A general test for SSD portfolio efficiency

Abstract

We develop and implement a Linear Programming test to analyze whether a given investment portfolio is efficient in terms of second-order stochastic dominance relative to all possible portfolios formed from a set of base assets. In case of efficiency, the primal model identifies a sub-gradient vector of a utility function that rationalizes the evaluated portfolio. In case of inefficiency, the dual model identifies a second, efficient portfolio that dominates the evaluated portfolio. The test gives a general necessary and sufficient condition, and can deal with general linear portfolio restrictions, inefficiency degree measures, and scenarios with unequal probabilities. We also develop a compact version of the test that substantially reduces computational burden at the cost of losing information about the dual dominating portfolio in case of inefficiency. An application to US investment benchmark data qualifies a broad stock market index as significantly inefficient, and suggests that no risk-averse investor would hold the market index in the face of attractive premiums offered by some more concentrated investment portfolios.

Appendix

1.1 Proof of Lemma 1:

“$\Leftarrow$” We will first prove that a violation of Definition 1 implies a violation of Definition 1’. One possible violation occurs in the rare case when

$$\begin{aligned} \sum _{s=1}^{S}p_{s} \left( u(\mathbf {r}^{s}\varvec{\lambda }) - u(\mathbf {r}^{s}\varvec{\tau }) \right) = 0 \quad \forall u\in U_{2}. \end{aligned}$$

Since $V_{2}\subset U_{2}$, this implies a violation of Definition 1’. We can therefore restrict attention to the more common cases with

$$\begin{aligned} \sum _{s=1}^{S}p_{s} \left( u(\mathbf {r}^{s}\varvec{\lambda }) - u(\mathbf {r}^{s}\varvec{\tau }) \right) = \nu < 0 \end{aligned}$$

(18)

for some $u \in U_{2}$. Since $clo(V_{2}) = U_{2}$, for every $\varepsilon >$0 and $u\in U_{2}$ we find

$$\begin{aligned} | \sum _{s=1}^{S}p_{s} \left( u(\mathbf {r}^{s}\varvec{\lambda }) - u(\mathbf {r}^{s}\varvec{\tau }) \right) - \sum _{t=1}^{T} p_{s}\left( v(\mathbf {r}^{s}\varvec{\lambda }) - v(\mathbf {r}^{s}\varvec{\tau }) \right) | \le \varepsilon \end{aligned}$$

(19)

for some $v \in V_{2}$. Choosing $0<\varepsilon <-\nu$, (18) and (19) imply that

$$\begin{aligned} \sum _{t=1}^{T} p_{s}\left( v(\mathbf {r}^{s}\varvec{\lambda }) - v(\mathbf {r}^{s}\varvec{\tau }) \right) \le \varepsilon + \nu < 0 \end{aligned}$$

for some $v \in V_{2}$, a violation of Definition 1’.

“$\Rightarrow$” We will now prove that a violation of Definition 1’ implies a violation of Definition 1. The most common case is

$$\begin{aligned} \sum _{t=1}^{T} p_{s}\left( v(\mathbf {r}^{s}\varvec{\lambda }) - v(\mathbf {r}^{s}\varvec{\tau }) \right) < 0. \end{aligned}$$

for some $v\in V_{2}$. Since $V_{2}\subset U_{2}$, it is clear that Definition 1 is violated. What remains is the less common case with

$$\begin{aligned} \sum _{s=1}^{S}p_{s} \left( v(\mathbf {r}^{s}\varvec{\lambda }) - v(\mathbf {r}^{s}\varvec{\tau }) \right) = 0 \end{aligned}$$

