0, ψ [u]s : = −λ Z s us − xH x dx + c[us − H s] 0 3.3 for all s ∈ 0, b, then u can be uniquely extended to a solution on [0, b], and 3.3 also holds for s = b. Since 3.1 and 3.3 imply u ′ 0 on 0, b, the existence of ub : = lim s →b us with 0 ≤ ub 1 and thus of the limit ψ[u]b : = lim s →b ψ [u]s follows. We only have to prove ub 0 and ψ[u]b 0, since 3.1 then provides the existence of u ′ b ∈ −∞, 0, so that the extended u indeed solves 3.1 on [0, b]. First, assume for contradiction that ub = 0. Then, lim s →b 1us = +∞, so some sequence s k → b exists such that lim k →∞ 1u ′ s k = +∞. Since, due to 3.1, 1 u ′ = − u ′ u 2 = − 1 2ψ[u] , C. Hipp, M. Plum Insurance: Mathematics and Economics 27 2000 215–228 221 we have lim k →∞ ψ [u]s k = 0 and thus, using ub = 0, −λ Z b ub − xH x dx − cHb = 0, which contradicts λ 0, u 0 on [0, b, c 0, H ≥ 0, H 6= 0 on [0, b]. Therefore, ub 0. Assuming ψ[u]b = 0 for contradiction, we obtain from 3.1 and 3.3, and ub 0 that lim s →b u ′ s = −∞. Since H has a locally bounded derivative, this implies lim s →b ψ [u] ′ s = −∞, contradicting 3.3 and ψ [u]b = 0. c The third step is to prove that each solution u ∈ C 1 0, b] ∩ C[0, b] of 3.1 on [0, b] for some b 0, satisfying 3.3 for all s ∈ 0, b], can be uniquely extended to a solution on [0, b + η] for some η 0, and 3.3 holds for all s ∈ 0, b + η]. The proof is by Banach’s Fixed-Point Theorem again, this time using the Banach space C[b, b + η], k · k ∞ , the closed subset D η,ρ : = {v ∈ C[b, b + η] : vb = ub, kv − ubk ∞ ≤ ρ}, and the operator T vs : = ub + Z s b vx 2 2 ˆ ψ [v]x dx, v ∈ D η,ρ , s ∈ [b, b + η], where ˆ ψ [v]s : = −λ Z b uxH s − x dx + Z s b vxH s − x dx + c[vs − H s]. Elementary calculations show that T maps D η,ρ into itself and is a contraction on D η,ρ if both η and ρ are sufficiently small. The unique fixed point v ∈ D η,ρ of T obtained from Banach’s Fixed-Point Theorem provides the desired unique extension of u. If η is chosen sufficiently small, 3.3 holds on 0, b + η]. d To prove the existence of a solution u on [0, ∞ with the properties asserted in the theorem, let u denote some solution on [0, ε 2 ] provided by a. Due to u ′ 0 and 3.1, u satisfies 3.3 on 0, ε 2 ]. Now denote by b ∗ , the supremum of all b ∈ [ε 2 , ∞ such that there exists a unique extension of u to a solution u b of 3.1 on [0, b], which satisfies 3.3 on 0, b]. The uniqueness requirement yields u b = u ˜b | [0,b] for ε 2 ≤ b ≤ ˜b b ∗ . Therefore, defining u ∈ C 1 0, b ∗ ∩ C[0, b ∗ by u | [0,b] : = u b , b ∈ [ε 2 , b ∗ , we obtain a unique extension of u to a solution on [0, b ∗ , which satisfies 3.3 and thus, u ′ 0 on 0, b ∗ . Assuming b ∗ ∞, we obtain from b and c that u can be uniquely extended to a solution on [0, b ∗ + η] satisfying 3.3 on 0, b ∗ + η]. This contradicts the supremum property of b ∗ . Thus, b ∗ = ∞, which proves the desired existence statement. e To show the additional assertion, let H have a finite integral over [0, ∞, and let K := R ∞ H s ds. Since H is moreover a decreasing function, standard results provide sHs → 0 for s → ∞. 3.4 Since u 0, u ′ 0 on 0, ∞, 3.1 provides with ψ[u] defined in 3.3 −ψ[u]s = − us 2 2u ′ s = 1 2 1 1u ′ s 0 s ∞. 3.5 We will prove i us → 0, ii sus → 0, iii s 32 us → 0 for s → ∞, and iii then provides that u has a finite integral over [0, ∞. To show i, we observe that u 0, u ′ 0 yield the existence of u ∞ : = lim s →∞ us ≥ 0 and of a sequence s k → ∞ such that lim k →∞ u ′ s k = 0. Assuming 222 C. Hipp, M. Plum Insurance: Mathematics and Economics 27 2000 215–228 u ∞ 0, we obtain from 3.1 that lim k →∞ ψ [u]s k = −∞, contradicting the fact that ψ[u] ≥ −λK − c. Thus, u ∞ = 0. To prove ii, we estimate, assuming that u and H are decreasing, −ψ[u]s ≤ λ Z s2 us − xH x dx + λ Z s s2 us − xH x dx + cHs ≤ λu 1 2 sK + 1 2 λsH 1 2 s + cHs, so that i and 3.4 yield lim s →∞ ψ [u]s = 0. From de l’Hospital’s rule and 3.5, we therefore obtain lim s →∞ sus = lim s →∞ s 1us = lim s →∞ 1 1u ′ s = −2 lim s →∞ ψ [u]s = 0. To show iii, we estimate similarly − √ sψ [u]s ≤ λ √ s Z s − √ s us − xH x dx + λ √ s Z s s − √ s us − xH x dx + c √ sH s ≤ λ √ su √ sK + λsHs − √ s + c √ sH s, so that ii and 3.4 yield lim s →∞ √ sψ [u]s = 0. De l’Hospital’s rule and 3.5 now provide lim s →∞ s 32 us = lim s →∞ s 32 1us = 3 2 lim s →∞ √ s 1u ′ s = −3 lim s →∞ √ sψ [u]s = 0.

4. Unboundedness of As