Week 11 Activity

Section 11.4 Week 11 Activity

Subsection 11.4.1 Leslie Matrices

Activity 11.4.1.

This is a Leslie matrix for a four-stage population model. Determine the long term behaviour by analyzing the eigenvalues and eigenvectors. Describe the eventual age ratios and determine whether or not the population is viable.

\begin{equation*} \left( \begin{matrix} 0 \amp 0 \amp 0.4 \amp 0.9 \amp \\ 0.8 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.4 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.9 \amp 0 \\ \end{matrix} \right) \end{equation*}

Solution.

Here is the dominant eigenvalue and the matching eigenvector (calculated by computer).

\begin{align*} \amp \lambda = 0.77 \amp \amp v = \left( \begin{matrix} 1.61 \\ 1.66 \\ 0.86 \\ 1 \end{matrix} \right) \end{align*}

The population is not viable, since the eigenvalue is less than one. The age ratios will approach \(1.61:1.66:0.86:1\) from youngest to oldest as the population decays.

Activity 11.4.2.

In the previous question, how much would you need to change the first fecundity coefficient to make the population viable.

Solution.

I mostly need to guess and test. Let me change the coefficient into \(0.7\)

\begin{equation*} \left( \begin{matrix} 0 \amp 0 \amp 0.7 \amp 0.9 \amp \\ 0.8 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.4 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.9 \amp 0 \\ \end{matrix} \right) \end{equation*}

The new dominant eigenvalue is \(0.82\text{.}\) This is not sufficient. Let me change the coefficient into \(1.0\text{.}\)

\begin{equation*} \left( \begin{matrix} 0 \amp 0 \amp 1.0 \amp 0.9 \amp \\ 0.8 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.4 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.9 \amp 0 \\ \end{matrix} \right) \end{equation*}

The new dominant eigenvalue is \(0.85\text{,}\) which is still not sufficient. Let me change the coefficient into \(2.0\text{.}\)

\begin{equation*} \left( \begin{matrix} 0 \amp 0 \amp 2.0 \amp 0.9 \amp \\ 0.8 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.4 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.9 \amp 0 \\ \end{matrix} \right) \end{equation*}

The new dominant eigenvalue is \(0.97\text{.}\) This is still not sufficient, but getting better. Let me change the coefficient into \(3.0\text{.}\)

\begin{equation*} \left( \begin{matrix} 0 \amp 0 \amp 3.0 \amp 0.9 \amp \\ 0.8 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.4 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.9 \amp 0 \\ \end{matrix} \right) \end{equation*}

The new dominant eigenvalue is \(1.06\text{,}\) so the population is now viable. I’ve overshot here by a bit, but this is good enough.

Activity 11.4.3.

\begin{equation*} \left( \begin{matrix} 0 \amp 15 \amp 200 \amp 100\\ 0.07 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.21 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.12 \amp 0 \\ \end{matrix} \right) \end{equation*}

Solution.

Here is the dominant eigenvalue and the matching eigenvector (calculated by computer).

\begin{align*} \amp \lambda = 1.69 \amp \amp v = \left( \begin{matrix} 2731.2 \\ 113.2 \\ 14.1 \\ 1 \end{matrix} \right) \end{align*}

The population is viable. The age ratios will approach \(2731.2:113.2:14.1:1\) from youngest to oldest.

Activity 11.4.4.

In the previous question, how much would you have to reduce the first survival coefficient to make the population no longer viable.

Solution.

I mostly need to guess and test. I’ll first set the survival to \(0.05\text{.}\)

\begin{equation*} \left( \begin{matrix} 0 \amp 15 \amp 200 \amp 100\\ 0.05 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.21 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.12 \amp 0 \\ \end{matrix} \right) \end{equation*}

The new dominant eigenvector is \(\lambda = 1.49\) This is still not enough. Now I’ll set the survival to \(0.01\text{.}\)

\begin{equation*} \left( \begin{matrix} 0 \amp 15 \amp 200 \amp 100\\ 0.01 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.21 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.12 \amp 0 \\ \end{matrix} \right) \end{equation*}

The new dominant eigenvector is \(\lambda = 0.83\text{.}\) The popuation is no longer viable. I may have overshot a bit, so I’ll also try \(0.02\text{.}\)

\begin{equation*} \left( \begin{matrix} 0 \amp 15 \amp 200 \amp 100\\ 0.02 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.21 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.12 \amp 0 \\ \end{matrix} \right) \end{equation*}

The new dominant eigenvector is \(\lambda = 1.06\text{,}\) so perhaps I didn’t overshoot that much. I could go to coefficients between \(0.01\) and \(0.02\text{,}\) but this is sufficient for now.

