Appendix D
Total Escape Probability for Chain Graphs, ΣβCN

So long as there is at least one escape route from CN the total escape probability must be unity:

ΣβCN = j=1N jCN 𝒮j,βCN = 1, (D.1)

otherwise, if Adj[CN] is the empty set, we have

ΣβCN = j=1N𝒮 j,βCN = 1. (D.2)

For example, to show that the formulae in Appendix C are consistent with the first result we expand

jCN = 1 Pj1,j Pj+1,j (D.3)

and define

P0,1 = PN+1,N = 0 (D.4)

for convenience, so that

ΣβCN =𝒮1,βCN P0,1𝒮1,βCN P2,1𝒮1,βCN + 𝒮2,βCN P1,2𝒮2,βCN P3,2𝒮2,βCN + 𝒮N1,βCN PN2,N1𝒮N1,βCN PN,N1𝒮N1,βCN + 𝒮N,βCN PN1,N𝒮N,βCN PN+1,N𝒮N,βCN. (D.5)

Using the recursion relations in Equation C.2 (which assume that there is an escape route from CN) we can show that

𝒮j,βCN 𝒮j1,βCN Pj,j1 𝒮j+1,βCN Pj,j+1 = 0, (D.6)

for jβ. We can now group together terms in Equation D.5 into sets of three that sum to zero. The terms that do not immediately cancel are as follows. From the first and second lines of Equation D.5 we have

𝒮1,βCN P2,1𝒮2,βCN = 0 (D.7)


𝒮1,βCN = 𝒮2,βCN P2,1L1 = 𝒮2,βCN P2,1. (D.8)

Similarly, on the last two lines we find

𝒮N,βCN PN,N1𝒮N1,βCN = 0 (D.9)


𝒮N,βCN = PN,N1𝒮N1,βCN RN = PN,N1𝒮N1,βCN . (D.10)

The final remaining terms are:

𝒮β,βCN 𝒮β1,βCNPβ,β1 𝒮β+1,βCNPβ,β+1 =𝒮β,βCN 𝒮β,βCN 1 1 Lβ 𝒮β,βCN 1 1 Rβ =𝒮β,βCN 1 Lβ + 1 Rβ 1 =𝒮β,βCN Lβ + Rβ LβRβ LβRβ =1, (D.11)

which proves Equation D.1.