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Acceptability maximization

  • * Corresponding author: Gabriela Kováčová

    * Corresponding author: Gabriela Kováčová 

IC acknowledges partial support from the National Science Foundation (US) grant DMS-1907568

Abstract Full Text(HTML) Figure(3) / Table(5) Related Papers Cited by
  • The aim of this paper is to study the optimal investment problem by using coherent acceptability indices (CAIs) as a tool to measure the portfolio performance. We call this problem the acceptability maximization. First, we study the one-period (static) case, and propose a numerical algorithm that approximates the original problem by a sequence of risk minimization problems. The results are applied to several important CAIs, such as the gain-to-loss ratio, the risk-adjusted return on capital and the tail-value-at-risk based CAI. In the second part of the paper we investigate the acceptability maximization in a discrete time dynamic setup. Using robust representations of CAIs in terms of a family of dynamic coherent risk measures (DCRMs), we establish an intriguing dichotomy: if the corresponding family of DCRMs is recursive (i.e. strongly time consistent) and assuming some recursive structure of the market model, then the acceptability maximization problem reduces to just a one period problem and the maximal acceptability is constant across all states and times. On the other hand, if the family of DCRMs is not recursive, which is often the case, then the acceptability maximization problem ordinarily is a time-inconsistent stochastic control problem, similar to the classical mean-variance criteria. To overcome this form of time-inconsistency, we adapt to our setup the set-valued Bellman's principle recently proposed in [23] applied to two particular dynamic CAIs - the dynamic risk-adjusted return on capital and the dynamic gain-to-loss ratio. The obtained theoretical results are illustrated via numerical examples that include, in particular, the computation of the intermediate mean-risk efficient frontiers.

    Mathematics Subject Classification: 91G10, 93E20, 93E35, 49L20.

    Citation:

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  • Figure 1.  Efficient frontiers (black) of the mean-risk problems and elements with the highest mean-to-risk ratio (green). All frontiers are depicted in the $ (\rho, \mathbb{E}) $ plane for the returns $ v_T - v_t $ with $ v_t = 1 $

    Figure 2.  Efficient frontiers for returns over time. The mean-risk profiles and the corresponding values of dRAROC are depicted for three trading strategies: the time consistent mean-risk strategy in one state $ \omega $ (yellow triangle), the switching strategy (red diamond) and the myopic strategy (magenta square). The element of the frontier with the highest dRAROC is also depicted at each time (green circle)

    Figure 3.  Efficient frontiers (black) of the problems (17) depicted for wealth $ V_t = 0. $ All frontiers are depicted in the $ (\mathbb{E}_t(V_T^-), \mathbb{E}_t (V_T)) $ plane. The corresponding highest value of $ dGLR $ (the slope of the frontier) is given

