Additive portfolio improvement and utility-efficient payoffs

Abstract

How can individual financial contracts be improved in an additive manner, such that any portfolio comprising improved contracts is at least as attractive as the portfolio of original contracts? We show that any additive procedure that improves contracts for all expected utility maximizers is a conditional expectation operator. Improved contracts are also attractive under robust Savage preferences. Furthermore, we generalize Bondarenko’s definition of ‘statistical arbitrage’ and show that the improved contracts do not admit this kind of arbitrage.

Appendix: Proofs

1.1 Proofs from Section 2

Proof

of Lemma 1

We show that any additive and cautious function is linear and monotonic, and leaves the constant functions invariant. The reverse implication holds a fortiori.

Additivity implies $A(0) = A(0+0) = A(0) + A(0)$ and therefore $A(0) = 0$.

Furthermore, $0 = A(0) = A(X - X) = A(X) + A(-X)$ and therefore $A(-X) = -A(X)$.

Additivity also implies $A(n X) = n\, A(X)$ for all natural numbers n. Thus, $m\, A((n/m) X) = n\, A(X)$ and therefore $A((n/m) X) = (n/m)\, A(X)$ for all positive rational numbers (n / m). That extends to all rational numbers due to $A(-X) = -A(X)$.

Being cautious in particular implies positivity $X \ge 0 \implies A(X) \ge 0$ (i.e., that positive payoffs should be mapped to positive payoffs, because improving a payoff should not introduce the risk of losses, if such a risk did not exist before). Moreover, being cautious implies that deterministic payoffs are not modified, because if $X=c$, we also have $X \le c \implies -X \ge -c \implies -A(X) = A(-X) \ge -c \implies A(X) \le c$.

Given additivity and its consequences, being cautious is equivalent to being both positive and not modifying constants.

Additivity and positivity further imply monotonicity, because if $X \ge Y$, we have $0 \le A(X - Y) = A(X) - A(Y)$ and therefore $A(X) \ge A(Y)$.

Due to monotonicity, we may approximate the positive part $c X^+$ and the negative part $c X^-$ from above and below with rational numbers, and due to additivity, $A((n/m) X) = (n/m)\, A(X)$ and $A(-X) = -A(X)$; we may assert that $A(c X) = c A(X)$ for all real constant numbers c. $\square$

The proof of Theorem 1 is prepared by a lemma.

Lemma 3

The amelioration A is additive (A1), cautious (A2), idempotent (A3) and does not increase expected loss (A4) if and only if it is a linear, monotonic, and contractive projection that leaves constant functions invariant.

Proof

of Lemma 3

First, we show that an additive, cautious and idempotent operator that does not increase expected loss is always a linear, monotonic, and contractive projection that leaves constant functions invariant.

A is contractive (non-expansive), in the sense that ${\mathbb {E}}_{}\left[ {|A(X)|}\right] \le {\mathbb {E}}_{}\left[ {|X|}\right]$, because ${\mathbb {E}}_{}\left[ {A(X)^-}\right] \le {\mathbb {E}}_{}\left[ {X^-}\right]$ implies ${\mathbb {E}}_{}\left[ {A(X)^+}\right] = {\mathbb {E}}_{}\left[ {(-A(X))^-}\right] ={\mathbb {E}}_{}\left[ {A(-X)^-}\right] \le {\mathbb {E}}_{}\left[ {(-X)^-}\right] = {\mathbb {E}}_{}\left[ {X^+}\right]$, and adding the two inequalities yields ${\mathbb {E}}_{}\left[ {|A(X)|}\right] \le {\mathbb {E}}_{}\left[ {|X|}\right]$.

We already know that additive and cautious mappings are always linear, monotonic and leave the constant functions invariant. Since idempotent operators (linear functions) are called projections—and we have just shown that the operator is contractive—the implication holds.

