# Trying to find long run behaviour of a dynamical system

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1. Nov 4, 2016

### Razberryz

1. The problem statement, all variables and given/known data
In any given year a person may or may not get the flu. Past records show that if a person has the flu one year then (due to a build up of antibodies) there is a 85% chance that they will not get the flu in the following year. If they don't have a flu in a given year then there is a 25% chance that they will get the flu the following year.

(a) If a person has the flu one year, what is the probability that they will also have the flu 2 years later?
(b) In the long run, what proportion of years does a person not have the flu?

2. Relevant equations
x(t) = Ptx(0)

3. The attempt at a solution
\begin{bmatrix}5/17 & -1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix}1^t & 0 \\ 0 & -0.1^t \end{bmatrix}\begin{bmatrix}5/17 & 0.1 \\ 1 & -0.1 \end{bmatrix}\begin{bmatrix}x_1(0) \\ x_2(0) \end{bmatrix}

first matrix is the eigenvectors

second in the diagonal

third is the multiplication of the first and second

4th is the original population I guess, which I don't know

According to an example from my notes, I think I'm supposed to multiply all these 4 matrices together which I did, and I got

(\begin{bmatrix}25/289 & 1/34 \\ 5/17 & 0.1 \end{bmatrix})*(\begin{bmatrix}x_1(0) \\ x_2(0) \end{bmatrix})

I don't know where to go from here. Any help appreciated.

2. Nov 4, 2016

### Razberryz

By the way, the transition matrix is:

\begin{bmatrix}0.15 & 0.25 \\ 0.85 & 0.75 \end{bmatrix}

3. Nov 4, 2016

### Ray Vickson

So, are you using the convention that the columns sum to 1? If so, you should state that, since it is not the most common convention. Most treatments I have seen--in every single book on the subject that I have ever seen, and in almost all of the papers I have ever read--use the opposite convention, that the rows sum to 1.

For (a), just use matrix multiplication. If $P$ is your 2x2 transition matrix (the last matrix you wrote above) then the 2-step transition matrix is $P^2 = P \cdot P$. For (b), just use the standard equilibrium (or limiting) equations. Using Eigenvalues and eigenvectors is a complete waste of time in this problem. They could be useful if you wanted explicit dynamic behavior $x(n) = P^n x(0)$ as a function of $n$, but for one or two values such as $n=2$ or $n=3$ the faster way is direct matrix multiplication.

4. Nov 4, 2016

### Razberryz

Thanks for the reply. What do you mean by "standard equilibrium (or limiting) equations"? I haven't come across these terms.

5. Nov 4, 2016

### Ray Vickson

Do you have a textbook or course notes? Surely they must be presented therein. If not, Google something like 'markov chain equilibrium distribution'.

6. Nov 4, 2016

### Razberryz

Searched on google, and we haven't learned it, so I'm guessing we can solve this without equilibrium distribution.

7. Nov 4, 2016

### Ray Vickson

Sure; you can just go ahead and evaluate $P^n x(0)$ for large $n$ using a matrix multiplication package, or using an eigenvalue expansion, then try to take the limit as $n \to \infty$. Or, you could do it the easy way and just solve a simple 2x2 linear system of equations.

Surely your Google search must have turned up something; I saw dozens of related articles, for example,
https://www.stat.auckland.ac.nz/~fewster/325/notes/ch9.pdf
The first 2 or 3 pages have all you need.

8. Nov 4, 2016

### Razberryz

Yes, I looked at that exact article, but it seems way beyond what we've learned. I'm not even able to decipher it unfortunately.

As for your suggestion, well, the entire problem is that I do not know x(0). Will a 2x2 (P*P) give me the same answer as P2?

9. Nov 4, 2016

### Ray Vickson

To do part (a) you DO know x(0). You have enough information to do part (b) without knowing the exact x(0).

BTW: I told you about the usual convention for Markov chains because maybe you did not realize there is even an issue. It is OK to use your convention within the confines of your course, but when going outside (such as to a Forum to post questions) you really do need to state your convention explicitly, because it it the opposite of what almost everybody else will be using, and they will often mis-read your posts.

10. Nov 4, 2016

### Razberryz

Sorry, I didn't know what the normal convention was. I simply posted what I was taught. Thanks, will do so in the future.

11. Nov 4, 2016

### Ray Vickson

I guessed as much; that is why I told you about it, so you would know.

As I said, it is perfectly OK to post what you have been taught if you are more comfortable doing that: just as long as you inform your audience of your convention.