Portfolio immunization under cone restrictions

Proof of Theorem 3.1.

(Sufficiency) If 𝔛⁢(𝟙B⁢𝐞j)∈𝒞ℝN for every B∈ℬ and j∈{1,…,n}, then, by properties of cones and since 𝔛 is a linear operator, we have that 𝔛⁢(∑m=1kbm⁢𝟙Bm⁢𝐞j)∈𝒞ℝN for all B1,…,Bk∈ℬ and bm≥0, as well. Moreover, for any random variable g such that P(g≥0)=1, there exists a sequence of nonnegative functions (gk) such that g=limk→∞⁡gk and gk=∑m=1kbm⁢𝟙Bm, where bm are nonnegative constants. Since 𝔛 is continuous,

Hence, 𝔛⁢(g⁢𝐞j)∈𝒞ℝN for every j=1,…,n, because 𝒞ℝN is closed. Every random vector 𝐟∈𝒞ℋ can be represented as the sum

where 𝐟j=gj⁢𝐞j, j=1,…,n, and gj are some random variables such that P(gj≥0)=1. Since 𝔛 is linear, we have that

and 𝔛⁢(𝐟)∈𝒞ℝN. So, 𝔛⁢(𝒞ℋ+)⊆𝒞ℝN.

(Necessity) Assume that 𝔛 is monotonic. Since for any B∈ℬ and j=1,…,n, we have that 𝟙B⁢𝐞j∈𝒞ℋ+, we obtain that 𝔛⁢(𝟙B⁢𝐞j)∈𝒞ℝN, by the monotonicity of 𝔛. ∎

Proof of Theorem 3.3.

(i) (Necessity) Assume that 𝔛 is monotonic and let 𝐥∈𝒞ℋ. Then 𝔛⁢(𝐥)∈𝒞ℝN, which means that 𝔛i⁢(𝐥)≥0 for all i=1,…,N. Equivalently, 𝔛i∈𝒞ℋ* for every i=1,…,N.

(Sufficiency) Assume now that 𝔛i∈𝒞ℋ* for every i=1,…,N. This means that 𝔛i⁢(𝐥)≥0 for every 𝐥∈𝒞ℋ and i=1,…,N. Hence, 𝔛⁢(𝐥)∈𝒞ℝN, so 𝔛⁢(𝒞ℋ)⊆𝒞ℝN.

(ii) First, let us note that if, for all i=1,…,N, 𝔛i is uncorrelated with every 𝐥∈𝒞ℋ, then

(Necessity) Assume that 𝔛 is monotonic, that is, 〈𝔛i,𝐥〉≥0 for all i=1,…,n and 𝐥∈𝒞ℋ. Let 𝐲∈E⁢𝒞ℋ. Then there exists 𝐥𝐲∈𝒞ℋ such that 𝐲=E⁢𝐥𝐲 and, by (A.1), we have

which is nonnegative. Hence, E⁢𝔛i∈(E⁢𝒞ℋ)⋄ for i=1,…,N.

(Sufficiency) Assume that E⁢𝔛i∈(E⁢𝒞ℋ)⋄ for i=1,…,N, that is, 〈E⁢𝔛i,𝐲〉≥0 for all 𝐲∈E⁢𝒞ℋ. Let 𝐥∈𝒞ℋ. By (A.1), we have

as E⁢𝐥∈E⁢𝒞ℋ. Hence, 𝔛⁢(𝐥)∈𝒞ℝN, so 𝔛⁢(𝒞ℋ)⊆𝒞ℝN. ∎

Proof of Lemma 3.6.

Let 𝐱∈E⁢(E⁢𝒞ℋ)*. This means that there exists 𝐱*∈ℋ such that 〈𝐱*,E⁢𝐥〉≥0 for all 𝐥∈𝒞ℋ and 𝐱=E⁢𝐱*. Hence, for every 𝐥∈𝒞ℋ, we have

so 𝐱∈(E⁢𝒞ℋ)⋄, which means that E⁢(E⁢𝒞ℋ)*⊆(E⁢𝒞ℋ)⋄.

Let 𝐱∈(E⁢𝒞ℋ)⋄. This means that for every 𝐥∈𝒞ℋ, we have 〈𝐱,E⁢𝐥〉≥0. Since 𝐱∈ℝn⊂ℋ, it follows that 𝐱∈(E⁢𝒞ℋ)* and 𝐱∈E⁢(E⁢𝒞ℋ)* because 𝐱 is nonrandom. Hence, (E⁢𝒞ℋ)⋄⊆E⁢(E⁢𝒞ℋ)*.

