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Stochastic Processes and

Advanced Mathematical Finance

Dunbar, Fall 2010

This problem is adapted from Stochastic Calculus and Financial Applications by J. Michael Steele, Springer, New York, 2001, Chapter 1, Section 1.6, page 9. This problem suggests how results on biased random walks can be worked into more realistic models.

Consider a naive model for a stock that has a support level
of $20/share because of a corporate buy-back program. Suppose
also that the stock price moves randomly with a downward bias
when the price is above 20,andrandomlywithanupwardbiaswhenthepriceisbelow20.
To make the problem concrete, we let Y _{n}
denote the stock price at time n, and we express our stock support
hypothesis by the assumptions that

We then reflect the downward bias at price levels above $20 by requiring that for k > 20:

We then reflect the upward bias at price levels below $20 by requiring that for k < 20:

Calculate the expected time for the stock to fall from $25 through the support level all the way down to $18.

I don’t believe that there is any way to solve this problem using formulas. Instead you will have to go back to basic principles to solve the problem.

The first thing to notice is that there is no natural upper
boundary for this problem. In effect, the stock situation is
like playing against an infinitely rich adversary. Therefore,
we will set an artificial boundary at 25 + M, where M is
some positive integer, and then at the end of the problem, we
will let M go to infinity. Let
D_{z} be the duration of the game, that is,
the expected number of moves until the stock goes from price
25 to price 18. Second, we note that the conditioning by
expectation, or one-step analysis, give the following set of
equations:

The lower boundary condition says that the walk will have no
duration if the walk starts at the absorbing barrier D_{18}. Likewise, there is another boundary
condition that says that the walk will have no duration if it
starts at the other ”artificial” absorbing
barrier, D_{25+M} = 0.
Third, note that the equations have a nice sameness about
them for all n > 20. This
would make them easy to solve with well-understood
techniques. Unfortunately the pattern does not hold for the
equations for D_{20}, D_{19}
and boundary condition D_{18}.
So what we must do is back-solve, starting at D_{1}8,
and successively eliminate D_{18},
and then D_{19}, leaving a relation between D_{20} and D_{21}.
That relation will become the new boundary condition for the
regular set of equations for n
> 20. Actually doing the back-solving, the new
boundary condition is easily seen to be:

Fourth, the general solution to the difference equation

is D_{n} = K_{1} +
K_{2}2^{n} + 3n.
Now substitute in the boundary conditions D_{25+M} = 0, and 27D_{21} +
33 = 28D_{20} to get two equations in the two
constants K_{1} and K_{2}.
The solutions are:

(This solution was done with Maple which accounts for the large integers in the exact solution, instead of floating point approximations.)

Therefore the general solution at n = 25 is:

Finally let M go infinity, and the limiting value is: 129. So, the expected time for the stock to go from 25 through the support level and down to 18 is 129 steps.

Back to main section with problem statement.

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