(20)

for some $v \in V_{2}$. This case arises naturally if $\mathbf {r}^{s}\varvec{\lambda }= \mathbf {r}^{s}\varvec{\tau }$, $s=1,2,...,S$, or, more generally, if the values for $\mathbf {r}^{s}\varvec{\lambda }$ are a permutation of the values for $\mathbf {r}^{s}\varvec{\tau }$, a situation when Definition 1 is clearly violated. In all other situations, choose a scenario $t \in \{1,2,...,S\}$ such that $\mathbf {r}^{t}\varvec{\lambda }\ne \mathbf {r}^{s}\varvec{\tau }$, $s=1,2,...,S$. Consider a small neighborhood of $\mathbf {r}^{t}\varvec{\lambda }$: $\mathcal {N} = (\mathbf {r}^{t}\varvec{\lambda }- a, \mathbf {r}^{t}\varvec{\lambda }+ a)$, a $>$ 0, such that $\mathbf {r}^{s}\varvec{\tau }\notin \mathcal {N}$, $s=1,2,...,S$ and $\mathbf {r}^{s}\varvec{\lambda }\notin \mathcal {N}$ for all $s$ satisfying $\mathbf {r}^{s}\varvec{\lambda }\ne \mathbf {r}^{s}\varvec{\tau }$. For $v\in V_{2}$ satisfying (20) let w(x) be equal to v(x) outside $\mathcal {N}$ and a linear approximation of v(x) inside $\mathcal {N}$:

$$\begin{aligned} w(x)&= v(x), \, x \notin \mathcal {N} \\&= v(\mathbf {r}^{t}\varvec{\lambda }- a) + \textstyle \frac{(x - \mathbf {r}^{t}\varvec{\lambda }+ a)}{2a}\left( v(\mathbf {r}^{t}\varvec{\lambda }+ a) - v(\mathbf {r}^{t}\varvec{\lambda }- a)\right) \!, \, x \in \mathcal {N}. \end{aligned}$$

By construction, $w \in U_{2}$. Since v(x) is strictly concave, $w(\mathbf {r}^{s}\varvec{\lambda }) < v(\mathbf {r}^{s}\varvec{\lambda })$ for all $s$ satisfying $\mathbf {r}^{s}\varvec{\lambda }= \mathbf {r}^{s}\varvec{\tau }$. Moreover, the definition of $\mathcal {N}$ gives $w(\mathbf {r}^{s}\varvec{\tau }) = v(\mathbf {r}^{s}\varvec{\tau })$, $s=1,2,...,S$ and $w(\mathbf {r}^{s}\varvec{\lambda }) = v(\mathbf {r}^{s}\varvec{\lambda })$ for all $s$ that satisfy $\mathbf {r}^{s}\varvec{\lambda }\ne \mathbf {r}^{s}\varvec{\tau }$. Hence,

$$\begin{aligned} \sum _{s=1}^{S}p_{s} \left( w(\mathbf {r}^{s}\varvec{\lambda }) - w(\mathbf {r}^{s}\varvec{\tau }) \right) < \sum _{s=1}^{S}p_{s} \left( v(\mathbf {r}^{s}\varvec{\lambda }) - v(\mathbf {r}^{s}\varvec{\tau }) \right) , \end{aligned}$$

and (20) implies

$$\begin{aligned} \sum _{s=1}^{S} p_{s}\left( w(\mathbf {r}^{s}\varvec{\lambda }) - w(\mathbf {r}^{s}\varvec{\tau }) \right) < 0, \end{aligned}$$

a violation of Definition 1.$\square$

1.2 Proof of Lemma 2:

“$\Rightarrow$” We will first prove that SSD dominance requires $\omega (\varvec{\lambda },\varvec{\tau }|\mathbf {w})>0$. Using Definition 1’ and (3), if portfolio $\varvec{\lambda }$ does not dominate portfolio $\varvec{\tau }$, then

$$\begin{aligned} T^{-1}\sum _{t=1}^{T} \left( v((X\varvec{\lambda })^{[t]}) - v(\mathbf {x}^{t}\varvec{\tau })\right) \le 0 \end{aligned}$$