Now, let 𝐱∈(E⁢𝒞ℋ)⋄, that is, 〈𝐱,E⁢𝐥〉≥0 for all 𝐥∈𝒞ℋ. Then, for every 𝐥∈𝒞ℋ,

as 𝐱∈ℝn. So, 𝐱∈𝒞ℋ* and 𝐱∈E⁢𝒞ℋ* because 𝐱 is nonrandom. This means that (E⁢𝒞ℋ)⋄⊆E⁢𝒞ℋ*. ∎

Proof of Lemma 3.8.

(Necessity) Assume that (E⁢𝒞ℋ)⋄=E⁢𝒞ℋ*. This means that 〈E⁢𝐚,E⁢𝐥〉≥0 for every 𝐚∈𝒞ℋ* and 𝐥∈𝒞ℋ. Equivalently, 〈E⁢𝐚,𝐥〉≥0, so E⁢𝐚∈𝒞ℋ* for every 𝐚∈𝒞ℋ*. Hence, E⁢𝒞ℋ*⊆𝒞ℋ*.

(Sufficiency) Assume that E⁢𝒞ℋ*⊆𝒞ℋ*. This means that, for all 𝐚∈𝒞ℋ* and 𝐥∈𝒞ℋ, 〈E⁢𝐚,𝐥〉≥0. Equivalently, 〈𝐚,E⁢𝐥〉≥0 for all 𝐚∈𝒞ℋ* and 𝐥∈𝒞ℋ, so 𝒞ℋ*⊆(E⁢𝒞ℋ)*. By elementary properties of cones and Lemma 3.6, we have E⁢𝒞ℋ*⊆E⁢(E⁢𝒞ℋ)*=(E⁢𝒞ℋ)⋄. The opposite inclusion follows from Lemma 3.6. Hence, E⁢𝒞ℋ*=(E⁢𝒞ℋ)⋄. ∎

Proof of Lemma 3.9.

First, we will show that E⁢𝒞ℋ*⊆(𝒞ℋ∩ℝn)⋄. Let 𝐱∈E⁢𝒞ℋ*. Then there exists 𝐚*∈𝒞ℋ* such that 𝐱=E⁢𝐚*. For any 𝐥∈𝒞ℋ∩ℝn, we have that

as 𝐚*∈𝒞ℋ*. This means that 𝐱∈(𝒞ℋ∩ℝn)⋄. Hence, E⁢𝒞ℋ*⊆(𝒞ℋ∩ℝn)⋄.

If, additionally, E⁢𝒞ℋ⊆𝒞ℋ, then E⁢𝒞ℋ=𝒞ℋ∩ℝn. Indeed, an easy observation yields 𝒞ℋ∩ℝn⊆E⁢𝒞ℋ. On the other hand, E⁢𝒞ℋ⊆ℝn and E⁢𝒞ℋ⊆𝒞ℋ, by assumption, so E⁢𝒞ℋ⊆𝒞ℋ∩ℝn.

Let us note that assumption E⁢𝒞ℋ⊆𝒞ℋ implies that E⁢𝒞ℋ*⊆𝒞ℋ*. Indeed, let 𝐚∈𝒞ℋ*. For every 𝐥∈𝒞ℋ, we have that 〈𝐚,𝐥〉≥0 and 〈𝐚,E⁢𝐥〉≥0, since E⁢𝐥∈𝒞ℋ. Hence,

for every 𝐥∈𝒞ℋ, which means that E⁢𝐚∈𝒞ℋ* and E⁢𝒞ℋ*⊆𝒞ℋ*. By Lemma 3.8, we have E⁢𝒞ℋ*=(E⁢𝒞ℋ)⋄, which, together with E⁢𝒞ℋ=𝒞ℋ∩ℝn, implies that E⁢𝒞ℋ*=(𝒞ℋ∩ℝn)⋄. ∎

Proof of Lemma 3.10.

Let 𝐛∈(𝒞ℋ∩ℝn)⋄⊆ℝn, that is, 〈𝐛,𝐥〉≥0 for all 𝐥∈𝒞ℋ∩ℝn. By the Riesz extension theorem, there exists 𝐛*∈ℋ such that 𝐛*|ℝn=𝐛 and 〈𝐛*,𝐥〉≥0 for all 𝐥∈𝒞ℋ. Let 𝐱∈ℝn. We have

which means that 〈E⁢𝐛*-𝐛,𝐱〉=0 for all 𝐱∈ℝn. Hence, E⁢𝐛*=𝐛 and 𝐛∈E⁢𝒞ℋ*. This means that

The opposite inclusion follows from the first assertion in Lemma 3.9. ∎