118 CHAPTER 3. FIRST STEP ANALYSIS FOR STOCHASTIC PROCESSES
expected duration equation(which is +1) arises because the game will always
be at least 1 step longer after the first step. In the current equation, theδsk
non-homogeneous term arises because the number of visits to levelk after
the first step will be 1 more ifk=sbut the number of visits to levelkafter
the first step will be 0 more ifk 6 =s.
For the ruin probabilities, the difference equation was homogeneous, and
we only needed to find the general solution. For the expected duration,
the difference equation was non-homogeneous with a non-homogeneous term
which was the constant 1, making the particular solution reasonably easy to
find. Now the non-homogeneous term depends on the independent variable,
so solving for the particular solution will be more involved.
First we find the general solutionWskhto the homogeneous linear difference
equation
Wskh =1
2
Wsh− 1 ,k+1
2
Wsh+1,k.This is easy, we already know that it isWskh =A+Bs.
Then we must find a particular solutionWskp to the non-homogeneous
equation
Wskp =δsk+1
2
Wsp− 1 ,k+1
2
Wsp+1,k.For purposes of guessing a plausible particular solution, temporarily re-write
the equation as
− 2 δsk=Wsp− 1 ,k− 2 Wskp +Wsp+1,k.
The expression on the right is a centered second difference. For the prior
expected duration equation, we looked for a particular solution with a con-
stant centered second difference. Based on our experience with functions it
made sense to guess a particular solution of the formC+Ds+Es^2 and then
substitute to find the coefficients. Here we seek a function whose centered
second difference is 0 except atkwhere the second difference jumps to 1.
This suggests the particular solution is piecewise linear, say
Wskp ={
C+Ds ifs≤k
E+Fs ifs > k.In the exercises, we verify that the solution of this set of equations is
Wskp ={
0 ifs < k
2(k−s) ifs≥k.