Day 36 in math modeling

• There will be a talk this Friday, and the next Friday, at 3:05 in MEP461:

  • This Friday: Mathematical Modeling Contest Solutions

  • Next: Greatest Hits of the KYMAA

• Today's climate news:

  The crazy:

  South Florida Nuclear Site to Expand Despite Sea Level Rise Projections

  In 2010, when the plant first applied for the license to build these two new reactors, the NRC wanted to know how the plant would stave off sea level rise in future decades. FPL did not mention climate change and used a 1-foot-per-century sea level rise projection in its calculation -- far less than NOAA's 5.6-foot worst-case scenario for 2100.

  Here's the raw satellite data:

  Here's the data corrected and modeled, from Climate-change-driven accelerated sea-level rise detected in the altimeter era, Nerem, et al. (2018):

  

  "If sea level continues to change at this rate and acceleration, sea-level rise by 2100 ($\approx 65$ cm) will be more than double the amount if the rate was constant at 3 mm/y."

• Today's stereopticon of the day: same as it ever was....

• The final project, updates:
  • Great work, data quality team! The data is back in Togo, where it will be examined by the Togolese Meteorologists. They have suggested that they may have additional data that they can use to fix up the problems you identified.

    For the moment, however, we push on, using only the data that you have confidence in.

  • You have a homework due Friday (evaluating our current model for your fair city).

  • You have a homework due today: respond to answers to our questions, with comments and new questions (one page, submitted electronically to both me and to Maria McMahon).

Let me start with a "where we are" with respect to these chains.

• Markov chains are models of Markov ("memoryless") processes -- the next state is only dependent on the state in which you find yourself now -- not how you got there.

• The matrices that arise from the Markov chain are stochastic: their rows (made up of only non-negative values) add to 1. So we think of the entries as probabilities of transitioning from one state to another.

• We can think of the Markov process like our original SIR models, where the total population was fixed. Everyone is somewhere, but they're always moving around, with those probabilities.

• One thing that's different, however, is that an infection can take off even if the initial infectious population is 0! That's because the rate of infectivity is constant, and not proportional to the size of the infected population.

• We found that, for regular Markov chains (that is, those whose matrices $M$ have all non-zero values for some power of $M$) there is a single fixed "equilibrium" distribution of states, given by the row of what we might call "$M^\infty$". If we iterate forever, the population of states will settle into this distribution. We should realize, however, that, just like the SIR model which reaches an equilibrium, the people in the system are always moving around! They're getting infected, recovering, losing immunity, and doing it all again (depending on the disease, of course).

• While a regular Markov process will equilibrate, we can't tell by the equilibrium vector what the rates are -- I demonstrated this last time.

  I talked about this in relation to the Galla people. Suppose we study a society whose class structure is Markovian. If we look at the census over time, we should see that the class distribution is stable -- with minor perturbation, due to small numbers, wars, etc. We can't tell exactly what its transition probabilities are from that, because different Markov matrices will have the same equilibrium distribution.

  Last time we looked at a Mathematica demonstration of that; I've also made an InsightMaker version -- let's take a quick look at that.

  ---

  Now I'd like you to demonstrate that it's true, in the two-by-two case. Assume Markov transition matrix \[ M=\left [ \begin{array}{cc} {a}&{1-a}\cr {1-b}&{b} \end{array} \right ] \] and we want the Markov process to equilibrate to \[ \underline{v}=\left [ \begin{array}{c} {\frac{1}{3}}\cr {\frac{2}{3}} \end{array} \right ] \] Find conditions that $a$ and $b$ must satisfy so that $\underline{v}$ is the equilibrium distribution of $M$.

  ---

  However, if we can determine the rates, as Hoffmann indicated we might, by census, then we should see that, over time, the system oscillates (due to these stochastic effects) about that equilibrium distribution of states.

• An interesting example application of Markov chains is to physical growth: this paper describes how the "architecture" of an apple tree can be modeled with fractals (e.g. the L-System) and a "hidden" Markov chain.

Now we're going to shift gears a little, and allow for those "wastebasket" states that we talked about last time: states from which, once the system enters them, there is no return.... Sounds pretty ominous! Let's call them absorbing states.

• Actually our first chain was an absorbing one: an SIR, where the recovered don't become susceptible again!

  Eventually everyone ended up in the recovered class. This would therefore be the absorbing state.

  The Markov music maker was also absorbing: there were certain notes, or chords, that led to "fin" -- the end. So if you hit those states, and roll the die just right, you're done composing! Your composition has been absorbed....

• So now we want to look at examples of absorbing Markov chains: those which contain these absorbing states. We distinguish

  • Absorbing states ($m$ of these), and

  • Transient states ($n$ of these).

• Last time we began with an extremely simple example: the case of one absorbing state, and one transient state.

  • The state graph

  • The transition matrix, and

  • The steady state distribution of states.
  Let's quickly review that. As I told you last time, this is good preparation for the matrix case: we're going to generalize it.

• As I told you last time, we want to turn the $r$ and $q$ (and even the $1$!) into matrices, and we'll extend the simple case to $m$ absorbing states and $n$ transient states. So what's that going to look like?

  \[ \left [ \begin{array}{cc} {I_{m \times m}}&{0_{m \times n}}\cr {R_{n \times m}}&{Q_{n \times n}} \end{array} \right ] \]

• Once we have written out those matrices, we can deduce interesting properties from them:

  1. Every absorbing Markov process eventually reaches some absorbing state (says that $M^\infty$ will eventually look like \[ \left [ \begin{array}{cc} {I_{m \times m}}&{0_{m \times n}}\cr {(I-Q)^{-1}R}&{0_{n \times n}} \end{array} \right ] \equiv \left [ \begin{array}{cc} {I_{m \times m}}&{0_{m \times n}}\cr {NR}&{0_{n \times n}} \end{array} \right ] \equiv \left [ \begin{array}{cc} {I_{m \times m}}&{0_{m \times n}}\cr {B}&{0_{n \times n}} \end{array} \right ] \]

  2. The matrix $N=(I-Q)^{-1}$ is called the fundamental matrix of the Markov chain. Its entries $N_{ij}$ are the average number of times the process is in the $j^{th}$ state if it starts in the $i^{th}$ transient state.

  3. The sum of the $i^{th}$ row of $N$ is the average number of steps before the process enters an absorbing state if it starts in the $i^{th}$ transient state.

  4. The matrix $B=NR$ has entries $B_{ij}$ which represent the probability of eventually entering the $j^{th}$ absorbing state if the process starts in the $i^{th}$ transient state.

• Now let's take a look at an interesting application where Markov chains prove to be pretty successful for Olinick: tennis matches.

  We'll apply these ideas, and actually test the Markovian nature of tennis tournament finals!