(21)

for some $v\in V_2$. Using (6), we can assume a strictly positive lower bound $\varepsilon >0$ for the sub-gradient decrements: $v'(z_{t}) - v'(z_{t+1}) > \varepsilon$, $t=1,2,...,T-1$ and $v'(z_{T}) >0$. Let $\vartheta = \max _{t}w_{t}/\varepsilon$ and $g(x) = \vartheta v(x)$. For this function, we have $g'(z_{t}) - g'(z_{t+1}) > \max _{t}w_{t}$ and $g'(z_{T}) >\max _{t}w_{t}$ which obeys (7.1) and (7.2) and therefore $g'(z_{t})$, $t=1,2,...,T$ is a feasible solution of (7). Using (5) and (21), we find a non-positive value for the objective function (7) when $\beta _t = g'(z_{t})$, $t=1,2,...,T$:

$$\begin{aligned} T^{-1}\sum _{t=1}^{T} \left( g'(z_{t})((X\varvec{\lambda })^{[t]} - \mathbf {x}^{t}\varvec{\tau })\right) = T^{-1}\sum _{t=1}^{T} \left( g((X\varvec{\lambda })^{[t]}) - g(\mathbf {x}^{t}\varvec{\tau })\right) \le 0. \end{aligned}$$

“$\Leftarrow$” We will now prove that SSD relation implies $\omega (\varvec{\lambda },\varvec{\tau }|\mathbf {w})>0$.

If $\omega (\varvec{\lambda },\varvec{\tau }|\mathbf {w})\le 0$ then

$$\begin{aligned} T^{-1}\sum _{t=1}^{T} \beta _{t}\left( (X\varvec{\lambda })^{[t]} - \mathbf {x}^{t}\varvec{\tau }\right) \le 0 \end{aligned}$$

for some $\beta _{t}$, $t=1,2,...,T$ satisfying (7.1) and (7.2). Consider $\alpha _t = \beta _{t}$ if $t \notin \Theta (\varvec{\lambda },\varvec{\tau })$ and

$$\begin{aligned} \alpha _t = \textstyle \frac{\sum ^{t_{2}+1}_{t=t_{1}}\beta _{t}}{t_{2}-t_{1}} \end{aligned}$$

for $t\in \Theta (\varvec{\lambda },\varvec{\tau })$ and $t_{1} = \min _{s} \{s\in \Theta (\varvec{\lambda },\varvec{\tau }) : (X\varvec{\lambda })^{[s]} = (X\varvec{\lambda })^{[t]} \wedge \mathbf {x}^{s}\varvec{\tau }= \mathbf {x}^{t}\varvec{\tau }\}$, $t_{2} = \max _{s}\{s\in \Theta (\varvec{\lambda },\varvec{\tau }):(X\varvec{\lambda })^{[s]} = (X\varvec{\lambda })^{[t]} \wedge \mathbf {x}^{s}\varvec{\tau }= \mathbf {x}^{t}\varvec{\tau }\}$. By construction, $\alpha _{t}$ obeys the required ranking of (6) and

$$\begin{aligned} T^{-1}\sum _{t=1}^{T} \alpha _{t}\left( (X\varvec{\lambda })^{[t]} - \mathbf {x}^{t}\varvec{\tau }\right) = T^{-1}\sum _{t=1}^{T} \beta _{t}\left( (X\varvec{\lambda })^{[t]} - \mathbf {x}^{t}\varvec{\tau }\right) . \end{aligned}$$

(22)

Therefore, there exists a utility function $v\in V_{2}$ and points $z_t =\varphi _{t}(X\varvec{\lambda })^{[t]} + (1-\varphi _{t})\mathbf {x}^{t}\varvec{\tau }$, $0\le \varphi _{t} \le 1$, $t=1,2,...,T$, such that $v'(z_{t}) = \alpha _t$ satisfy (5) and (6), and, in addition

$$\begin{aligned} T^{-1}\sum _{t=1}^{T} \left( v((X\varvec{\lambda })^{[t]}) - v(\mathbf {x}^{t}\varvec{\tau })\right) = T^{-1}\sum _{t=1}^{T} \alpha _{t}\left( (X\varvec{\lambda })^{[t]} - \mathbf {x}^{t}\varvec{\tau }\right) \le 0 \end{aligned}$$

that is, portfolio $\varvec{\lambda }$ does not dominate portfolio $\varvec{\tau }$. The last inequality follows from (22). $\square$

