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
AT =









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 · · · P2 −E 0
0 0 0 0 · · · 0 Q2 −E









and Q = Q2Q2 · · · Q1Q1 = (Q2)
T
2
(Q1)
T
2
. But AT is a block-wise lower diagonal
matrix, and so AT 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









⎪πT



−(l)
=
ψ
ϕ + ψ
+
ϕ ψ
ϕ + ψ
·
(1 ϕ
ψ)l
1 + (1 ϕ ψ)
T
2
,
πT
+(l)
=
ϕ
ϕ + ψ

ϕ ψ
ϕ + ψ
·
(1 ϕ ψ)l
1 + (1 ϕ ψ)
T
2
,
−(l
+
T
2
) = πT
+(l),
πT
+(l
+
T
2
) = πT
−(l),
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 πT

(0) + πT
+
(0) = 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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