What are the Long-Term Trends of a Matrix in Algebra?

pocketlint59
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hi, first-time user here

although this question deals with algebra, I feel that it pertains to matricies and probability more

basically, how do i find long term trends of a matrix with algebra?

ex. 2 competing soft drink companines cola A, and cola B presently have 1/3 and 2/3 shares of the market, respectively.

92% of cola A drinmiers will buy cola A again , the remainder switch to cola B
85% of Cola B drinkers will buy cola B again, and the remainder will switch brands

What is the market share after 2 weeks?
please help me and thanks in advance
 
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What is the market share after 2 weeks?
This question is impossible to answer. You haven't told us whether "92% of cola A drinmiers will buy cola A again , the remainder switch to cola B" applies every day, week, 2 seconds, 5 years, etc. If this information is for a 2 week period, then you just need to apply the matrix once. If it's for a 1 week period, you have to apply it twice, etc. so as you can tell, we can't answer the question unless you tell us how many times the matrix would apply, which depends on what length of time the given information pertains to. Anyways, the initial market share is a 1x2 vector (or 2x1, it doesn't matter too much), v = (1/3 2/3). Your matrix A will be:

0.92 .0.08
0.15 .0.85

After one time period, the market share will be represented by vA, after n time periods the market share will be represented by vAn. If your time period is x weeks (if your time period is 1 day, then x = 1/7), then n = 2/x.

To find the long-term trends, you need to compute the limit as n approaches infinity of vAn. You will need to diagonalize A to make this simple. Do you know where to go from here?
 
time = weekly basis

initial matrix
A B
[0.33 0.67]​
transition matrix
A B
A [0.92 0.08]
B [0.15 0.85]​
it's finding the steady-state vector using algebra

i hope this helps
 
Last edited:
The steady state vector is that vector v that satisfies Av = v. What I told you about finding the limit may have been wrong. It will be much easier to find the v if you look at it like this:

Av = v
Av = Iv
Av - Iv = 0
(A - I)v = 0

You'll see that A-I is a rather nice matrix, and it will be easy to see what v should be from there.
 
is V in place of b because of a typo?
 
No, v and A are the matrices I said they were in post 2.
 
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