On corrected phase-type approximations of the time value of ruin with heavy tails

Remark 1.

With i∈{1,2}, suppose qi are positive constants and that the functions αi and βi from [0,∞) to [0,∞) are eventually positive. If α1=o⁢(α2) and α2=O⁢(β2), then

satisfy α=O⁢(β).

Proof of Proposition 2.

First consider δ>0. We must show pϵ∈𝒮d⁢(0). Following pϵ⁢(u)∼ϵ⁢c⁢(u), pϵ∈ℒd⁢(0) results from c∈ℒd⁢(0). As probability densities, c and pϵ are globally integrable. Hence, having assumed that c∈𝒮d⁢(0), we conclude pϵ∈𝒮d⁢(0) also. Because ωϵ⁢(u)∼ϵ⁢ωc⁢(u), ωc∈ℒd⁢(α) implies ωϵ∈ℒd⁢(α) (for all α≥0). Then, for δ>0 and α>0,

The first asymptotic relation follows from [18, Corollary 3.2 (2)].

Next consider δ>0 and α=0. We observed that ωϵ⁢(u)∼ϵ⁢ωc⁢(u). Because we assumed ωb directly Riemann integrable and ωc locally integrable, the assumption ωc∈𝒮d⁢(0) implies the condition ωϵ∈𝒮d⁢(0). Recall the relations that b=o⁢(c) and ωb=o⁢(ωc). Following Remark 1 with q1=1-ϵ and q2=ϵ, the relation c=O⁢(ωc) implies pϵ=O⁢(ωϵ). Likewise, ωc=O⁢(c) implies ωϵ=O⁢(pϵ). Therefore,

The first asymptotic relation follows from [18, Corollary 3.2 (3)].

Now let δ=0. The density pϵ is eventually non-increasing because we assume so for c, and b as a phase-type density is asymptotically Erlangian. We need to have P¯ϵ∈𝒮d⁢(0); first, C¯∈ℒd⁢(0) implies P¯ϵ∈ℒd⁢(0) because P¯ϵ⁢(u)∼ϵ⁢C¯⁢(u). Next, having assumed ηϵ and ηc both finite, the densities P¯ϵ and C¯ are globally integrable, and locally as well. Finally, C¯∈𝒮d⁢(0) implies P¯ϵ∈𝒮d⁢(0). Similarly to δ>0, we find ωϵ∈ℒd⁢(α) for α>0. Then, for δ=0 and α>0,

Here, we first used [18, Corollary 3.2 (5)], and then of course P¯ϵ⁢(u)∼ϵ⁢C¯⁢(u).

Lastly, consider δ=α=0. Because we assumed Ω¯c∈ℒd⁢(0), we have Ω¯ϵ∈ℒd⁢(0) because Ω¯b=o⁢(Ω¯c). Clearly, ωϵ is eventually non-increasing. From Remark 1, the relation C¯=O⁢(Ω¯c) implies P¯ϵ=O⁢(Ω¯ϵ) because B¯=o⁢(C¯). Likewise, Ω¯c=O⁢(C¯) implies Ω¯ϵ=O⁢(P¯ϵ) because Ω¯b=o⁢(Ω¯c). Finally, Ω¯b=o⁢(Ω¯c) with ωc globally integrable and Ω¯b dRi implies that Ω¯ϵ∈𝒮d⁢(0) when Ω¯c∈𝒮d⁢(0). By [18, Lemma 5.2 (2)], global integrability of ωc implies that Ω¯c is locally integrable. Therefore, for δ=α=0,

The first asymptotic relation is justified since we established Assumption 1 implies the conditions of [18, Corollary 3.2 (6)]. ∎

Lemma 2.

Let ρ≥0 and α≥0; assume that h∈Ld⁢(α). We have: if α∨ρ>0, then Tρ⁡h⁢(u)∼1ρ+α⁢h⁢(u) and moreover Tρ⁡h∈Ld⁢(α).

Proof.

The dominated convergence theorem is justified by [18, Lemma 4.1 (1)] and shows the first assertion after applying h∈ℒd⁢(α). The second assertion follows automatically from the first, namely that h∈ℒd⁢(α) implies Tρ⁡h⁢(u-y)∼eα⁢y⁢Tρ⁡h⁢(u) for all real y. ∎

Lemma 3.

Let p be a probability density in the class Ld⁢(γ) for some γ>0. If a∈(0,1), then

for any γ~∈(0,γ).

Proof.

We use induction. For the inductive step, letting γ~∈(0,γ), suppose p∗n⁢(x)=O⁢(e-γ~⁢x), and analyze p∗(n+1)⁢(x)=∫0xp∗n⁢(x-y)⁢p⁢(y)⁢dy. By [18, Lemma 4.1 (2)], for our γ~ there exist a1,x1>0 such that x≥x1 implies p⁢(x)≤a1⁢e-γ~⁢x. By the inductive hypothesis, we may choose a2,x2>0 for which x≥x2 implies p∗n⁢(x)≤a2⁢e-γ~⁢x. Now set x0=x1∨x2 and let x≥2⁢x0. Split p∗(n+1)⁢(x) into

and

For I1⁢(x), we use the variable change z=x-y. Observing that x-z≥x-x0≥x0, we find

For I2⁢(x), note that the limits of integration imply x-y≥x0. It follows that

By [18, Lemma 4.1 (3)], p^⁢(-γ~) is finite. Then for x≥2⁢x0, we have the inequality p∗(n+1)⁢(x)≤(A1+A2)⁢e-γ~⁢x. Hence, by induction p∗n⁢(x)=O⁢(e-γ~⁢x) for all n∈ℕ+. By a simpler induction argument, ∑i=1nai⁢p∗i⁢(x)=O⁢(e-γ~⁢x) for all n∈ℕ+. Letting n→∞ establishes the lemma. ∎

Lemma 4.

