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A note on the solution equivalence to general models for RIM quantifier problems
JID:FSS AID:7192 /COR
[m3SC+; v1.255; Prn:31/03/2017; 11:06] P.1 (1-4)
Available online at www.sciencedirect.com
ScienceDirect Fuzzy Sets and Systems ••• (••••) •••–••• www.elsevier.com/locate/fss
Correspondence
A note on the solution equivalence to general models for RIM quantifier problems Dug Hun Hong ∗ Department of Mathematics, Myongji University, Yongin Kyunggido 449-728, South Korea Received 14 July 2016; received in revised form 28 February 2017; accepted 26 March 2017
Abstract This note presents a counterexample that illustrates why the lack of absolute continuity destroys the correctness of a main result of Liu (2008) and provides a modified model for the minimax RIM quantifier problem and the correct formulation of Liu’s result. © 2017 Elsevier B.V. All rights reserved.
Keywords: Fuzzy sets; OWA operator; RIM quantifier; Absolute continuity; Generating function; Minimal variability; Minimax disparity; Solution equivalence
Recently, Hong [1] studied the relationship between the minimum variance and minimax disparity RIM quantifier problems and investigated the equivalence of their solutions to minimum variance and minimax disparity RIM quantifier problems. Hong [2] also studied the relationship between the maximum entropy and minimax ratio RIM quantifier problems and investigated the equivalence of their solutions to the maximum entropy and minimax ratio RIM quantifier problems. Liu [3] proposed a general RIM quantifier determination model, solved it analytically using the optimal control technique and investigated the solution equivalence to the minimax problem for the RIM quantifier. We note that the result of Liu [3] is correct, only if the generating function is absolutely continuous, an assumption that seems to be taken for granted in the paper. But in general, a generating function is not assumed to be absolutely continuous. In this note, we present a counterexample showing that the result of Liu [3] is no longer valid if only the continuity of f is assumed, and provide a modified model for the minimax RIM quantifier problem and an improved correct formulation of Liu’s result. Liu [3] considered a general model for the minimax RIM quantifier problem as follows:
Minimize
Mf = maxr∈(0,1) |F (f (r))f (r)|
* Fax: 82 31 336 9528.
E-mail address: [email protected]. http://dx.doi.org/10.1016/j.fss.2017.03.014 0165-0114/© 2017 Elsevier B.V. All rights reserved.
JID:FSS AID:7192 /COR
[m3SC+; v1.255; Prn:31/03/2017; 11:06] P.2 (1-4)
D.H. Hong / Fuzzy Sets and Systems ••• (••••) •••–•••
2
1 rf (r)dr = α,
subject to
0 < α < 1,
(1)
0
1 f (r)dr = 1, 0
f (r) > 0, where the generating functions are continuous and F is a strictly convex function on [0, ∞), that is differentiable to at least the 2nd order. Liu [3] proved the following: Theorem 1 ([3, Theorem 13]). There is a unique optimal solution for (1), and the optimal solution has the form −1 ∗ (F ) (a r + b∗ ), if (F )−1 (a ∗ r + b∗ ) ≥ 0, f ∗ (r) = 0 elsewhere, where a ∗ and b∗ are determined by the constraints: ⎧ 1 ∗ ⎪ ⎨ 0 rf (r)dr = α 1 ∗ f (r)dr = 1, ⎪ ⎩ 0∗ f (r) ≥ 0, which is the same as the solution of the general minimum RIM quantifier problem. Theorem 1 is invalid if only the continuity of f is assumed. To show this, we consider the following example. Example 1. Let F (x) = (1/2)x 2 , then F (x) = x and F (x) = 1. Let C(x) be a Cantor function that is non-decreasing 1 1 continuous with C (x) = 0, a.e. (see [4]). Let