The reverse implication is shown as follows. Linear and monotonic projection that leaves constant functions invariant is a fortiori additive, cautious, and idempotent. Furthermore, monotonicity implies $|A(X^-)| = A(X^-) \ge -A(X)$, therefore $|A(X^-)| \ge \max \{-A(X), 0\} = A(X)^-$. Since the operator is contractive, we get ${\mathbb {E}}_{}\left[ {X^-}\right] = {\mathbb {E}}_{}\left[ {|X^-|}\right] \ge {\mathbb {E}}_{}\left[ {|A(X^-)|}\right] \ge {\mathbb {E}}_{}\left[ {A(X)^-}\right]$. $\square$

Proof

of Theorem 1

We have shown that 1. implies 2. in Lemma 3, the converse holds a fortiori. [1] asserts that 2. implies 3., and the converse holds because of the well-known properties of the conditional expectation. The equivalence of 2. and 3. is known as Douglas’s theorem, or the Ando–Douglas–Seever theorem (see e.g., Proposition 4.9. of [12] and the recent generalization by [18]). Additionally, [28] offers a measure theoretic proof.

Thus far, we know that $A(X)={\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right]$ for some sigma algebra $\mathcal {G}$. Let $\mathcal {A}$ denote the sigma algebra generated by all ameliorated payoffs A(X). $\mathcal {G} \supseteq \mathcal {A}$ because all ameliorated payoffs ${\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right]$ are $\mathcal {G}$-measurable. $\mathcal {G} \subseteq \mathcal {A}$ because every $\mathcal {G}$-measurable payoff X is its own amelioration $A(X)={\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right] = X$. Combined, we have $\mathcal {G} = \mathcal {A}$ and $A(X)={\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {A}}\right]$.

On $L^2$, conditional expectations are orthogonal projections.

Claim 4. holds for all conditional expectations, because of Jensen’s inequality. Further, 4. implies 1. because (A5) implies (A2) almost surely and (A4). Indeed, for the concave and increasing function $u(x) = \min \{0, x-d\}$, assumption (A5) implies ${\mathbb {E}}_{}\left[ {\min \{0, X-d\}}\right] \le {\mathbb {E}}_{}\left[ {\min \{0, A(X)-d\})}\right]$. If $X\ge d$, the expectation on the left-hand side vanishes, implying that the expectation on the right-hand side will also vanish, and therefore $A(X)\ge d$ almost surely. Moreover, (A5) implies ${\mathbb {E}}_{}\left[ {\min \{0, X\}}\right] \le {\mathbb {E}}_{}\left[ {\min \{0, A(X)\})}\right]$, and therefore, ${\mathbb {E}}_{}\left[ {-\min \{0, X\}}\right] \ge {\mathbb {E}}_{}\left[ {-\min \{0, A(X)\})}\right]$ (A4). $\square$

Proof

of Proposition 1

Given (A1), (A2), (A3), and (A4), we know from Theorem 1 that A is a conditional expectation and Jensen’s inequality yields (A8) and (A5). Obviously, (A5) implies (A6). We have seen that additivity implies $A(-X) = -A(X)$, and therefore (A6) implies (A7). It remains to be shown that (A7) implies (A4). Monotonicity yields $A(X^-) \ge \max \{-A(X), 0\} = A(X)^-$, and keeping the mean yields ${\mathbb {E}}_{}\left[ {X^-}\right] = {\mathbb {E}}_{}\left[ {A(X^-)}\right] \ge {\mathbb {E}}_{}\left[ {A(X)^-}\right]$. $\square$

1.2 Proof from Section 3.1

Proof

of Proposition 2

1. implies 2.: Let the random variable X be integrable with respect to both measures and let V be bounded and $\mathcal {A}$-measurable. Then,

$$\begin{aligned}&{\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ { ({\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right] - {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right] ) V}\right] = {\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ {X V}\right] - {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ { Z\, {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right] \, V}\right] \\&= {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ {Z X V}\right] - {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ { {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {Z} \right| {\mathcal {A}}\right] \, X V}\right] = {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ {(Z-{\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {Z} \right| {\mathcal {A}}\right] ) \, X V}\right] \end{aligned}$$

If Z is $\mathcal {A}$-measurable, the right-hand side vanishes. Since the left-hand side also vanishes and V was an arbitrary $\mathcal {A}$-measurable bounded random variable, the $\mathcal {A}$-measurable random variable ${\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right] - {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right]$ has to vanish $\mathbb {Q}_{}$-almost surely.

