scholarly journals Improvement of Bobrovsky–Mayor–Wolf–Zakai Bound

Entropy ◽  
2021 ◽  
Vol 23 (2) ◽  
pp. 161
Author(s):  
Ken-ichi Koike ◽  
Shintaro Hashimoto

This paper presents a difference-type lower bound for the Bayes risk as a difference-type extension of the Borovkov–Sakhanenko bound. The resulting bound asymptotically improves the Bobrovsky–Mayor–Wolf–Zakai bound which is difference-type extension of the Van Trees bound. Some examples are also given.

2018 ◽  
Vol 38 (2) ◽  
pp. 429-440
Author(s):  
Rafał Wieczorek ◽  
Hanna Podsędkowska

The entropic upper bound for Bayes risk in a general quantum case is presented. We obtained generalization of the entropic lower bound for probability of detection. Our result indicates upper bound for Bayes risk in a particular case of loss function – for probability of detection in a pretty general setting of an arbitrary finite von Neumann algebra. It is also shown under which condition the indicated upper bound is achieved.


2019 ◽  
Vol 485 (2) ◽  
pp. 142-144
Author(s):  
A. A. Zevin

Solutions x(t) of the Lipschitz equation x = f(x) with an arbitrary vector norm are considered. It is proved that the sharp lower bound for the distances between successive extremums of xk(t) equals π/L where L is the Lipschitz constant. For non-constant periodic solutions, the lower bound for the periods is 2π/L. These estimates are achieved for norms that are invariant with respect to permutation of the indices.


10.37236/1188 ◽  
1994 ◽  
Vol 1 (1) ◽  
Author(s):  
Geoffrey Exoo

For $k \geq 5$, we establish new lower bounds on the Schur numbers $S(k)$ and on the k-color Ramsey numbers of $K_3$.


10.37236/1748 ◽  
2003 ◽  
Vol 10 (1) ◽  
Author(s):  
Nagi H. Nahas

The best lower bound known on the crossing number of the complete bipartite graph is : $$cr(K_{m,n}) \geq (1/5)(m)(m-1)\lfloor n/2 \rfloor \lfloor(n-1)/2\rfloor$$ In this paper we prove that: $$cr(K_{m,n}) \geq (1/5)m(m-1)\lfloor n/2 \rfloor \lfloor (n-1)/2 \rfloor + 9.9 \times 10^{-6} m^2n^2$$ for sufficiently large $m$ and $n$.


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