16 1. HEURISTICS

The matrix AT of one-step transition probabilities of

Zε

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

.