An improved distance-based total uncertainty measure in belief function theory

Abstract

Uncertainty quantification of mass functions is a crucial and unsolved issue in belief function theory. Previous studies have mostly considered this problem from the perspective of viewing the belief function theory as an extension of probability theory. Recently, Yang and Han have developed a new distance-based total uncertainty measure directly and totally based on the framework of belief function theory so that there is no switch between the frameworks of belief function theory and probability theory in that measure. However, we have found some obvious deficiencies in Yang and Han’s uncertainty measure which could lead to counter-intuitive results in some cases. In this paper, an improved distance-based total uncertainty measure has been proposed to overcome the limitations of Yang and Han’s uncertainty measure. The proposed measure not only retains the desired properties of original measure, but also possesses higher sensitivity to the change of evidences. A number of examples and applications have verified the effectiveness and rationality of the proposed uncertainty measure.

Appendix

Proof of Property 1 Given a BPA m over the FOD Θ, let the belief interval of element 𝜃 _i in Θ be [Bel(𝜃 _i ),Pl(𝜃 _i )]. Since $0 \le {d_{E}^{I}} \left ({[Bel(\theta _{i}), Pl(\theta _{i})],[0 ,1 ]} \right ) \le 1$, in (21) the upper bound of iTU ^I(m), which is n, is obtained when ${d_{E}^{I}} \left ({[Bel(\theta _{i}), Pl(\theta _{i})],[0 ,1 ]} \right ) = 0$, ∀𝜃 _i ∈Θ. The condition ${d_{E}^{I}} \left ({[Bel(\theta _{i}), Pl(\theta _{i})],[0 ,1 ]} \right ) = 0$, ∀𝜃 _i ∈Θ, can uniquely be satisfied in the vacuous BPA m(Θ)=1.

Similarly, the lower bound of iTU ^I(m), which is 0, is obtained when ${d_{E}^{I}} \left ({[Bel(\theta _{i}), Pl(\theta _{i})],[0 ,1 ]} \right ) = 1$, ∀𝜃 _i ∈Θ, which corresponds to [Bel(𝜃 _i ),Pl(𝜃 _i )]=[0,0] or [Bel(𝜃 _i ),Pl(𝜃 _i )]=[1,1]. It is easily known that the condition of iTU ^I getting its lower bound is associated with a BPA m(𝜃 _i )=1 and m(A)=0 ∀A≠𝜃 _i ,A⊆Θ.

For each 𝜃 _i ∈Ω, assume the belief interval is [Bel(𝜃 _i ),Pl(𝜃 _i )]. Since BPA m is initially defined on Θ, we have [Bel(𝜃 _i ),Pl(𝜃 _i )]=[0,0] if 𝜃 _i ∈Φ and 𝜃 _i ∉Θ. Therefore,

$$\begin{array}{@{}rcl@{}} iTU_{\Omega}^{I} (m) &=& \sum\limits_{\theta_{i} \in {\Theta} } {\left[ {1 - {d_{E}^{I}} ([Bel(\theta_{i} ),Pl(\theta_{i} )],[0,1])} \right]}\\ &&+ \sum\limits_{\theta_{i} \notin {\Theta} ,\theta_{i} \in {\Phi} } {\left[ {1 - {d_{E}^{I}} ([0,0],[0,1])} \right]} \\ &=& \sum\limits_{\theta_{i} \in {\Theta} } {\left[ {1 - {d_{E}^{I}} ([Bel(\theta_{i} ),Pl(\theta_{i} )],[0,1])} \right]}\\ &&+ \sum\limits_{\theta_{i} \notin {\Theta} ,\theta_{i} \in {\Phi} } 0 \\ &=& \sum\limits_{\theta_{i} \in {\Theta} } {\left[ {1 - {d_{E}^{I}} ([Bel(\theta_{i} ),Pl(\theta_{i} )],[0,1])} \right]} \\ &=& iTU_{\Theta}^{I} (m) \\ \end{array} $$