    Table Algorithm 1.  Approximating maximal acceptability α* via risk minimization

     | Show Table
    DownLoad: CSV

    Table 1.  Algorithm 1 for AIT, GLR and RAROC in a toy market model

    Panel A: Return matrix R in the toy market model (two assets and four states of the world).
    AIT GLR RAROC
    Iter $x_L$ $x_U$ $x$ $p(x)$ Iter $x_L$ $x_U$ $x$ $p(x)$ Iter $x_L$ $x_U$ $x$ $p(x)$
    Step 1 Step 1 Step 1
    $1$ $0$ $\infty$ $2$ + $1$ $0$ $\infty$ $2$ $-$ $1$ $0$ $\infty$ $2$ +
    $2$ $0$ $2$ $1$ + $2$ $2$ $\infty$ $4$ + $2$ $0$ $2$ $1$ +
    $3$ $0$ $1$ $0.5$ $-$ Step 2 $3$ $0$ $1$ $0.5$ $-$
    Step 2 $1$ $2$ $4$ $3$ $-$ Step 2
    $1$ $0.5$ $1$ $0.75$ $-$ $2$ $3$ $4$ $3.5$ + $1$ $0.5$ $1$ $0.75$ $-$
    $2$ $0.75$ $1$ $0.875$ + $3$ $3$ $3.5$ $3.25$ + $2$ $0.75$ $1$ $0.875$ +
    $3$ $0.75$ $0.875$ $0.8125$ + $4$ $3$ $3.25$ $3.125$ $-$ $3$ $0.75$ $0.875$ $0.8125$ $-$
    $4$ $0.75$ $0.8125$ $0.7813$ + $5$ $3.125$ $3.25$ $3.1875$ + $4$ $0.8125$ $0.875$ $0.8438$ +
    $5$ $0.75$ $0.7813$ $0.7656$ + $6$ $3.125$ $3.1875$ $3.1563$ + $5$ $0.8125$ $0.8438$ $0.8281$ +
    $6$ $0.75$ $0.7656$ $0.7578$ $-$ $7$ $3.125$ $3.1563$ $3.1406$ $-$ $6$ $0.8125$ $0.8281$ $0.8203$ $-$
    $7$ $0.7578$ $0.7656$ $0.7617$ $-$ $8$ $3.1406$ $3.1563$ $3.1484$ + $7$ $0.8203$ $0.8281$ $0.8242$ +
    $8$ $0.7617$ $0.7656$ $0.7637$ $-$ $9$ $3.1406$ $3.1484$ $3.1445$ + $8$ $0.8203$ $0.8242$ $0.8223$ +
    $9$ $0.7637$ $0.7656$ $0.7647$ $-$ $10$ $3.1406$ $3.1445$ $3.1426$ $-$ $9$ $0.8203$ $0.8223$ $0.8213$ $-$
    $10$ $0.7647$ $0.7656$ $0.7651$ $-$ $11$ $3.1426$ $3.1445$ $3.1436$ + $10$ $0.8213$ $0.8223$ $0.8218$ +
    $11$ $0.7651$ $0.7656$ $0.7654$ + $12$ $3.1426$ $3.1436$ $3.1431$ + $11$ $0.8213$ $0.8218$ $0.8215$ +
    $12$ $0.7651$ $0.7654$ $0.7653$ $-$ $13$ $3.1426$ $3.1431$ $3.1428$ $-$ $12$ $0.8213$ $0.8215$ $0.8214$ $-$
    $13$ $0.7653$ $0.7654$ $0.7653$ $-$ $14$ $3.1428$ $3.1431$ $3.1429$ + $13$ $0.8214$ $0.8215$ $0.8215$ +
    $0.76532$ $0.76538$ $15$ $3.1428$ $3.1429$ $3.1429$ + $0.82141$ $0.82147$
    $h^\epsilon = (55.17\%, 44.83\%)$ $3.14282$ $3.14288$ $h^\epsilon = (93.75\%, 6.25\%)$
    $h^\epsilon = (73.33\%, 26.67\%)$
    Panel B: Iterations of Algorithm 1 with input parameters $x_0 = 2, \epsilon = 10^{-4}$ and $\bar{M} = 15$. The last two rows give, respectively, the bounds $x_L$ and $x_U$ on the maximal acceptability, and an $\text{V@R}epsilon$-optimal portfolio.
     | Show Table
    DownLoad: CSV

    Table 2.  Iterations of the modified, the mixed and the zero-level version of Algorithm 1 for $\text{GLR}$ in the market model from Table 1, Panel A ($ \alpha^* = 3.1428 $) with the tolerance $ \epsilon = 10^{-4} $. In the modified version the bisection is performed on the parameter $ q = \frac{1}{2+x} \in [0, 0.5] $ after verifying the signs of $ p(0) $ and $ p(\infty) $. The termination criterion is set on the parameter $ x $ to guarantee an $ \epsilon $-solution is obtained. With the termination criterion on the parameter $ q $ the algorithm would finish after $ 13 $ iteration of Step 2, however, the interval for maximal acceptability would have length 1.6e-03. The mixed version switches to a bisection on the parameter $ x $ as soon as a finite upper bound $ x_U $ is obtained. -the zero-level version computes after each iteration the level $ y $ for which the portfolio solving the risk minimization problem has zero risk. This level is used as a lower bound. The algorithm is run with initial parameters $ x_0 = 2, \epsilon = 10^{-4} $ and $ \bar{M} = 15 $

