It is well known that the linear extension majority (LEM) relation of a poset of size $n\ge 9$ can contain cycles. In this paper we are interested in obtaining minimum cutting levels ${\alpha }_m$ such that the crisp relation obtained from the mutual rank probability relation by setting to $0$ its elements smaller than or equal to ${\alpha }_m$, and to $1$ its other elements, is free from cycles of length $m$. In a first part, theoretical upper bounds for ${\alpha }_m$ are derived using known transitivity properties of the mutual rank...

Given a fixed dependency graph $G$ that describes a Bayesian network of binary variables $X_1,\cdots ,X_n$, our main result is a tight bound on the mutual information $I_c(Y_1,\cdots ,Y_k)={\sum }_{j=1}^{k}H\left(Y_j\right)/c-H(Y_1,\cdots ,Y_k)$ of an observed subset $Y_1,\cdots ,Y_k$ of the variables $X_1,\cdots ,X_n$. Our bound depends on certain quantities that can be computed from the connective structure of the nodes in $G$. Thus it allows to discriminate between different dependency graphs for a probability distribution, as we show from numerical experiments.

The regression model X(t),Y(t);t=1,...,n with random explanatory variable X is transformed by prescribing a partition $S_1,...,S_k$ of the given domain S of X-values and specifying$X\left(1\right),...,X\left(n\right)\cap S_i=X_{i1},...,X_{i\alpha \left(i\right)},i=1,...,k.$Through the conditioning$\alpha \left(i\right)=a\left(i\right),i=1,...,k,X_{i1},...,X_{i\alpha \left(i\right)};i=1,...,k=x_{11},...,x_{ka\left(k\right)}$the initial model with i.i.d. pairs (X(t),Y(t)),t=1,...,n, becomes a conditional fixed-design $(x_{11},...,x_{ka\left(k\right)})$ model$Y_{ij},i=1,...,k;j=1,...,a\left(i\right)$where the response variables $Y_{ij}$ are independent and distributed according to the mixed conditional distribution $Q(·,x_{ij})$ of Y given X at the observed value $x_{ij}$.Afterwards, we investigate the case$\left(Q\right)E\left(Y^{\text{'}}\right|x)={\sum }_{i=1}^k{b}_i\left(x\right){\theta }_i{I}_{S_i}\left(x\right),\left(Q\right)D\left(Y\right|x)={\sum }_{i=1}^k{d}_i\left(x\right){\Sigma }_i{I}_{S_i}\left(x\right)$which...