2. implies 1.: Conversely, if ${\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right] = {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right]$ $\mathbb {Q}_{}$-a.s., the left-hand side of the chain of equations vanishes, and therefore, the right-hand side also vanishes. Setting $V=1$ and noting that all bounded random variables (even those that are not $\mathcal {A}$-measurable) can take the place of X, this implies that $Z = {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {Z} \right| {\mathcal {A}}\right]$ $\mathbb {P}_{}$-a.s., and therefore that Z is $\mathcal {A}$-measurable.

2. implies 3.: Thanks to the tower property, ${\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ {{\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right] }\right] = {\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ {{\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right] }\right] = {\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ {X}\right]$.

3. implies 2.: Consider the random variable $Y = X V$ for an arbitrary bounded and $\mathcal {A}$-measurable random variable V. Applying 3. to Y implies that ${\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ {X V}\right] = {\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ {{\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {X V} \right| {\mathcal {A}}\right] }\right] = {\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ { {\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right] V}\right]$. Since this holds for all bounded and $\mathcal {A}$-measurable random variables V and since ${\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right]$ is $\mathcal {A}$-measurable, ${\mathbb {E}}_{\mathbb {P}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right] ={\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ \left. {X} \right| {\mathcal {A}}\right]$ $\mathbb {Q}_{}$-almost surely. $\square$

1.3 Proof from Section 4.1

Proof

of Proposition 3

Suppose conditioning on a sigma algebra that does not contain all information revealed by the pricing kernel. Choose a strictly decreasing differentiable function f of the pricing kernel Z and define a payoff f(Z). This payoff is a strictly decreasing function of the pricing kernel and is therefore optimal for an expected utility maximizer with a marginal utility function $u'$ proportional to the inverse function of f because then $u'(f(Z)) \propto Z$; it is, however, modified by the amelioration operator (because, as the kernel, this payoff is not measurable with respect to the sigma algebra). Conversely, no risk-averse investor would object to conditioning on the pricing kernel because optimal payoffs are not modified and even non-optimal payoffs are at least as attractive after conditioning. Doing so after conditioning on a sigma algebra that is strictly larger than the sigma algebra generated by the pricing kernel yields the same result as direct conditioning on the pricing kernel, and so the direct way is preferable. $\square$

Remark 6

In the proof of Proposition 3 we define a payoff that is a decreasing function of the pricing kernel. Knowing this function, the construction of a marginal utility function such that this payoff maximizes expected utility is straightforward. Readers may wish to learn from [3] that even beyond this construction cost-efficent payoffs can be obtained via utility maximization.

1.4 Proof from Section 5.2

Proof

of Proposition 4

Any distribution preserving operator satisfies (A2), (A3) and (A4). If such an operator is additive (A1), Theorem 1 says it is a conditional expectation. A random variable has the same distribution as its conditional expectation if and only if it is a.s. equal to its conditional expectation (if and only if it is measurable with respect to the conditioning sigma algebra). The latter property of conditional expectations is e.g. shown as Exercise 1.12 in [10]. $\square$