f ∗ (x) = 2C(x) since 0 C(r)dr = 1/2 and let 0 (1 − r)f ∗ (r)dr = α0 . Then we have maxr∈(0,1) |F (f ∗ (r))f ∗ (r)| = 0. Hence f ∗ (x) = 2C(x) is the optimal solution of problem (2) under a given orness level, α0 . This example shows that Theorem 1 is incorrect. Indeed, f ∗ (x) = 2C(x) differs from the optimal solution in Theorem 1. If f (x) is continuous, f (x) may not exist in general. We do consider the Lebesgue measure [4] in this problem and modify the general model for the minimax RIM quantifier problem as follows: Minimize
Mf = ess supr∈(0,1) |F (f (x))f (x)| 1 rf (r)dr = α,
subject to
0 < α < 1,
(2)
0
1 f (r)dr = 1, 0
f (r) > 0. Now, we give an improved correct formulation of Liu’s result. Theorem 2. Suppose that the generating functions are absolutely continuous, F is a strictly convex function on [0, ∞), and F is absolutely continuous, then there is a unique optimal solution for (2), and that optimal solution has the form f ∗ (r) = max{(F )−1 (a ∗ r + b∗ ), 0}
(3)
JID:FSS AID:7192 /COR
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D.H. Hong / Fuzzy Sets and Systems ••• (••••) •••–•••
3
where a ∗ and b∗ are determined by the constraints: ⎧ 1 ∗ ⎪ ⎨ 0 rf (r)dr = α 1 ∗ ⎪ 0 f (r)dr = 1 ⎩ f ∗ (r) ≥ 0. Proof. If a ∗ = 0, then (f ∗ ) (x) = 0. Hence f ∗ (x) = 1 which is the optimal solution for α = 1/2. Without loss of generality, we can assume that α ∈ (0, 1/2). We first show f ∗ , defined by (3), is an optimal solution. Since F (x) is strictly increasing, we have that a ∗ < 0. Hence, without loss of generality, we may assume that {r|(F )−1 (a ∗ r + b∗ ) < 0} = (t, 1] for some t ∈ (0, 1] and F (f ∗ (r)) = a ∗ r + b∗ , r ∈ [0, t]. Then |F (f ∗ (x))(f ∗ ) (x)| = |a ∗ | on (0, t) and 0, otherwise. This shows that ess supr∈(0,1) |F (f ∗ (r))f ∗ (r)| = |a ∗ |. Let f be a nonnegative absolutely continuous function satisfying 1 = that
1 0
f (r)dr and
1 0
rf (r)dr = α and suppose
ess supr∈(0,1) |F (f (r))f (r)| < |a ∗ |. Since a ∗ < 0, we have that ess inf r∈(0,1) F (f (r))f (r) > a ∗ . Then we should have that F (f (0)) < F (f ∗ (0)) = b∗ . Indeed, if this assertion does not hold, then we have F (f (0)) ≥ F (f ∗ (0)) = b∗ . If R(x) = F (f (x)), then, noting x that since both F (x) and f (x) are absolutely continuous, R(x) = 0 R (r)dr + R(0) and we have for x ∈ (0, t), x
F (f (x)) = R(x) =
x
F (f (r))f (r)dr + R(0)
R (r)dr + R(0) = 0
0
x >
(4)
a ∗ dr + b∗ = a ∗ x + b∗
0
= F (f ∗ (x)).
t 1 Since F is strictly increasing, we have f (x) > f ∗ (x) and therefore 1 = 0 f ∗ (r)dr < 0 f (r)dr which is a contradiction. It follows that F (f (0)) < F (f ∗ (0)) = b∗ . Now, let H (r) = F (f (r)) − F (f ∗ (r)). Then, since H (x) = F (f (r))f (r) − a ∗ > 0 a.e. for r ∈ (0, t) and H (0) < 0, and H (r) ≥ 0, r ∈ [t, 1], there exists t0 ∈ (0, t] such that H (t0 ) = F (f (t0 )) − F (f ∗ (t0 )) = 0. Therefore h(r) = f (r) − f ∗ (r) < 0, r ∈ (0, t0 ) and h(r) = f (r) − f ∗ (r) ≥ 0, r ∈ (t0 , 1), since F is strictly increasing. Then we have 1
t0 rh(r)dr =
0
1 rh(r)dr +
t0 rh(r)dr ≥
1 rh(r)dr +
0
t0
0
t0
t0
t0
rh(r)dr −
= 0
> 0, which is contradictory to
t0 h(r)dr = 0
1 0
t0 h(r)dr
(5)
t0
(r − t0 )h(r)dr 0
rh(r)dr = 0 by our initial assumption. It follows that
Min{ess supr∈(0,1) |F (f (r))f (r)|} = |a ∗ |. Hence f ∗ (x) is an optimal solution. We now demonstrate its uniqueness. Let f be a nonnegative absolutely continuous 1 1 function satisfying 0 f (r)dr = 1 and 0 rf (r)dr = α, and suppose that ess supr∈(0,1) |F (f (r))f (r)| = |a ∗ |. Then we can easily show that F (f (0)) ≤ F (f ∗ (0)) = b∗ using a similar inequality to that in (4). We can also show that F (f (0)) ≥ F (f ∗ (0)) = b∗ using a similar inequality to that in (5). Thus we have that F (f (0)) = F (f ∗ (0)) = b∗ . Now, since ess inf r∈(0,1) F (f (r))f (r) = a ∗ , for x ∈ [0, t] and F is absolutely continuous, then