    Modified algorithm for GLR Mixed algorithm for GLR Zero-level algorithm for GLR
    Iter $q_L$ $q_U$ $q$ $x$ $p(x)$ Iter $q_L$ $q_U$ $q$ $x$ $p(x)$ Iter $x_L$ $x_U$ $x$ $y$ $p(x)$
    Step 1 Step 1 Step 1
    $0$ $\infty$ + $0$ $\infty$ + $1$ $0$ $\infty$ $2$ $3.1429$ $-$
    $0.5$ $0$ $-$ $0.5$ $0$ $-$ $2$ $3.1429$ $\infty$ $6.2857$ $3.1429$ +
    Step 2 Step 2 Step 2
    $1$ $0$ $0.5$ $0.25$ $2$ $-$ $1$ $0$ $0.5$ $0.25$ $2$ $-$ $1$ $3.1429$ $6.2857$ $4.7143$ $3.1429$ +
    $2$ $0$ $0.25$ $0.125$ $6$ + $2$ $0$ $0.25$ $0.125$ $6$ + $2$ $3.1429$ $4.7143$ $3.9286$ $3.1429$ +
    $3$ $0.125$ $0.25$ $0.1875$ $3.3333$ + Iter $x_L$ $x_U$ $x$ $p(x)$ $3$ $3.1429$ $3.9286$ $3.5357$ $3.1429$ +
    $4$ $0.1875$ $0.25$ $0.2188$ $2.5714$ $-$ $3$ $2$ $6$ $4$ + $4$ $3.1429$ $3.5357$ $3.3393$ $3.1429$ +
    $5$ $0.1875$ $0.2188$ $0.2031$ $2.9231$ $-$ $4$ $2$ $4$ $3$ $-$ $5$ $3.1429$ $3.3393$ $3.2411$ $3.1429$ +
    $6$ $0.1875$ $0.2031$ $0.1953$ $3.1200$ $-$ $5$ $3$ $4$ $3.5$ + $6$ $3.1429$ $3.2411$ $3.1920$ $3.1429$ +
    $7$ $0.1875$ $0.1953$ $0.1914$ $3.2245$ + $6$ $3$ $3.5$ $3.25$ + $7$ $3.1429$ $3.1920$ $3.1674$ $3.1429$ +
    $8$ $0.1914$ $0.1953$ $0.1934$ $3.1717$ + $7$ $3$ $3.25$ $3.125$ $-$ $8$ $3.1429$ $3.1674$ $3.1551$ $3.1429$ +
    $9$ $0.1934$ $0.1953$ $0.1943$ $3.1457$ + $8$ $3.125$ $3.25$ $3.1875$ + $9$ $3.1429$ $3.1551$ $3.1490$ $3.1429$ +
    $10$ $0.1943$ $0.1953$ $0.1948$ $3.1328$ $-$ $9$ $3.125$ $3.1875$ $3.1563$ + $10$ $3.1429$ $3.1490$ $3.1459$ $3.1429$ +
    $11$ $0.1943$ $0.1948$ $0.1946$ $3.1393$ $-$ $10$ $3.125$ $3.1563$ $3.1406$ $-$ $11$ $3.1429$ $3.1459$ $3.1444$ $3.1429$ +
    $12$ $0.1943$ $0.1946$ $0.1945$ $3.1425$ $-$ $11$ $3.1406$ $3.1563$ $3.1484$ + $12$ $3.1429$ $3.1444$ $3.1436$ $3.1429$ +
    $13$ $0.1943$ $0.1945$ $0.1944$ $3.1441$ + $12$ $3.1406$ $3.1484$ $3.1445$ + $13$ $3.1429$ $3.1436$ $3.1432$ $3.1429$ +
    $14$ $0.1944$ $0.1945$ $0.1944$ $3.1433$ + $13$ $3.1406$ $3.1445$ $3.1426$ $-$ $14$ $3.1429$ $3.1432$ $3.1430$ $3.1429$ +
    $15$ $0.1944$ $0.1945$ $0.1944$ $3.1429$ + $14$ $3.1426$ $3.1445$ $3.1436$ + $15$ $3.1429$ $3.1430$ $3.1430$ $3.1429$ +
    $16$ $0.1944$ $0.1945$ $0.1945$ $3.1427$ $-$ $15$ $3.1426$ $3.1436$ $3.1431$ + $(x_L, x_U) = (3.14286, 3.14295)$
    $17$ $0.1944$ $0.1945$ $0.1944$ $3.1428$ $-$ $16$ $3.1426$ $3.1431$ $3.1428$ $-$ $h^\epsilon = (73.33\%, 26.67\%)$
    $18$ $0.1944$ $0.1944$ $0.1944$ $3.1429$ $-$ $17$ $3.1428$ $3.1431$ $3.1429$ +
    $(x_L, x_U) = (3.14285, 3.14290)$ $18$ $3.1428$ $3.1429$ $3.1429$ +
    $q_U - q_L =$ 1.9e-06, $x_U - x_L =$ 5.0e-05 $(x_L, x_U) = (3.14282, 3.14288)$
    $h^\epsilon = (73.33\%, 26.67\%)$ $h^\epsilon = (73.33\%, 26.67\%)$
     | Show Table
    DownLoad: CSV

