观点

Proposition 11 ([Common blocks]). The common blocks exist indefinitely, and their pairs of slab lists are iid. Their widths K_i have a common law satisfying M≤K_i<∞.s., μ:= E K_i<∞. The expected sum of the radii of all slabs on both sides …

Proposition 11 ([Common blocks]). The common blocks exist indefinitely, and their pairs of slab lists are iid. Their widths Ki have a common law satisfying M≤Ki<∞.s., μ:= E Ki<∞. The expected sum of the radii of all slabs on both sides of one common block is finite. In particular, if W'0=0, W'i=∑j=1iKj, then W'i/i→μ almost surely.

证明 1证否 0作为观点被引用 1链条图发起推演

发起推演1

证明
Let K be the first positive common width, allowing K=∞ for now. For each fixed pair of finite prefixes that realize K=k , the fact that k is their first common width is decided by those prefixes. Their unused tails are independent tapes with the initial laws. Enumerating these prefix pairs therefore gives the renewal recursion c_0=1, c_n=∑_ k=1 ^n P(K=k)c_ n-k (n≥1), c_n= u_n^+ u_n^-. Set f(t)=∑_ k≥1 P(K=k)t^k for 0<t<1 . Nonnegative power-series summation gives ∑_ n≥0 c_nt^n=(1)/(1-f(t)). Equation implies c_n≥ξ^2p_+p_->0 for every n≥M . Hence ∑c_nt^n≥ξ^2p_+p_-t^M/(1-t) , and consequently (1-f(t))/(1-t) stays bounded as t1 . First this forces f(t)→1 , proving K<∞ almost surely. We can then write (1-f(t))/(1-t) = E(1-t^K)/(1-t) E K, which proves finite mean. The same prefix factorization at each common renewal proves that the successive pairs of lists are iid. The lower bound K≥M follows from the minimum slab width. The strong law gives the assertion about W'_i . Let N_+ be the number of positive words in the first common block. Since every width is at least one, N_+≤K . The event N_+≥i is determined by the first i-1 positive widths and the entire negative tape: none of those positive width sums may be a negative width sum. It is therefore independent of the i th positive word and its radius R_i^+ . Tonelli's theorem gives E∑_ i=1 ^ N_+ R_i^+ =∑_ i≥1 P(N_+≥i)E_ ν_+ R = E N_+\,E_ ν_+ R<∞. The identical argument for the negative tape proves the claimed integrability, using Proposition 6 ([Radius moment under coexistence]).