JID:FSS AID:7192 /COR
[m3SC+; v1.255; Prn:31/03/2017; 11:06] P.4 (1-4)
D.H. Hong / Fuzzy Sets and Systems ••• (••••) •••–•••
4
x
F (f (x)) =
x
F (f (r))f (r)dr + F (f (0)) ≥ 0
a ∗ dr + b∗
0 ∗
∗
∗
= a x + b = F (f (x)). f (r) ≥ f ∗ (r)
Then for r ∈ [0, 1]. Therefore we have that f (r) = f ∗ (r) noting that both functions are continuous and 1 1 ∗ 0 f (r)dr = 0 f (r)dr. The proof is complete. Acknowledgements This research was supported by the Korea Research Foundation Grant funded by Korean Government (KRF-2008313-C00111). References [1] D.H. Hong, The relationship between the minimum variance and minimax disparity RIM quantifier problems, Fuzzy Sets Syst. 181 (2011) 50–57. [2] D.H. Hong, The relationship between the maximum entropy and minimax ratio RIM quantifier problems, Fuzzy Sets Syst. 202 (2012) 110–117. [3] X. Liu, A general model of parameterized OWA aggregation with given orness level, Int. J. Approx. Reason. 48 (2008) 598–627. [4] R.L. Wheeden, A. Zygmund, Measure and Integral: An Introduction to Real Analysis, Marcel Dekker, Inc., 1977.
[m3SC+; v1.255; Prn:31/03/2017; 11:06] P.1 (1-4)
Available online at www.sciencedirect.com
ScienceDirect Fuzzy Sets and Systems ••• (••••) •••–••• www.elsevier.com/locate/fss
Correspondence
A note on the solution equivalence to general models for RIM quantifier problems Dug Hun Hong ∗ Department of Mathematics, Myongji University, Yongin Kyunggido 449-728, South Korea Received 14 July 2016; received in revised form 28 February 2017; accepted 26 March 2017
Abstract This note presents a counterexample that illustrates why the lack of absolute continuity destroys the correctness of a main result of Liu (2008) and provides a modified model for the minimax RIM quantifier problem and the correct formulation of Liu’s result. © 2017 Elsevier B.V. All rights reserved.
Keywords: Fuzzy sets; OWA operator; RIM quantifier; Absolute continuity; Generating function; Minimal variability; Minimax disparity; Solution equivalence
Recently, Hong [1] studied the relationship between the minimum variance and minimax disparity RIM quantifier problems and investigated the equivalence of their solutions to minimum variance and minimax disparity RIM quantifier problems. Hong [2] also studied the relationship between the maximum entropy and minimax ratio RIM quantifier problems and investigated the equivalence of their solutions to the maximum entropy and minimax ratio RIM quantifier problems. Liu [3] proposed a general RIM quantifier determination model, solved it analytically using the optimal control technique and investigated the solution equivalence to the minimax problem for the RIM quantifier. We note that the result of Liu [3] is correct, only if the generating function is absolutely continuous, an assumption that seems to be taken for granted in the paper. But in general, a generating function is not assumed to be absolutely continuous. In this note, we present a counterexample showing that the result of Liu [3] is no longer valid if only the continuity of f is assumed, and provide a modified model for the minimax RIM quantifier problem and an improved correct formulation of Liu’s result. Liu [3] considered a general model for the minimax RIM quantifier problem as follows:
Minimize
Mf = maxr∈(0,1) |F (f (r))f (r)|
* Fax: 82 31 336 9528.