Activity 11.4.5.

\begin{equation*} \left( \begin{matrix} 0 \amp 0.32 \amp 0.36 \amp 0.29 \\ 0.95 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0.8 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.9 \amp 0 \\ \end{matrix} \right) \end{equation*}

Solution.

Here is the dominant eigenvalue and the matching eigenvector (calculated by computer).

\begin{align*} \amp \lambda = 0.92 \amp \amp v = \left( \begin{matrix} 1.12 \\ 1.16 \\ 1.02 \\ 1 \end{matrix} \right) \end{align*}

The population is not viable, since the dominant eigenvalue is less than 1. The age ratios will approach \(1.12:1.16:1.02:1\) from youngest to oldest.

Activity 11.4.6.

In the previous example, how can you change the survival to make the population viable. (You may change any of the three survival coefficients).

Solution.

The survival coefficients here are already quite high. The maximum they can be is \(1\text{,}\) since no more than \(100\%\) of an age category can survive. I’m first going to check what happens when I set all the survival coefficients equal to \(1\text{.}\)

\begin{equation*} \left( \begin{matrix} 0 \amp 0.32 \amp 0.36 \amp 0.29 \\ 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ \end{matrix} \right) \end{equation*}

The new dominant eigenvalue is \(0.989\text{.}\) Even with the maximum possible survival, the population isn’t viable. The fecundity rates are simply too low for any hope for this population.

Subsection 11.4.2 Markov Chains for Probability Graphs

Activity 11.4.7.

Figure 11.4.1. Probability Graph 1

Figure 11.4.1 shows a probability graph. Calculate the long term probability of ending up in each vertex.

Solution.

First, I need to write down the stochastic matrix.

\begin{equation*} \begin{pmatrix} 0.6 \amp 0.05 \amp 0 \\ 0.2 \amp 0.55 \amp 0.7 \\ 0.2 \amp 0.4 \amp 0.3 \end{pmatrix} \end{equation*}

Then I calculate (by computer) the dominant eigenvector.

\begin{equation*} \begin{pmatrix} 0.0722 \\ 0.5773 \\ 0.3505 \end{pmatrix} \end{equation*}

This gives me the long-term probability distribution. There is roughly a \(7.2\%\) chance of ending on the first vertex; a \(57.7\%\) change of ending on the second vertex; and a \(35.1\%\) chance of ending on the third vertex.

Activity 11.4.8.

Figure 11.4.2. Probability Graph 1

Figure 11.4.2 shows a probability graph. Calculate the long term probability of ending up in each vertex. (Assume any possibile path between vertices that doesn’t have an edge in the graph has a probability of zero).

Solution.

First, I need to write down the stochastic matrix.

\begin{equation*} \begin{pmatrix} 0.3 \amp 0.2 \amp 0 \amp 0 \amp 0.5 \\ 0.1 \amp 0.1 \amp 0.4 \amp 0 \amp 0.5 \\ 0.4 \amp 0 \amp 0.1 \amp 0.8 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0.2 \amp 0.7 \amp 0 \amp 0.2 \amp 0 \end{pmatrix} \end{equation*}

Then I calculate (by computer) the dominant eigenvector.

\begin{equation*} \begin{pmatrix} 0.2375 \\ 0.2418 \\ 0.1900 \\ 0.0950 \\ 0.2358 \end{pmatrix} \end{equation*}

This gives me the long-term probability distribution. There is roughly a \(23.7\%\) change of ending on vertex \(1\text{;}\) a \(24.2\%\) change of ending on vertex \(2\text{;}\) a \(19.0\%\) change of ending on vertex \(3\text{;}\) a \(9.5\%\) change of ending on vertex \(4\text{;}\) and a \(23.6\%\) change of ending on vertex \(5\text{.}\)

Activity 11.4.9.

Figure 11.4.3. Probability Graph 3

Figure 11.4.3 shows a probability graph. Calculate the long term probability of ending up in each vertex.

Solution.

First, I need to write down the stochastic matrix.

\begin{equation*} \begin{pmatrix} 0.7 \amp 0.4 \amp 0 \amp 0 \amp 0 \\ 0.3 \amp 0 \amp 0.9 \amp 0 \amp 0 \\ 0 \amp 0.6 \amp 0 \amp 0.2 \amp 0 \\ 0 \amp 0 \amp 0.1 \amp 0 \amp 0.6 \\ 0 \amp 0 \amp 0 \amp 0.8 \amp 0.4 \end{pmatrix} \end{equation*}

Then I calculate (by computer) the dominant eigenvector.

\begin{equation*} \begin{pmatrix} 0.3529 \\ 0.2647 \\ 0.1764 \\ 0.0882 \\ 0.1176 \end{pmatrix} \end{equation*}

This gives me the long-term probability distribution. There is roughly a \(35.3\%\) change of ending on vertex \(1\text{;}\) a \(26.5\%\) change of ending on vertex \(2\text{;}\) a \(17.6\%\) change of ending on vertex \(3\text{;}\) a \(8.8\%\) change of ending on vertex \(4\text{;}\) and a \(11.8\%\) change of ending on vertex \(5\text{.}\)

Activity 11.4.10.

