Below I propose a solution to the difference equation
ft+1(z)=p12ft(z/A)+p21ft(z/B)+p22ft(z/(A+B)),
where the pij's are positive, p12+p21+p22le1
and ft is a pdf.
By integrating both sides of the given equation from −infty to infty we obtain after appropriate change of variables in the right hand side integrals
1=Ap12+Bp21+(A+B)p22.
Now we look for a solution of our initial problem in the form
ft(z)=suminftyn=0qn(t)zn.
Substituting the above ansatz into our equation yields after elementary manipulation
qn(t+1)=left(p12A−n+p21B−n+(A+B)−nright)qn(t).
For fix n, the last equation is a linear difference equation that can be easily solved to produce
an(t)=bntp12A−n+p21B−n+(A+B)−n
,
where an is independent of t i.e. it's a pure constant.
Finally we obtain the closed-form solution
ft(z)=suminftyn=0bntp12A−n+p21B−n+(A+B)−nzn.
Note that f1(z)=suminftyn=0bnzn.
Thus our solution is completely specified given the initial pdf f1(z).
What is left is to tackle the issue of convergence and possible look for alternative representation of the solution.
No comments:
Post a Comment