1.3 Proof of Theorem 1:

Let $\zeta (\varvec{\tau }|\mathbf {w}) = \max _{\varvec{\lambda }\in \Lambda }\omega (\varvec{\lambda },\varvec{\tau }|\mathbf {w})$. Lemma 2 implies that portfolio $\varvec{\tau }$ is efficient if and only if $\zeta (\varvec{\tau }|\mathbf {w}) =0$. Our strategy is to prove that $\zeta (\varvec{\tau }|\mathbf {w}) = \xi (\varvec{\tau }|\mathbf {w})$. $\zeta (\varvec{\tau }|\mathbf {w})$ has a max–min structure that minimizes expected utility over utility for every portfolio. An alternative min–max formulation maximizes expected utility over portfolios for every utility function.

$$\begin{aligned} \zeta (\varvec{\tau }|\mathbf {w})&= \min _{\gamma _{t,s}} \max _{\varvec{\lambda }\in \Lambda } T^{-1} \sum \limits ^{T}_{t=1} \sum \limits ^{T}_{s=1}\gamma _{t,s}\mathbf {x}^{t}\varvec{\lambda }- T^{-1}\sum \limits ^{T}_{t=1} \sum \limits ^{T}_{s=t}\textstyle \frac{1}{s}\sum ^{T}_{k=1} \gamma _{k,s}\mathbf {x}^{t}\varvec{\tau }\nonumber \\&\mathrm {s.t.} \ \ (8.1),\, (8.2),\,(3.3). \end{aligned}$$

(23)

Both formulations are equivalent because the optimization is over two convex sets; see Sion (1958). Since the portfolio set (1) is a polytope, the embedded LP maximization over the portfolio space

$$\begin{aligned} \max _{\varvec{\lambda }}&T^{-1}\sum ^{T}_{t=1} \sum ^{T}_{s=1}\gamma _{t,s}\mathbf {x}^{t}\varvec{\lambda }\\ \mathrm {s.t.}&\sum ^{N}_{j=1}a_{ij}\lambda _{j}\le b_{i}, \ \ i=1,2,...,M \nonumber \end{aligned}$$

involves the following LP dual minimization problem:

$$\begin{aligned} \min _{\theta _{s}}&\sum ^{M}_{i=1}b_{i}\theta _{i} \\ \mathrm {s.t.}&\sum ^{M}_{i=1}a_{ij}\theta _{i} - T^{-1}\sum ^{T}_{t=1} \sum ^{T}_{s=1}\gamma _{t,s}x_{j}^{t} = 0, \ \ j=1,2,...,N \nonumber \\&\theta _{i} \ge 0, \ \ i=1,2,...,M. \nonumber \end{aligned}$$

Making the substitution in (23) gives (9).$\Box$

1.4 Proof of Theorem 2:

The cumulative returns $\Omega (\varvec{\lambda },s)$ can be expressed as follows:

$$\begin{aligned} \Omega (\varvec{\lambda },s) = \min _{\alpha _{k}}T^{-1}\sum _{k=1}^{T}\alpha _{k}\mathbf {x}^{k}\varvec{\lambda }\\ \mathrm {s.t.} \ \ \sum _{t=1}^{T}\alpha _{t} = s \\ 0 \le \alpha _{t}\le 1,&t = 1,2,...,T \end{aligned}$$

The optimal values $\alpha _{k}^{*}$ are equal to one if $\mathbf {x}^{k}\varvec{\lambda }\le (X\varvec{\lambda })^{[s]}$ and are equal to zero otherwise. Taking the dual of this program, we find

$$\begin{aligned} \Omega (\varvec{\lambda },s) = \max _{z_{s},\nu _{t,s}}\left( sz_{s} - \sum _{t=1}^{T}\nu _{t,s} \right) \\ \mathrm {s.t.} \ \ z_{s} - \nu _{t,s} \le T^{-1}\mathbf {x}^{t}\varvec{\lambda }&t = 1,2,...,T \nonumber \\ \nu _{t,s} \ge 0&t = 1,2,...,T. \nonumber \end{aligned}$$

(24)

