Math 489/889

Stochastic Processes and

Homework 6

Steve Dunbar

Due Monday, October 13, 2010

Problem 1

 (1.1)

Part a

For \$ T_0 = 10 \$  and \$ a = 20 \$ , draw a graph of theduration as a function of the probability \$ q \$ .

Part b

For \$ a = 20 \$  and \$ q = 0.55 \$  draw a graph of the expected duration as a function of \$ T_0 \$ .

Part c

For \$ a = 20 \$  and \$ q = 0.45 \$  draw a graph of the expected duration as a function of \$ T_0 \$ .

Problem 2

The boundary condition at state 26 says that the duration of the "game" from 26 down to 18 is the sum of hte duration of the subsequent games from 26 down to 25, and then from 25 to 18.  We can compute the duration of the first sub-game using Corollary 1.  Then the set of 9 first-step equations in 9 unknowns is:

 (2.1)

 (2.2)

 (2.3)

Problem 3

Insert the trial solution

into the difference equation.   For  s < kthe substitution yields so the equation is satisfied by the trial particular solution.

For the difference equation becomes and again the equation is satisfied identically.

For , the difference equation becomes    Therefore, the particular solution is

= .

Problem 4

 (4.1)

 (4.2)

 (4.3)

 (4.4)

 (4.5)

 (4.6)

 (4.7)

 (4.8)

 (4.9)

 (4.10)

Problem 5

 (5.1)

Part a

 (5.1.1)

Part b

 (5.2.1)

Part c

 (5.3.1)

 (5.3.2)

Inspect carefully, one solution set with is real, the other terms have complex factors and .

We can also solve the partial derivative eqations "by hand".

Set the derivative equal to 0, then clear denominators.

 (5.3.3)

 (5.3.4)

Substitute this into the derivative with respect to .

 (5.3.5)

 (5.3.6)

Now this derivative is when .  Note that and are not realistic values of , that is, the values and are not in the domain so the denominator is never in the domain.  Likewise, and are not .  Then substitute back into the expression for

 (5.3.7)

 (5.3.8)

 (5.3.9)