From Eq. A < δ is needed to ensure a well-defined E[U (πr (Q))], and this leads to r δ + s α > w which not only implies that only a positive unit profit can prompt the retailer to join the game, it also r − w gives a tolerable maximal temporary shrinkage r − w α = r − s and a tolerable maximal permanent shrinkage β = ; clearly, α > β . Therefore, the permanent shrinkage is more intolerant in compared with the temporary shrinkage. To maximize E[U (πr (Q))], we take the first- and second-order derivatives with respect to Q as follows. dE[U (πr (Q))] = rδ + sα − w − (r − s)δ F (δ Q) − (λ −1)(w − s(1− β ))F ( AQ) dQ
From Eq. (5.1), transport through the system is then described by the following equation J(z) = G V˜3 , (5.3) 2J 1/2 where 3 (z) 4 V˜4 11 12 2N1 − T (0) + 2N2 − T (0) + T (0) + T (0) 13 (0) (0) T (0) − T (0) 00 (x) 00
From Eq. 13), the actual purchase probability for the nth consumer should be P = ⟨ exp hn(Sn−1, . . . , Sn−Nr ) ⟩, (4.1) [exp hn (Sn−1 n−Nr ) + exp —hn (Sn−1 n−Nr )] The ⟨⟩ represents an average of all possible 2Nr outcomes of the sample configu- rations. For a small sample memory, an average over a binomial distribution is required. As Nr increases, tails of the probability distribution will decrease rapidly and all possible configurations will centralize around its expectation value. In this case, the central limit theorem could be applied and a Gaussian distribution will be a good approximation. ΣNr For simplicity, define the random variable ξ = (1/N ) S , then rewrite i r j=1 i−j the social cue field in terms of ξi as Jξi. The average of this variable ⟨ξi⟩ and its variance σ2 = ⟨ξ2⟩ — ⟨ξi⟩2 can be calculated as i i ⟨ξi⟩ = " XXx XXx Xx−j 1 (4.2) ξi 2 = (4/N2) Pi−j(1 — Pi−j). Using the Gaussian approximation for ξn, with parameters given in Eq.(4.2), gives for the probability distribution of ξ0: PGauss,i(ξi) = q exp " # — 2 (ξi — ⟨ξi⟩)2 (2σ ) (4.3) 2πσξi ξi
From Eq the Dyson equation, Gα can be written in terms of the free-space GF G0, Gα = G0 + G0TαG0, (2.33) where Tα is the total T-matrix for the system, which describes scattering by both the impurity potential vα and the coherent potential vc. Alternatively, Gα can be expressed in terms of the GF for the coherent potential lattice Gc, Gα = Gc + GcTα,0Gc (2.34)
From Eq the rate of emission of ML pulses per unit time is given by stress versus stress curve of a fluorescent N- acetylanthranilic acid and hexaphenylcarbodiphosphorane crystals, respectively. It is seen that the ML appears concurrently with the steps occurring in the stress-strain curve of the crystals. As the steps in the stress-strain curve correspond to the movement of a crack in the crystal, it seems that the creation of new surfaces is responsible for the ML emission. The ML-strain and stress-strain curves of the sample were determined by Xxxxxxx and Xxxx (1980a,b) by slowly compressing the crystals at a fixed strain rate
From Eq we can find that the sources (X, Y˜ , Z) satisfy X ↔ Y˜ ↔ Z. Furthermore, from Eq. (22), we can also find that X ↔ Y˜ ↔ Y , which implies R(X, Y, Z) ⊂ R(X, Y˜ , Z). x|u
From Eq we can find that the sources (X, Y˜ , Z) satisfy X ↔ Y˜ ↔ Z. Furthermore, from Eq. (22), we can also find that X ↔ Y˜ ↔ Y , which implies R(X, Y, Z) ⊂ R(X, Y˜ , Z). x|u Thus, it suffice to show that Eq. (20) holds for any (Rp, Rk) ∈ R(X, Y˜ , Z). In steps 2 and 3, we will show that I(U ; X|Y˜ ) = I(U ; X) − I(U ; Y˜ ), I(U ; Y˜ |Z) = I(U ; Y˜ ) − I(U ; Z), I(U ; X|Z) = I(U ; X) − I(U ; Z), I(U ; X|Y˜ ) = I(U ; X|Z) − I(U ; Y˜ |Z) By noting the relations Σxz Σx x|u Rk − µRp ≤ I˜k(Σ∗ ) − µI˜p(Σ∗ ) (25) for random variables satisfying U ↔ X ↔ Y˜ ↔ Z, we have where U is not necessarily Gaussian. By using the conditional I˜k (Σ x|u ) − µI˜p(Σ x|u) version of EPI [26], we have = I(U ; Y˜ |Z) − µI(U ; X|Y˜ ) = (1 + µ)I(U ; Y˜ |Z) − µI(U ; X|Z) h(X|U, Z) − γh(X + N3|U, Z) (32) ≤ h(X|U, Z) = [(1 + µ)h(Y˜ |Z) − µh(X|Z)] +[µh(X|U, Z) − (1 + µ)h(Y˜ |U, Z)] − γm log . exp Σ Σ h(X|U, Z) + exp Σ h(N3) = [(1 + µ)h(Y˜ |Z) − µh(X|Z)] − log |Ky˜xKT | ≤ f h(N3) − log(γ − 1); h(N3) 2 y˜x +µ[h(X|U, Z) − γh(X + K−1N2|U, Z)] = [(1 + µ)h(Y˜ |Z) − µh(X|Z)] − 1 + µ log |K y˜x
From Eq we must have λ ≤ x (1−γ)−ε . From Eq.