1.5 Proofs from Section 6

Proof

of Proposition 5

Jensen’s inequality yields

$$\begin{aligned} {\mathbb {E}}_{}\left[ {Z'\, u({\mathbb {E}}_{}\left[ \left. {X} \right| {\sigma (\mathcal {D}')}\right] )}\right]&\ge {\mathbb {E}}_{}\left[ {Z'\, {\mathbb {E}}_{}\left[ \left. {u(X)} \right| {\sigma (\mathcal {D}')}\right] }\right] \\&= {\mathbb {E}}_{}\left[ {{\mathbb {E}}_{}\left[ \left. {Z'\, u(X)} \right| {\sigma (\mathcal {D}')}\right] }\right] = {\mathbb {E}}_{}\left[ {Z'\, u(X)}\right] \end{aligned}$$

for every $Z' \in \mathcal {D}' \supseteq \mathcal {D}$ and therefore $U({\mathbb {E}}_{}\left[ \left. {X} \right| {\sigma (\mathcal {D}')}\right] ) \ge U(X).$ $\square$

Proof

of Corollary 5

The conditional expectations in the first sentence coincide because of Proposition 2. Proposition 5 yields the claim; both representations are equivalent for $\mathcal {D}'' = \{Z\, Z' : Z' \in \mathcal {D}'\}$, because ${\mathbb {E}}_{\mathbb {Q}_{}{}}\left[ {Z'\, u(X)}\right] = {\mathbb {E}}_{}\left[ {Z\, Z'\, u(X)}\right]$. $\square$

1.6 Proof from Section 7

Proof

of Proposition 6

(a) Suppose that ${\mathbb {E}}_{}\left[ {Z\, X}\right] = 0$ and that almost surely ${\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right] \ge 0$.

For the direction from left to right, assume that Z is $\mathcal {G}$-measurable. Then, $0 = {\mathbb {E}}_{}\left[ {Z\, X}\right] = {\mathbb {E}}_{}\left[ {Z {\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right] }\right]$. This, the strict positivity of the kernel $Z>0$, and the assumption ${\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right] \ge 0$ imply ${\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right] = 0$ almost surely. In particular ${\mathbb {E}}_{}\left[ {X}\right] = {\mathbb {E}}_{}\left[ {{\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right] }\right] = 0$ and there is no statistical arbitrage with respect to $\mathcal {G}$.

For the reverse direction, assume that there is no statistical arbitrage with respect to $\mathcal {G}$. Consider $X = {\mathbb {E}}_{}\left[ \left. {Z} \right| {\mathcal {G}}\right] /Z - 1$; verify that ${\mathbb {E}}_{}\left[ {Z\, X}\right] = 0$, and proceed as follows:

Jensen’s inequality for conditional expectations and with the strictly convex function $x\mapsto 1/x$ says that ${\mathbb {E}}_{}\left[ \left. {{\mathbb {E}}_{}\left[ \left. {Z} \right| {\mathcal {G}}\right] /Z} \right| {\mathcal {G}}\right] \ge 1/ {\mathbb {E}}_{}\left[ \left. {Z/{\mathbb {E}}_{}\left[ \left. {Z} \right| {\mathcal {G}}\right] } \right| {\mathcal {G}}\right] = 1/1 = 1$ and equality holds if and only if Z is $\mathcal {G}$-measurable.

The absence of statistical arbitrage yields $0 = {\mathbb {E}}_{}\left[ {X}\right] = \mathbb {E}[\mathbb {E}[\mathbb {E}[Z|{\mathcal {G}}]/Z|\mathcal {G}] - 1]$. The nonnegative integrand must vanish, ${\mathbb {E}}_{}\left[ {{\mathbb {E}}_{}\left[ \left. {Z} \right| {\mathcal {G}}\right] /Z|\mathcal {G}] = 1}\right]$, equality holds in Jensen’s inequality, and Z is $\mathcal {G}$-measurable.

(b) Since ${\mathbb {E}}_{}\left[ \left. {A(X)} \right| {\mathcal {G}}\right] = {\mathbb {E}}_{}\left[ \left. {{\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right] } \right| {\mathcal {G}}\right] = {\mathbb {E}}_{}\left[ \left. {X} \right| {\mathcal {G}}\right] = A(X)$, the absence of statistical arbitrage with respect to $\mathcal {G}$ is equivalent to the absence of the usual arbitrage, and this condition holds in the presence of a strictly positive kernel. $\square$