Appendix C
Pathway Sums for Chain Graphs, 𝒮α,βCN

To obtain the total probability of leaving the chain CN via node α if started from node β, i.e. ℰαCN𝒮α,βCN, we must calculate the pathway sum 𝒮α,βCN. We start with the case α = β and obtain 𝒮β,βCN. Consider any path that has reached node N − 1. The probability factor due to all possible N − 1 to N recrossings is simply RN−1 = 1∕(1 − PN−1,NPN,N−1). We need to include this factor every time we reach node N − 1 during recrossings of N − 2 to N − 1. The corresponding sum becomes

RN−2 = ∑ m=0∞(P N−2,N−1PN−1,N−2RN−1)m = 1 1 − PN−2,N−1PN−1,N−2RN−1. (C.1)

Similarly, we can continue summing contributions in this way until we have recrossings of β to β + 1, for which the result of the nested summations is Rβ = 1∕(1 − Pβ,β+1Pβ+1,βRβ+1). Hence, Rβ is the total transition probability for pathways that return to node β and are confined to nodes with index greater than β without escape from CN.

We can similarly calculate the total probability for pathways returning to β and confined to nodes with indices smaller than β. The total probability factor for recrossings between nodes 1 and 2 is L2 = 1∕(1 − P1,2P2,1). Hence, the required probability for recrossings between nodes 2 and 3 including arbitrary recrossings between 1 and 2 is L3 = 1∕(1 − P2,3P3,2L2). Continuing up to recrossings between nodes β − 1 and β we obtain the total return probability for pathways restricted to this side of β as Lβ = 1∕(1 − Pβ−1,βPβ,β−1Lβ−1). The general recursive definitions of Lj and Rj are:

Lj = 1, j = 1, 1∕(1 − Pj−1,jPj,j−1Lj−1),j > 1, andRj = 1, j = N, 1∕(1 − Pj,j+1Pj+1,jRj+1),j < N. (C.2)

We can now calculate 𝒮β,βCN as

𝒮β,βCN =∑ m=0∞∑ n=0m n! m!(n − m)! Pβ−1,βPβ,β−1Lβ−1 n P β,β+1Pβ+1,βRβ+1 m−n =∑ m=0∞(P β−1,βPβ,β−1Lβ−1 + Pβ,β+1Pβ+1,βRβ+1)m =(1 − Pβ−1,βPβ,β−1Lβ−1 − Pβ,β+1Pβ+1,βRβ+1)−1 = 1 − Lβ − 1 Lβ − Rβ − 1 Rβ −1 = LβRβ Lβ − LβRβ + Rβ, (C.3)

where we have used Equation C.2 and the multinomial theorem [242].

We can now derive 𝒮α,βCN as follows. If α > β we can write

𝒮α,βCN = 𝒮α−1,βCN Pα,α−1Rα. (C.4)

𝒮α−1,βCN gives the total transition probability from β to α − 1, so the corresponding probability for node α is 𝒮α−1,βCN times the branching probability from α − 1 to α, i.e. Pα,α−1, times Rα, which accounts for the weight accumulated from all possible paths that leave and return to node α and are restricted to nodes with indexes greater than α. We can now replace 𝒮α−1,βCN by 𝒮α−2,βCNPα−1,α−2Rα−1 and so on, until 𝒮α,βCN is expressed in terms of 𝒮β,βCN. Similarly, if α < β we have

𝒮α,βCN = 𝒮α+1,βCN Pα,α+1Lα, (C.5)

and hence

𝒮α,βCN = 𝒮β,βCN ∏ i=αβ−1P i,i+1Li,α < β, 𝒮β,βCN ∏ i=β+1αP i,i−1Ri,α > β. (C.6)