    Table 3.  The behavior of Algorithm 1 for various input parameters in a market model with $ d = 10 $ assets with short-selling constraints

    Panel A: $\text{AIT}$, maximal acceptability $\alpha^* = 25.45$.
    $x_0$ $\epsilon$ $M$ Step 1 Step 2 Run time
    Iter $[x_L, x_U]$ Iter $x_U - x_L$ (s)
    $2$ $10^{-4}$ $15$ 5 $[16, 32]$ 18 6.1e-05 3.78
    $20$ $10^{-4}$ $15$ 2 $[20, 40]$ 18 7.6e-05 3.40
    $200$ $10^{-4}$ $15$ 4 $[25, 50]$ 18 9.5e-05 3.56
    $2$ $10^{-8}$ $15$ 5 $[16, 32]$ 31 7.5e-09 6.22
    $2^{20}$ $10^{-4}$ $15$ 15 $[0, 64]$ no Step 2 1.88
    $2^{20}$ $10^{-4}$ $30$ 17 $[16, 32]$ 18 6.1e-05 4.67
    $2^{-10}$ $10^{-4}$ $15$ 15 $[16, \infty]$ no Step 2 4.61
    $2^{-10}$ $10^{-4}$ $30$ 16 $[16, 32]$ 18 6.1e-05 7.15
    Panel B: $\text{GLR}$, maximal acceptability $\alpha^* = 279.62$.
    $x_0$ $\epsilon$ $M$ Step 1 Step 2 Run time
    Iter $[x_L, x_U]$ Iter $x_U - x_L$ (s)
    $2$ $10^{-4}$ $15$ 9 $[256,512]$ 22 6.1e-05 21.53
    $20$ $10^{-4}$ $15$ 5 $[160,320]$ 21 7.6e-05 17.27
    $200$ $10^{-4}$ $15$ 2 $[200,400]$ 21 9.5e-05 15.74
    $2$ $10^{-8}$ $15$ 9 $[256,512]$ 35 7.5e-09 30.40
    $2^{25}$ $10^{-4}$ $15$ 15 $[0, 2048]$ no Step 2 6.50
    $2^{25}$ $10^{-4}$ $30$ 18 $[0.5,1]$ 22 6.1e-05 23.91
    $2^{-10}$ $10^{-4}$ $15$ 15 $[16, \infty]$ no Step 2 13.41
    $2^{-10}$ $10^{-4}$ $30$ 20 $[256,512]$ 22 6.1e-05 30.27
    Panel C: $\text{RAROC}$, maximal acceptability $\alpha^* = 279.62$.
    $x_0$ $\epsilon$ $M$ Step 1 Step 2 Run time
    Iter $[x_L, x_U]$ Iter $x_U - x_L$ (s)
    $2$ $10^{-4}$ $15$ 2 $[2, 4]$ 15 6.1e-05 7.20
    $20$ $10^{-4}$ $15$ 4 $[2.5, 5]$ 15 7.6e-05 9.41
    $200$ $10^{-4}$ $15$ 8 $[1.56, 3.13]$ 14 9.4e-05 12.84
    $2$ $10^{-8}$ $15$ 2 $[2, 4]$ 28 7.5e-09 11.07
    $2^{20}$ $10^{-4}$ $15$ 15 $[0,64]$ no Step 2 10.55
    $2^{20}$ $10^{-4}$ $30$ 20 $[2, 4]$ 15 6.1e-05 19.59
    $2^{-15}$ $10^{-4}$ $15$ 15 $[0.5, \infty]$ no Step 2 7.04
    $2^{-15}$ $10^{-4}$ $30$ 18 $[2, 4]$ 15 6.1e-05 13.41
     | Show Table
    DownLoad: CSV

    Table 4.  A comparison of the different versions of the algorithm in a market with $ d = 10 $ assets and $ \vert \Omega \vert = 1000 $ states of the world both with and without short-selling. A tolerance $ \epsilon = 10^{-4} $ is used for all algorithms, the original and zero-level version use $ x_0 = 2 $ and $ \bar{M} = 15 $. Obtaining the final approximation $ [x_L, x_U] $ is denoted in the table by $ \alpha^* $, values are listed to two decimal places