Since $\forall A \subseteq {\Theta } :[Bel_{m_{1} } (A),Pl_{m_{1} } (A)] \subseteq [Bel_{m_{2} } (A),Pl_{m_{2} } (A)]$ , there exists $\forall \theta _{i} \in {\Theta } :[Bel_{m_{1} } (\theta _{i}),Pl_{m_{1} } (\theta _{i})] \subseteq [Bel_{m_{2} } (\theta _{i}),Pl_{m_{2} } (\theta _{i})]$ . Then, we have ${d_{E}^{I}} \left ({[Bel_{m_{1} } (\theta _{i} ),Pl_{m_{1} } (\theta _{i} )],[0,1]} \right ) \ge {d_{E}^{I}} \left ({[Bel_{m_{2} } (\theta _{i} ),Pl_{m_{2} } (\theta _{i} )],[0,1]} \right )$, ∀𝜃 _i ∈Θ. Therefore, according to (21), iTU ^I(m ₁)≤iTU ^I(m ₂).

Let us first prove property 4.1.

In a bayesian BPA, since m(A)=0, ∀|A|≠1, we have m(𝜃 _i ) = Bel(𝜃 _i ) = Pl(𝜃 _i ), ∀𝜃 _i ∈Θ. Assume m(𝜃 _i ) = Bel(𝜃 _i ) = Pl(𝜃 _i ) = x _i for 𝜃 _i , so $\sum \limits _{i = 1}^{n} {x_{i} } = 1$, and

$$\begin{array}{@{}rcl@{}} {d_{E}^{I}} \left( {[Bel(\theta_{i} ),Pl(\theta_{i} )],[0,1]} \right) &=& \sqrt {\left( {x_{i} - 0} \right)^{2} + \left( {x_{i} - 1} \right)^{2} }\\ &=& \sqrt {2{x_{i}^{2}} - 2x_{i} + 1}. \end{array} $$

Then

$$iTU^{I} (m) = \sum\limits_{i = 1}^{n} {\left( {1 - \sqrt {2{x_{i}^{2}} - 2x_{i} + 1} } \right)}. $$

Therefore, we have the following Lagrangian function

$$L(m) = \sum\limits_{i = 1}^{n} {\left( {1 - \sqrt {2{x_{i}^{2}} - 2x_{i} + 1} } \right)} + \lambda \left( {\sum\limits_{i = 1}^{n} {x_{i} } - 1} \right) $$

where λ is the Lagrangian constant. By calculating the derivatives, we obtain

$$\left\{ \begin{array}{l} \frac{{\partial L(m)}}{{\partial x_{i} }} = \frac{{1 - 2x_{i}}}{{\sqrt {2{x_{i}^{2}} - 2x_{i} + 1} }} + \lambda \\ \frac{{\partial L(m)}}{{\partial \lambda }} = \sum\limits_{i = 1}^{n} {x_{i} } - 1 \\ \end{array} \right. $$

Then, by equaling these two functions to 0, we have

$$\left\{ \begin{array}{l} x_{i} = \frac{1}{2}\left( {1 \pm \sqrt {\frac{{\lambda^{2} }}{{2 - \lambda^{2} }}} } \right) \\ \sum\limits_{i = 1}^{n} {x_{i} } = 1 \\ \end{array} \right. $$

At the same time, since each maximum value satisfies

$$\frac{{1 - 2x_{i} }}{{\sqrt {2{x_{i}^{2}} - 2x_{i} + 1} }} + \lambda = 0,\quad \forall i \in \{ 1, {\cdots} ,n\} $$

we have

$$\frac{{1 + \lambda \sqrt {2{x_{i}^{2}} - 2x_{i} + 1} }}{2} = x_{i},\quad \forall i. $$

Hence

$$\sum\limits_{i = 1}^{n} {\frac{{1 + \lambda \sqrt {2{x_{i}^{2}} - 2x_{i} + 1} }}{2}} = \sum\limits_{i = 1}^{n} {x_{i} } = 1, $$

namely,

$$\frac{n}{2} + \frac{\lambda }{2}\sum\limits_{i = 1}^{n} {\sqrt {2{x_{i}^{2}} - 2x_{i} + 1} } = 1. $$

