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I have a final exam tomorrow and I am stuck on one question on my review sheet. Can someone help explain how to do this problem? Any help would be appreciated.

Here is the problem:

(a) There are two coastal cities that are both beautiful places to live and work. Despite being lovely places to live there is some movement in population between the two cities. Mysteriously every year exactly 4 percent of the inhabitants of cit A decide to move to city B, and 8 percent of city B dwellers decide to move to city A. If people don't move between city A and city B then they just stay in the city they began in. What is the transition matrix that describes this annual migration?

(b) What happens if in addition to people moving around like in part (a), you know that 2 percent of city A dwellers leave the coast for good, and 4 percent of city B dwellers leave the coast for good. What does the transition matrix look like?

I believe the answer to part (a) is:

[0.04, 0.92; 0.96, 0.08]

But I am having trouble with part (b).

Thank you,

Brian

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2 Answers 2

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Your answer to a) is wrong. Here's how a transition matrix works: for every $i$ and $j$, $P_{ij}$ (the entry in the $i$th row and $j$th column) is the probability of moving from state $i$ to state $j$.

For the first problem, there are two states: a person will be either in city A (state 1) or in city B (state 2). If you're in state 1, the probability of switching to state 2 is $.04$, and the probability of staying to state $1$ ("moving from state 1 to state 1") is $0.96$. So, $P_{11} = 0.96$ and $P_{12} = 0.04$.

Similarly, $P_{21} = 0.08$ and $P_{22} = 0.92$. All together, we have $$ P = \pmatrix{0.96 & 0.04 \\0.08 & 0.92} $$ For the second problem, we have three states: state 1 is "in city A". State 2 is "in city B". State 3 is "away from the coast for good". See if you can figure out what the transition matrix should look like. Hint: $P_{33} = 1$.

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  • $\begingroup$ Thanks so much! I think I understand now. Would the answer be [0.92, 0.04, 0.02; 0.08, 0.96, 0.04; 0, 0, 1] ? I assumed that someone in State 3 would not go back to city A or city B is zero. So P31 and P32 are zero. $\endgroup$
    – Brian
    Commented Dec 15, 2016 at 3:33
  • $\begingroup$ @Brian each row should add up to $1$; you've made some mistakes there. The third row is right, though $\endgroup$
    – Ben Grossmann
    Commented Dec 15, 2016 at 3:36
  • $\begingroup$ I need to change 0.92 to 0.94 and 0.96 to 0.88. These are the probabilities that someone stays in city A and city B correct? $\endgroup$
    – Brian
    Commented Dec 15, 2016 at 3:39
  • $\begingroup$ Yes, now everything is right. That is, you should have $$ P = \pmatrix{ 0.94 & 0.04 & 0.02\\ 0.08 & 0.88 & 0.04\\ 0&0&1 } $$ does that all make sense, now? $\endgroup$
    – Ben Grossmann
    Commented Dec 15, 2016 at 3:43
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    $\begingroup$ Yes it does! Thank you so much for explaining. $\endgroup$
    – Brian
    Commented Dec 15, 2016 at 3:44
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You're pretty much right for (a). For (b), consider having a three-state system that includes "Lives in A", "Lives in B", "Lives in neither". Then there's a transition from "Lives in A" to "Lives in neither" of 0.02, and one from "Lives in B" to "Lives in neither" of 0.04, while the transitions for "Lives in neither" to the three states is 0, 0, 1 (since it's implied that no-one new ever moves back to the coast).

Then, once you've got that transition matrix, just remove the "Lives in neither" parts, because the loss of people from the coast is actually already handled by the fact that the transitions between A and B don't sum to 1 any more.

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