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Math 489/Math 889
Stochastic Processes and
Advanced Mathematical Finance
Dunbar, Fall 2010


Problem Statement

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

Pr[Y = 21|Y = 20] = 9/10 n+1 n Pr[Yn+1 = 19|Yn = 20] = 1/10

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

 Pr[Yn+1 = k + 1|Yn = k] = 1/3 Pr[Y = k - 1|Y = k] = 2/3. n+1 n

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

Pr[Y = k + 1|Y = k] = 2/3 n+1 n Pr[Yn+1 = k - 1|Yn = k] = 1/3

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.

Solution

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 Dz 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:

D25 = (1/3)D26 + (2/3)D24 + 1 D24 = (1/3)D25 + (2/3)D23 + 1 D23 = (1/3)D24 + (2/3)D22 + 1 D22 = (1/3)D23 + (2/3)D21 + 1 D21 = (1/3)D22 + (2/3)D20 + 1 D20 = (1/10)D19 + (9/10)D21 + 1 D19 = (2/3)D20 + (1/3)D18 + 1 D18 = 0

The lower boundary condition says that the walk will have no duration if the walk starts at the absorbing barrier D18. Likewise, there is another boundary condition that says that the walk will have no duration if it starts at the other ”artificial” absorbing barrier, D25+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 D20, D19 and boundary condition D18. So what we must do is back-solve, starting at D18, and successively eliminate D18, and then D19, leaving a relation between D20 and D21. 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:

27D21 + 33 = 28D20.

Fourth, the general solution to the difference equation

Dn = (1/3)Dn+1 + (2/3)Dn - 1 + 1

is Dn = K1 + K22n + 3n. Now substitute in the boundary conditions D25+M = 0, and 27D21 + 33 = 28D20 to get two equations in the two constants K1 and K2. The solutions are:

 25+M 25+M K1 = 6 * (-340787200 - 13631488 *M + 9 * 2 )/(27262976 + 2 ) K2 = - 3 * (43 + M )/(27262976 + 225+M ).

(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:

 25+M 6*--340787200----13631488-*-M--+-9-*-2------100663296* --------43-+-M---------- 27262976 + 225+M 27262976 + 225+M ) + 75

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.


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