Substituting (24) for each s into (13), we find:

$$\begin{aligned} \xi (\varvec{\tau }|\mathbf {w}) = \max _{d_{s},z_{s},v_{t,s},\varvec{\lambda }} \sum ^{T}_{s=1}w_{s}d_{s}&\\ \mathrm {s.t.} \ \ sz_{s} - \sum ^{T}_{t=1}v_{t,s} -T^{-1}\sum _{k=1}^{s}\mathbf {x}^{k}\varvec{\tau }&\ge d_{s}, \ \ s=1,2,...,T \nonumber \\ z_{s}- v_{t,s} -T^{-1}\mathbf {x}^{t}\varvec{\lambda }&\le 0, \ \ t,s = 1,2,...,T \ \ \nonumber \\ v_{t,s}&\ge 0, \ \ t,s = 1,2,...,T \ \ \nonumber \\ d_{s}&\ge 0, \ \ s = 1,2,...,T \ \ \nonumber \\ \varvec{\lambda }&\in \Lambda .\nonumber \end{aligned}$$

(25)

Since we maximize the weighted average of $d_{s}$, the first set of inequality constraints must be fulfilled as equations. Hence,

$$\begin{aligned} z_{s} = \textstyle \frac{1}{s}\sum ^{T}_{t=1}v_{t,s} + T^{-1}\textstyle \frac{1}{s}\sum _{k=1}^{s}\mathbf {x}^{k}\varvec{\tau }+ \textstyle \frac{1}{s}d_{s}, \ \ s=1,2,...,T. \end{aligned}$$

Substituting these equalities into the second set of inequality constraints, we find (10). $\square$

1.5 Proof of Theorem 3:

Theorem 1 implies that a given portfolio $\varvec{\tau }$ is SSD efficient if and only if $\xi (\varvec{\tau }|\mathbf {w})=0$. If $\xi (\varvec{\tau }|\mathbf {w})>0$, then Theorem 2 shows that solution portfolio $\varvec{\lambda }^{*}$ SSD dominates $\varvec{\tau }$. Since $\varvec{\lambda }^{*}$ maximizes the weighted sum of cumulative returns among all dominating portfolios, using Definition 1”, $\varvec{\lambda }^{*}$ is SSD efficient.$\Box$

1.6 Proof of Theorem 4:

To simplify the proof, we say that portfolio $\varvec{\lambda }$ is co-monotone with portfolio $\varvec{\tau }$ if

$$\begin{aligned} (\mathbf {x}^{t_{1}}\varvec{\lambda }- \mathbf {x}^{t_{2}}\varvec{\lambda })(\mathbf {x}^{t_{1}}\varvec{\tau }- \mathbf {x}^{t_{2}}\varvec{\tau }) \ge 0 \quad \forall t_{1},t_{2}\in \{1,2,...,T\}. \end{aligned}$$

This definition means that the ranking of $X\varvec{\lambda }$ may differ from the ranking of $X\varvec{\tau }$ only inside $\Phi _k$.

If portfolio $\varvec{\tau }$ is SSD inefficient then there exists a portfolio $\varvec{\lambda }\in \Lambda$ that SSD dominates $\varvec{\tau }$. Moreover, every portfolio $\bar{\varvec{\lambda }}(\alpha ) = (1-\alpha )\varvec{\tau }+ \alpha \varvec{\lambda }$, $\alpha \in [0,1]$ must also SSD dominate $\varvec{\tau }$. Furthermore, for sufficiently small $\alpha \in [0,1]$ a portfolio $\bar{\varvec{\lambda }}(\alpha ) = (1-\alpha )\varvec{\tau }+ \alpha \varvec{\lambda }$ is co-monotone with portfolio $\varvec{\tau }$. Hence, there exists a portfolio $\bar{\varvec{\lambda }}(\alpha )$ such that $\bar{\omega }(\bar{\varvec{\lambda }}(\alpha ),\varvec{\tau }|\mathbf {w}) > 0$, where