When n≥2, we have λ≤0. Therefore,

$$\frac{{1 - 2x_{i} }}{{\sqrt {2{x_{i}^{2}} - 2x_{i} + 1} }} \ge 0 ,\quad \forall i $$

which implies $x_{i} \le \frac {1}{2}$, ∀i∈{1,⋯,n}. So, we obtain that

$$\left\{ \begin{array}{l} x_{i} = \frac{1}{2}\left( {1 - \sqrt {\frac{{\lambda^{2} }}{{2 - \lambda^{2} }}} } \right) \\ \sum\limits_{i = 1}^{n} {x_{i} } = 1 \\ \end{array} \right. $$

Namely, L(m) gets a maximum value at

$$x_{1} = x_{2} = {\cdots} = x_{n} = 1/n. $$

Moreover, we calculate the second-order derivative of the Lagrangian function L(m)

$$\begin{array}{l} \frac{{\partial^{2} L(m)}}{{\partial {x_{i}^{2}} }} = \frac{{(2x_{i} - 1)^{2} }}{{\left( {\sqrt {2{x_{i}^{2}} - 2x_{i} + 1} } \right)^{3} }} - \frac{2}{{\sqrt {2{x_{i}^{2}} - 2x_{i} + 1} }} \\ \quad \quad \quad = \frac{1}{{\sqrt {2{x_{i}^{2}} - 2x_{i} + 1} }}\left[ {\frac{{(2x_{i} - 1)^{2} }}{{2{x_{i}^{2}} - 2x_{i} + 1}} - 2} \right] \\ \quad \quad \quad = \frac{1}{{\sqrt {2{x_{i}^{2}} - 2x_{i} + 1} }} \cdot \frac{{ - 1}}{{2{x_{i}^{2}} - 2x_{i} + 1}} \\ \quad \quad \quad < 0 \\ \end{array} $$

which means that iTU ^I is strictly concave. Therefore, for the bayesian BPA m, iTU ^I(m) attains the global maximum value when m(𝜃 _i )=1/n, ∀𝜃 _i ∈Θ.

Then, let us first prove property 4.2.

Assume that there is a bayesian BPA m defined on FOD Θ with |Θ| = n. In terms of property 3.1, iTU ^I(m) attains the maximum value when m(𝜃 _i )=1/n, ∀𝜃 _i ∈Θ. Then we have

$$\begin{array}{l} f = \max \;iTU^{I} (m) = \sum\limits_{i = 1}^{n} {\left( {1 - \sqrt {(1/n)^{2} + (1/n - 1)^{2} } } \right)} \\ \quad \quad \quad \quad \quad \quad \quad = n - \sqrt {n^{2} - 2n + 2} \\ \end{array} $$

Here,

$$\begin{array}{l} f_{n}^{\prime} = 1 - {{(n - 1)} \mathord{\left/ {\vphantom {{(n - 1)} {\sqrt {n^{2} - 2n + 2} }}} \right. \kern-\nulldelimiterspace} {\sqrt {n^{2} - 2n + 2} }} \\ \quad = 1 - {{(n - 1)} \mathord{\left/ {\vphantom {{(n - 1)} {\sqrt {(n - 1)^{2} + 1} }}} \right. \kern-\nulldelimiterspace} {\sqrt {(n - 1)^{2} + 1} }} \\ \quad > 0 \\ \end{array} $$

which means f is monotonically increasing with respect to n, $n \in \mathbb {N}^ +$. Furthermore, $\underset {{n \to + \infty }}{\lim } f = 1$. Hence

$$\max \;iTU^{I} (m_{1} ) < \max \;iTU^{I} (m_{2} ) $$

for bayesian BPAs m ₁ defined on Θ₁ with |Θ₁| = n ₁ and m ₂ defined on Θ₂ with |Θ₂| = n ₂, where n ₁<n ₂.