E-mail address: [email protected]. http://dx.doi.org/10.1016/j.fss.2017.03.014 0165-0114/© 2017 Elsevier B.V. All rights reserved.
JID:FSS AID:7192 /COR
[m3SC+; v1.255; Prn:31/03/2017; 11:06] P.2 (1-4)
D.H. Hong / Fuzzy Sets and Systems ••• (••••) •••–•••
2
1 rf (r)dr = α,
subject to
0 < α < 1,
(1)
0
1 f (r)dr = 1, 0
f (r) > 0, where the generating functions are continuous and F is a strictly convex function on [0, ∞), that is differentiable to at least the 2nd order. Liu [3] proved the following: Theorem 1 ([3, Theorem 13]). There is a unique optimal solution for (1), and the optimal solution has the form −1 ∗ (F ) (a r + b∗ ), if (F )−1 (a ∗ r + b∗ ) ≥ 0, f ∗ (r) = 0 elsewhere, where a ∗ and b∗ are determined by the constraints: ⎧ 1 ∗ ⎪ ⎨ 0 rf (r)dr = α 1 ∗ f (r)dr = 1, ⎪ ⎩ 0∗ f (r) ≥ 0, which is the same as the solution of the general minimum RIM quantifier problem. Theorem 1 is invalid if only the continuity of f is assumed. To show this, we consider the following example. Example 1. Let F (x) = (1/2)x 2 , then F (x) = x and F (x) = 1. Let C(x) be a Cantor function that is non-decreasing 1 1 continuous with C (x) = 0, a.e. (see [4]). Let f ∗ (x) = 2C(x) since 0 C(r)dr = 1/2 and let 0 (1 − r)f ∗ (r)dr = α0 . Then we have maxr∈(0,1) |F (f ∗ (r))f ∗ (r)| = 0. Hence f ∗ (x) = 2C(x) is the optimal solution of problem (2) under a given orness level, α0 . This example shows that Theorem 1 is incorrect. Indeed, f ∗ (x) = 2C(x) differs from the optimal solution in Theorem 1. If f (x) is continuous, f (x) may not exist in general. We do consider the Lebesgue measure [4] in this problem and modify the general model for the minimax RIM quantifier problem as follows: Minimize
Mf = ess supr∈(0,1) |F (f (x))f (x)| 1 rf (r)dr = α,
subject to
0 < α < 1,
(2)
0
1 f (r)dr = 1, 0
f (r) > 0. Now, we give an improved correct formulation of Liu’s result. Theorem 2. Suppose that the generating functions are absolutely continuous, F is a strictly convex function on [0, ∞), and F is absolutely continuous, then there is a unique optimal solution for (2), and that optimal solution has the form f ∗ (r) = max{(F )−1 (a ∗ r + b∗ ), 0}
(3)
JID:FSS AID:7192 /COR
[m3SC+; v1.255; Prn:31/03/2017; 11:06] P.3 (1-4)
D.H. Hong / Fuzzy Sets and Systems ••• (••••) •••–•••
3
where a ∗ and b∗ are determined by the constraints: ⎧ 1 ∗ ⎪ ⎨ 0 rf (r)dr = α 1 ∗ ⎪ 0 f (r)dr = 1 ⎩ f ∗ (r) ≥ 0. Proof. If a ∗ = 0, then (f ∗ ) (x) = 0. Hence f ∗ (x) = 1 which is the optimal solution for α = 1/2. Without loss of generality, we can assume that α ∈ (0, 1/2). We first show f ∗ , defined by (3), is an optimal solution. Since F (x) is strictly increasing, we have that a ∗ < 0. Hence, without loss of generality, we may assume that {r|(F )−1 (a ∗ r + b∗ ) < 0} = (t, 1] for some t ∈ (0, 1] and F (f ∗ (r)) = a ∗ r + b∗ , r ∈ [0, t]. Then |F (f ∗ (x))(f ∗ ) (x)| = |a ∗ | on (0, t) and 0, otherwise. This shows that ess supr∈(0,1) |F (f ∗ (r))f ∗ (r)| = |a ∗ |. Let f be a nonnegative absolutely continuous function satisfying 1 = that
1 0
f (r)dr and
1 0
rf (r)dr = α and suppose
ess supr∈(0,1) |F (f (r))f (r)| < |a ∗ |. Since a ∗ < 0, we have that ess inf r∈(0,1) F (f (r))f (r) > a ∗ . Then we should have that F (f (0)) < F (f ∗ (0)) = b∗ . Indeed, if this assertion does not hold, then we have F (f (0)) ≥ F (f ∗ (0)) = b∗ . If R(x) = F (f (x)), then, noting x that since both F (x) and f (x) are absolutely continuous, R(x) = 0 R (r)dr + R(0) and we have for x ∈ (0, t), x
F (f (x)) = R(x) =
x
F (f (r))f (r)dr + R(0)
R (r)dr + R(0) = 0
0
x >
(4)
a ∗ dr + b∗ = a ∗ x + b∗
0
= F (f ∗ (x)).
