Day 37 in math modeling

Next Friday: more Greatest Hits of the KYMAA

The sad:

Ocean-Based Food Security Threatened in a High CO2 World: Togo ranks second in food insecurity in the world, due to climate change.

The bad:

The oceans' circulation hasn't been this sluggish in 1,000 years. That's bad news.

The Atlantic Ocean circulation that carries warmth into the Northern Hemisphere's high latitudes is slowing down because of climate change, a team of scientists asserted Wednesday, suggesting one of the most feared consequences is already coming to pass.

The Atlantic meridional overturning circulation has declined in strength by 15 percent since the mid-20th century to "a new record low," the scientists conclude in a peer-reviewed study published in the journal Nature. That's a decrease of 3 million cubic meters of water per second, the equivalent of nearly 15 Amazon rivers.

A North American Climate Boundary Has Shifted 140 Miles East Due to Global Warming

• Your homework (evaluating our current model for your fair city) is due today.

• Make sure that you're using the "correct version" of "$R^2$": recall that on day 11 I switched the formula for the value of $R^2$ in the non-linear (actually general) case, to one that reduced to the correct $R^2$ for the linear regression case, but still allowed us to get a reasonable comparative statistic in the non-linear case.

  \[ R^2=1-\frac{SSE}{SS_{Tot}}=1-\frac{\sum_{i=1}^n(y_i-\hat{y}_i)^2}{\sum_{i=1}^n(y_i-\overline{y})^2} \] From day 11: "I was abusing a property of the linear model to attempt to compute $R^2$. This version is essentially comparing SSE across models, only, and reduces to the $R^2$ for the linear model."

  The issue that gave rise to reconsidering this was that at least one group is finding negative values of $R^2$ in one town, when evaluating the global model. So "$R^2$" is a bit of a misnomer.... The maximum value of the $R^2$ above can never exceed 1 (and will be 1 when the model fits perfectly). It can be negative, however -- whenever the mean model $\overline{y}$ beats the alternative model, represented by $\hat{y}$.

• Maria R. and Terra have identified some periods in the data which they think are worth exploring.

  • Climate oscillations

  • Sunspot activity, with a period of around 12 years.

  • El Nino/Southern Oscillation: Southern Oscillation Index (SOI) since 1876

  • Shorter cycles: Madden-Julian oscillation (1-3 months)

• Questions for the Togolese

  Thank you, Maria M., for compiling those. I added a few of my own. Take a glance, and see of there are any that you feel have been missed. Then I'll get these on their way this afternoon.

• 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!

In place of that, however, I'd like to do an example of a classic Markov chain: the random walk. Random walks are extraordinarily important -- not because we randomly walk, but because molecules do, say, and pollen grains on the surface of a glass of water....

So I'll have you all take a crack at the following problem (you will probably want to reuse the Mathematica code for the Tennis example):

This is to be handed in next Friday, 4/20. You may work on it together.

1. Draw the State Graph.

2. What is the transition matrix? (Arrange it with absorbing states first, and then transients, in order from lowest to highest).

3. Starting from position 1, what is the probability that it will have been absorbed in three steps?

   1. Draw a tree diagram

   2. Compute the third power of the matrix, and check that you get the same answer!

4. If you start from 2,

   1. how long on average will the particle spend in transient states?

   2. what is the distribution of final states?

   3. Do these answers change if the probability of going left becomes 1/3, and right 2/3?