Question 3.8

[jensen]

I can understand why:

P[X_n+t = 0 | X_n = 0] = 1/4
P[X_n+t = 2 | X_n = 2] = 3/4

are boundary conditions of Model 5.1, but why

P[X_n+t = 1 | X_n = 0]
P[X_n+t = 1 | X_n = 2]

are boundaries, too? Is it because you cannot stay in State 1, therefore it's a reflecting state with two possible route?

Thanks.

 

[Robert Chadburn]

Hi
States 0 and 2 are both examples of a "mixed boundary". They are not fully reflecting (which would mean you would automatically bounce back out of the state the you way you had just come in, with a probabiity of 1), nor are they fully absorbing (in which case you would stay in the state once you entered it, for ever). It's a half-way house, with non-zero probabities of staying put and of bouncing back out again. So all mixed boundaries will have two probabilities associated with them (as defined on page 21, section 5.4, and you will see that the solution 3.8 takes exactly this form). You have to write both probabilities down because the state you can go back to is important in specifying how the boundary works.
Hope this helps.
Robert