    Panel A: $\text{AIT}$, the maximal acceptability with short-selling constraints ($h \geq 0$) is $\alpha^* = 25.45,$ without short-selling constraints ($h$ free) it is $\alpha^* =25.72$.
    lgorithm Step 1 Bisection on $q$ Bisection on $x$ $x_U - x_L$ Run time
    Iter $[x_L, x_U]$ Iter $[x_L, x_U]$ Iter $[x_L, x_U]$ (s)
    $h \geq 0$ Original 5 $[16, 32]$ 18 $\alpha^*$ 6.1e-05 3.32
    Modified 2 $[0, \infty]$ 23 $\alpha^*$ 8.3e-05 3.67
    Mixed 2 $[0, \infty]$ 5 $[15, 31]$ 18 $\alpha^*$ 6.1e-05 3.90
    Zero level 3 $[23.42, 46.84]$ 18 $\alpha^*$ 5.9e-05 2.96
    $h$ free Original 5 $[16, 32]$ 18 $\alpha^*$ 6.1e-05 4.89
    Modified 2 $[0, \infty]$ 23 $\alpha^*$ 8.5e-05 5.11
    Mixed 2 $[0, \infty]$ 5 $[15, 31]$ 18 $\alpha^*$ 6.1e-05 5.09
    Zero level 3 $[23.45, 46.89]$ 18 $\alpha^*$ 5.1e-05 4.36
    Panel B: $\text{GLR}$, the maximal acceptability with short-selling constraints ($h \geq 0$) is $\alpha^* = 279.62,$ without short-selling constraints ($h$ free) it is $\alpha^* =288.88$.
    Algorithm Step 1 Bisection on $q$ Bisection on $x$ $x_U - x_L$ Run time
    Iter $[x_L, x_U]$ Iter $[x_L, x_U]$ Iter $[x_L, x_U]$ (s)
    $h \geq 0$ Original 9 $[256,512]$ 22 $\alpha^*$ 6.1e-05 19.45
    Modified 2 $[0, \infty]$ 29 $\alpha^*$ 7.4e-05 18.42
    Mixed 2 $[0, \infty]$ 8 $[254,510]$ 22 $\alpha^*$ 6.1e-05 19.75
    Zero level 3 $[279.62,559.24]$ 22 $\alpha^*$ 6.7e-05 14.54
    $h$ free Original 9 $[256,512]$ 22 $\alpha^*$ 6.1e-05 39.56
    Modified 2 $[0, \infty]$ 29 $\alpha^*$ 7.9e-05 40.85
    Mixed 2 $[0, \infty]$ 8 $[254,510]$ 22 $\alpha^*$ 6.1e-05 41.66
    Zero level 3 $[288.88,577.76]$ 22 $\alpha^*$ 6.8e-05 32.17
    Panel C: $\text{RAROC}$, the maximal acceptability with short-selling constraints ($h \geq 0$) is $\alpha^* = 2.98,$ without short-selling constraints ($h$ free) it is $\alpha^* =3.08$.
    Algorithm Step 1 Bisection on $q$ Bisection on $x$ $x_U - x_L$ Run time
    Iter $[x_L, x_U]$ Iter $[x_L, x_U]$ Iter $[x_L, x_U]$ (s)
    $h \geq 0$ Original 2 $[2, 4]$ 15 $\alpha^*$ 6.1e-05 5.88
    Modified 2 $[0, \infty]$ 18 $\alpha^*$ 6.0e-05 5.88
    Mixed 2 $[0, \infty]$ 2 $[1, 3]$ 15 $\alpha^*$ 6.1e-05 5.62
    Zero level 2 $[2.98, 5.96]$ 15 $\alpha^*$ 9.1e-05 5.53
    $h$ free Original 2 $[2, 4]$ 15 $\alpha^*$ 6.1e-05 6.91
    Modified 2 $[0, \infty]$ 18 $\alpha^*$ 6.3e-05 8.00
    Mixed 2 $[0, \infty]$ 3 $[3, 7]$ 18 $\alpha^*$ 6.1e-05 8.12
    Zero level 2 $[3.08, 6.15]$ 15 $\alpha^*$ 9.4e-05 7.04
     | Show Table
    DownLoad: CSV
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