$$\begin{aligned}&\bar{\omega }(\bar{\varvec{\lambda }}(\alpha ),\varvec{\tau }|\mathbf {w}) = \min _{\beta _{t}}T^{-1}\sum _{t=1}^{T} \beta _{t}\left( \mathbf {x}^{t}\bar{\varvec{\lambda }}(\alpha ) - \mathbf {x}^{t}\varvec{\tau }\right) \\ \mathrm {s.t.}&\beta _{t} - \beta _{s} \ge w_{k}, \ \ \ \ t\in \Phi _k, \ \ s\in \Phi _{k+1}, \ \ k = 1,2,...,K-1 \nonumber \\&\beta _{t} \ge w_{K}, \ \ \ \ t\in \Phi _K \nonumber \end{aligned}$$

(26)

Since both $\mathbf {x}^{t}\bar{\varvec{\lambda }}(\alpha )$ and $\mathbf {x}^{t}\varvec{\tau }$ take the same values for all $t \in \Phi _k^{l_k}$, we may replace the variables $\beta _{t}, t \in \Phi _k^{l_k}$ by a single variable $\beta _k^{l_k}$ and rewrite (26) as follows:

$$\begin{aligned}&\bar{\omega }(\bar{\varvec{\lambda }}(\alpha ),\varvec{\tau }|\mathbf {w}) = \min _{ \beta _{k}^{l_k}} T^{-1}\sum ^{K}_{k=1}\sum ^{L_k}_{l_k=1}\beta _k^{l_k} \sum _{t \in \Phi _k^{l_k}} \left( \mathbf {x}^{t}\bar{\varvec{\lambda }}(\alpha ) - \mathbf {x}^{t}\varvec{\tau }\right) \\ \mathrm {s.t.}&\beta _{k}^{l_k} - \beta _{k+1}^{l_{k+1}} \ge w_{k}, \ \ k = 1,...,K-1, \ \ l_k = 1,...,L_k,\ \ l_{k+1} = 1,...,L_{k+1} \nonumber \\&\beta _{K}^{l_K} \ge w_{K}, \ \ l_K = 1,...,L_K \nonumber \end{aligned}$$

This means that $\kappa (\varvec{\tau }|\mathbf {w})>0$, where

$$\begin{aligned}&\kappa (\varvec{\tau }|\mathbf {w}) = \min _{ \beta _{k}^{l_k}}\max _{\varvec{\lambda }\in \Lambda }T^{-1}\sum ^{K}_{k=1}\sum ^{L_k}_{l_k=1}\beta _k^{l_k} \sum _{t \in \Phi _k^{l_k}} \left( \mathbf {x}^{t}\varvec{\lambda }- \mathbf {x}^{t}\varvec{\tau }\right) \\ \mathrm {s.t.}&\beta _{k}^{l_k} - \beta _{k+1}^{l_{k+1}} \ge w_{k}, \ \ k = 1,...,K-1, \ \ l_k = 1,...,L_k, \ \ l_{k+1} = 1,...,L_{k+1} \nonumber \\&\beta _{K}^{l_K} \ge w_{K}, \ \ l_K = 1,...,L_K \nonumber \end{aligned}$$

We used the Minimax theorem to reverse the order of the two optimization operations. Similar to the proof of Theorem 1, taking the dual to the embedded maximization problem

$$\begin{aligned} \max _{\varvec{\lambda }\in \Lambda }T^{-1}\sum ^{K}_{k=1}\sum ^{L_k}_{l_k=1}\beta _k^{l_k} \sum _{t \in \Phi _k^{l_k}}\mathbf {x}^{t}\varvec{\lambda }\end{aligned}$$

we find that $\upsilon (\varvec{\tau }|\mathbf {w}) = \kappa (\varvec{\tau }|\mathbf {w}) >0$.

If portfolio $\varvec{\tau }$ is SSD efficient then no co-monotone dominating portfolio exists and hence, $\kappa (\varvec{\tau }|\mathbf {w}) \le 0$. Since $\kappa (\varvec{\tau }|\mathbf {w})$ is always non-negative, $\upsilon (\varvec{\tau }|\mathbf {w})=\kappa (\varvec{\tau }|\mathbf {w}) = 0$.$\Box$