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					Department of Applied Mathematics and Statistics,
University of Copenhagen, Universitetsparken 5, 2100 Copenhagen Ø, Denmark, www.math.ku.dk/ams




                                                                                    Niels Richard Hansen
                                                                                    February 23, 2006




1     Exercises

This exercise is about the random walk ascending and descending ladder height distribution
when the increment distribution has finite support concentrated on a lattice. In this case
the two distributions can be computed by finding roots in a polynomial equation. The
solution is based on the Wiener-Hopf factorisation identity.
We need to introduce some notation. Let n, m ≥ 1 be given and assume that

                      P(X1 = k) = pk ,       k ∈ {−m, −m + 1, . . . , n − 1, n}
          n
where     k=−m pk   = 1 and p−m , pn > 0. Let

                          p+,k = P(Sτ+ = k, τ+ < ∞),               k = 1, 2, . . . , n

and
                       p−,k = P(Sτ− = k, τ− < ∞),                k = 0, −1, . . . , −m.
and define for z ∈ C
                                                       n
                                         p+ (z) =          p+,k z k
                                                     k=1

and
                                             m
                                  p− (z) =         p−,−k z −k ,       z = 0,
                                             k=0

together with
                                                 n
                                    p(z) =            pk z k ,    z = 0.
                                             k=−m

Question 1.1. Show that the Wiener-Hopf identity (Theorem 8.41) implies that

                                (1 − p+ (z))(1 − p− (z)) = 1 − p(z)                                    (1)

for z ∈ C\{0}.
Hint: First argue for |z| = 1 then use an extension argument.

Question 1.2. Argue that if the random walk has negative drift then |p− (z)| < 1 for
|z| > 1. Then argue that q(z) = z m (1 − p(z)) is a polynomial, and from the identity above
that q has precisely n roots with absolute value > 1, and that these roots are also the
roots of 1 − p+ (z).
2                                                                                   1. Exercises


The roots of q can be computed from q numerically, and we can single out those with
absolute value > 1. Let α1 , . . . , αn with |αi | > 1 denote these roots. The ascending ladder
height distribution given in terms of the point probabilities p+,1 , . . . , p+,n is on the other
hand not known.

Question 1.3. Show that p+,k , k = 1, . . . , n is a solution to the following system of linear
equations:
                                   n
                                        αk p+,k = 1,
                                         i              i = 1, . . . , n.
                                k=1

Under what conditions does there exist a unique solution to these equations?

Question 1.4. Argue that q has a total of n+m roots = 0 with m of them having absolute
value ≤ 1, that q− (z) = z m (1 − p− (z)) is a polynomial, and that the m roots of q with
absolute value ≤ 1 are precisely the roots of q− .

Let β1 , . . . , βm with |βi | ≤ 1 denote the roots of q with absolute value ≤ 1.

Question 1.5. Show that p−,k , k = 0, −1, . . . , −m is a solution to the following system
of linear equations:
                               m
                                       α−k p−,−k = 1,
                                        i                i = 1, . . . , n.
                              k=0

Under what conditions does there exist a unique solution to these equations?
Hint: By the negative drift you know the extra equation m p−,−k = 1.
                                                          k=0

				
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