16 1. HEURISTICS The matrix AT of one-step transition probabilities of is explicitly given by AT = 0 Q1 0 0 · · · 0 0 0 0 0 Q1 0 · · · 0 0 0 . . . . . . . . ... . . . . . . 0 0 0 0 · · · 0 Q2 0 0 0 0 0 · · · 0 0 Q2⎠ Q2 0 0 0 · · · 0 0 0 . The matrix AT has block structure. In this notation 0 means a 2×2-matrix with all entries equal to zero, Q1, and Q2 are the 2-dimensional matrices defined in (1.12). Applying some algebra we see that the equation (AT E)R = 0 is equivalent to AT R = 0, where A T = Q E 0 0 0 · · · 0 0 0 Q1 −E 0 0 · · · 0 0 0 . . . . . . . . ... . . . . . . 0 0 0 0 · · · −E 0 0 0 0 0 0 · · · P 2 −E 0 0 0 0 0 · · · 0 Q∗ 2 −E and Q = Q∗Q∗ 2 2 · · · Q∗Q∗ 1 1 = (Q∗) 2 T 2 (Q∗) 1 T 2 . But A T is a block-wise lower diagonal matrix, and so A T R = 0 can be solved in the usual way resulting in the following formulas. For every T 2N, the stationary distribution πT of Y ε with matrices of one- step probabilities defined in (1.12) is given by (1.13) ⎪πT ⎪π−(l (l) = ψ ϕ + ψ + ϕ ψ ϕ + ψ · (1 ϕ ψ)l 1 + (1 ϕ ψ) T 2 , π+(l) T = ϕ ϕ + ψ ϕ ψ ϕ + ψ · (1 ϕ ψ)l 1 + (1 ϕ ψ) T 2 , T + T 2 ) = π+(l), T π+(l T + T 2 ) = π−(l), T 0 l T 2 1. The proof of (1.13) is easy and instructive, and will be contained in the following arguments. Note that πT (0) satisfies the matrix equation (Q2) T 2 (Q1) T 2 E πT (0) = 0 with additional condition π−(0) T + π+(0) T = 1. To calculate (Q2) T 2 (Q1) T 2 , we use a formula for the k-th power of 2 × 2-matrices Q = 1 a a b 1 b , a, b R, proved in a straightforward way by induction on k which reads 1 a a b 1 b k = 1 a + b b a b a + (1 a b)k a + b a −a −b b .
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