Let δ≥0. Choose an R∈(0,γ) satisfying l∘⁢(-R)=0. Then mb′⁢(u) is in Ld⁢(R).

Proof.

By Lemma 2, g∘⁢(x)=λ⁢𝒯ρ∘⁡P∘⁢(x) is in ℒd⁢(γ) for all δ≥0. Because g∘ is defective, Lemma 3 implies that ∑n=1∞g∘∗n⁢(x)=O⁢(e-γ~⁢x) for all γ~∈(0,γ). So, choose γ~ such that γ~∈(R,γ). Now we examine δ>0 and δ=0 separately.

For δ>0, because γ>R, [18, Lemma 4.3 (1)] implies b∗m∘⁢(u)∼m∘⁢(u)⁢b^⁢(-R). Thus, b∗m∘∈ℒd⁢(R), and likewise b∗m∘-m∘∈ℒd⁢(R) by closure of ℒd⁢(R). Applying Lemma 2 to b∗m∘-m∘ shows us that λ⁢Tρ∘⁡(b∗m∘-m∘)∈ℒd⁢(R) also. By the definition of m∘ and the fact that ωb=ω∘, we see

Because ∑n=1∞g∘∗n⁢(x)=O⁢(e-γ~⁢x) for all γ~∈(R,γ), a second invocation of [18, Lemma 4.3 (1)] shows

Hence mb′∈ℒd⁢(R) because m∘∈ℒd⁢(R). For δ=0, analogously to b∗m∘ we have B¯∗m∘∈ℒd⁢(R). With B¯∗m∘ replacing b∗m∘-m∘, the reasoning for (A.1) demonstrates

Hence mb′ is in ℒd⁢(R) for δ=0 also. ∎

Proof of Theorem 3.

First let δ>0. We find that c∗m∘∈𝒮d⁢(0) as follows. Since m∘⁢(u)=O⁢(e-R⁢u) for some R>0, we see

from [18, Lemma 4.3 (1)]. For any y∈ℝ, similarly

But by the definition of ℒd⁢(0), c satisfies c⁢(u)∼c⁢(u-y) for any y∈ℝ. Then clearly c∗m∘⁢(u)∼c∗m∘⁢(u-y) such that c∗m∘∈ℒd⁢(0). Because c and m∘ are locally integrable and c∗m∘⁢(u)∼c⁢(u)⁢m^∘⁢(0), we see that c∈𝒮d⁢(0) implies c∗m∘∈𝒮d⁢(0). Likewise, since Lemma 2 shows that Tρ∘⁡c∗m∘⁢(u)∼1ρ∘⁢c∗m∘⁢(u), we observe Tρ∘⁡c∗m∘∈𝒮d⁢(0).

Consider the other summands in hc′. Since δ>0, Lemma 2 implies Tρ∘⁡ωc∈ℒd⁢(α) for all α≥0. Meanwhile, Tρ∘⁡m∘ is in ℒd⁢(R) and so o⁢(c∗m∘), such that hc′⁢(u)∼λρ∘⁢(ωc+c∗m∘)⁢(u). If α>0, then ωc is also o⁢(c∗m∘) and disappears from the right-hand side. Because m^∘⁢(0)=λδ⁢ρ∘⁢Ω¯^∘⁢(ρ∘), finally we get

The second line of (A.2) follows because ωc⁢(u)+c∗m∘⁢(u)∼ωc⁢(u)+c⁢(u)⁢m^∘⁢(0) when α=0. As u→∞,

For, c and ωc being in ℒd⁢(0) are eventually positive, so that ωc⁢(u)c⁢(u)⁢m^∘⁢(0)>0. The zero limit of course follows from c∗m∘⁢(u)∼c⁢(u)⁢m^∘⁢(0).

Next, hc′∈ℒd⁢(0) for α≥0 is locally integrable. The compound geometric term with g∘⁢(x) is O⁢(e-γ~⁢x). So, we may apply [18, Lemma 4.3 (1)]. We find

such that mc′⁢(u)∼ρ∘δ⁢hc′⁢(u). Combining (A.2) with (A.4) establishes Theorem 3 when δ>0.

Now let δ=0. For α≥0, our integrability assumptions on ωc imply local integrability of Ω¯c by [18, Lemma 5.2 (1) and (2)]. Hence hc′ is locally integrable also. Corresponding to (A.2), we find with

that

For (A.5), analogously to δ>0, C¯∗m∘∈𝒮d⁢(0) follows from having C¯∈𝒮d⁢(0) with ηc<∞ and m∘∈ℒd⁢(R) with R>0. When α>0, ωc∈ℒd⁢(α) implies Ω¯c∈ℒd⁢(α) by Lemma 2. Hence, Ω¯c is o⁢(C¯∗m∘). When α=0, the analogue of (A.3) is that Ω¯c⁢(u)+C¯∗m∘⁢(u)∼Ω¯c⁢(u)+C¯⁢(u)⁢m^∘⁢(0). Because the term hc′ is in ℒd⁢(0) for α≥0, the analogue of (A.4) holds. Namely, mc′⁢(u)∼11-ψ∘⁢(0)⁢hc′⁢(u). Combined with (A.5), this establishes Theorem 3 for δ=0. ∎