t 1 Since F is strictly increasing, we have f (x) > f ∗ (x) and therefore 1 = 0 f ∗ (r)dr < 0 f (r)dr which is a contradiction. It follows that F (f (0)) < F (f ∗ (0)) = b∗ . Now, let H (r) = F (f (r)) − F (f ∗ (r)). Then, since H (x) = F (f (r))f (r) − a ∗ > 0 a.e. for r ∈ (0, t) and H (0) < 0, and H (r) ≥ 0, r ∈ [t, 1], there exists t0 ∈ (0, t] such that H (t0 ) = F (f (t0 )) − F (f ∗ (t0 )) = 0. Therefore h(r) = f (r) − f ∗ (r) < 0, r ∈ (0, t0 ) and h(r) = f (r) − f ∗ (r) ≥ 0, r ∈ (t0 , 1), since F is strictly increasing. Then we have 1
t0 rh(r)dr =
0
1 rh(r)dr +
t0 rh(r)dr ≥
1 rh(r)dr +
0
t0
0
t0
t0
t0
rh(r)dr −
= 0
> 0, which is contradictory to
t0 h(r)dr = 0
1 0
t0 h(r)dr
(5)
t0
(r − t0 )h(r)dr 0
rh(r)dr = 0 by our initial assumption. It follows that
Min{ess supr∈(0,1) |F (f (r))f (r)|} = |a ∗ |. Hence f ∗ (x) is an optimal solution. We now demonstrate its uniqueness. Let f be a nonnegative absolutely continuous 1 1 function satisfying 0 f (r)dr = 1 and 0 rf (r)dr = α, and suppose that ess supr∈(0,1) |F (f (r))f (r)| = |a ∗ |. Then we can easily show that F (f (0)) ≤ F (f ∗ (0)) = b∗ using a similar inequality to that in (4). We can also show that F (f (0)) ≥ F (f ∗ (0)) = b∗ using a similar inequality to that in (5). Thus we have that F (f (0)) = F (f ∗ (0)) = b∗ . Now, since ess inf r∈(0,1) F (f (r))f (r) = a ∗ , for x ∈ [0, t] and F is absolutely continuous, then
JID:FSS AID:7192 /COR
[m3SC+; v1.255; Prn:31/03/2017; 11:06] P.4 (1-4)
D.H. Hong / Fuzzy Sets and Systems ••• (••••) •••–•••
4
x
F (f (x)) =
x
F (f (r))f (r)dr + F (f (0)) ≥ 0
a ∗ dr + b∗
0 ∗
∗
∗
= a x + b = F (f (x)). f (r) ≥ f ∗ (r)
Then for r ∈ [0, 1]. Therefore we have that f (r) = f ∗ (r) noting that both functions are continuous and 1 1 ∗ 0 f (r)dr = 0 f (r)dr. The proof is complete. Acknowledgements This research was supported by the Korea Research Foundation Grant funded by Korean Government (KRF-2008313-C00111). References [1] D.H. Hong, The relationship between the minimum variance and minimax disparity RIM quantifier problems, Fuzzy Sets Syst. 181 (2011) 50–57. [2] D.H. Hong, The relationship between the maximum entropy and minimax ratio RIM quantifier problems, Fuzzy Sets Syst. 202 (2012) 110–117. [3] X. Liu, A general model of parameterized OWA aggregation with given orness level, Int. J. Approx. Reason. 48 (2008) 598–627. [4] R.L. Wheeden, A. Zygmund, Measure and Integral: An Introduction to Real Analysis, Marcel Dekker, Inc., 1977.