Here is a model of a game of chance, in the same setup we used at the end of Section 11.3. Instead of looking at the long run probability of winning, what is the probability of still playing after 5 rounds when starting from \(1\text{,}\) \(2\) or \(3\) stakes. What about after 10 rounds?

Figure 11.4.4. Probability Graph 4

Solution.

First, I need to write down the stochastic matrix.

To get the probabilities when starting from \(1\) stake after \(5\) rounds, I calculate the matrix action five times on the vector that represents \(1\) stake.

\begin{equation*} \begin{pmatrix} 1 \amp 0.4 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0.6 \amp 0 \amp 0.7 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0.3 \amp 1 \end{pmatrix}^5 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.598 \\ 0 \\ 0.2535 \\ 0 \\ 0.1485 \end{pmatrix} \end{equation*}

The probability of still playing is the sum of the three middle probabilities, since the game stops when we are at \(0\) or \(4\) stakes. Here that sum is \(25.4\%\text{.}\)

To get the probabilities when starting from \(2\) stakes after \(5\) rounds, I calculate the matrix action five times on the vector that represents \(2\) stakes.

\begin{equation*} \begin{pmatrix} 1 \amp 0.4 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0.6 \amp 0 \amp 0.7 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0.3 \amp 1 \end{pmatrix}^5 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.33 \\ 0.21125 \\ 0 \\ 0.21125 \\ 0.2475 \end{pmatrix} \end{equation*}

The probability of still playing is the sum of the three middle probabilities, since the game stops when we are at \(0\) or \(4\) stakes. Here that sum is \(42.3\%\text{.}\)

To get the probabilities when starting from \(3\) stakes after \(5\) rounds, I calculate the matrix action five times on the vector that represents \(3\) stakes.

\begin{equation*} \begin{pmatrix} 1 \amp 0.4 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0.6 \amp 0 \amp 0.7 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0.3 \amp 1 \end{pmatrix}^5 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.231 \\ 0 \\ 0.29575 \\ 0 \\ 0.47325 \end{pmatrix} \end{equation*}

The probability of still playing is the sum of the three middle probabilities, since the game stops when we are at \(0\) or \(4\) stakes. Here that sum is \(29.6\%\text{.}\)

To get the probabilities when starting from \(1\) stake after \(10\) rounds, I calculate the matrix action five times on the vector that represents \(1\) stake.

\begin{equation*} \begin{pmatrix} 1 \amp 0.4 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0.6 \amp 0 \amp 0.7 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0.3 \amp 1 \end{pmatrix}^{10} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.68165 \\ 0.0535519 \\ 0 \\ 0.0535519 \\ 0.211241 \end{pmatrix} \end{equation*}

The probability of still playing is the sum of the three middle probabilities, since the game stops when we are at \(0\) or \(4\) stakes. Here that sum is \(10.7\%\text{.}\)

To get the probabilities when starting from \(2\) stakes after \(10\) rounds, I calculate the matrix action five times on the vector that represents \(2\) stakes.

\begin{equation*} \begin{pmatrix} 1 \amp 0.4 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0.6 \amp 0 \amp 0.7 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0.3 \amp 1 \end{pmatrix}^{10} \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.505125 \\ 0 \\ 0.116029 \\ 0 \ 0.378845 \end{pmatrix} \end{equation*}

The probability of still playing is the sum of the three middle probabilities, since the game stops when we are at \(0\) or \(4\) stakes. Here that sum is \(11.6\%\text{.}\)

To get the probabilities when starting from \(3\) stakes after \(10\) rounds, I calculate the matrix action five times on the vector that represents \(3\) stakes.

\begin{equation*} \begin{pmatrix} 1 \amp 0.4 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0.6 \amp 0 \amp 0.7 \amp 0 \\ 0 \amp 0 \amp 0.5 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0.3 \amp 1 \end{pmatrix}^{10} \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.328598 \\ 0.0624772 \\ 0 \\ 0.0624772 \\ 0.546448 \end{pmatrix} \end{equation*}

The probability of still playing is the sum of the three middle probabilities, since the game stops when we are at \(0\) or \(4\) stakes. Here that sum is \(12.5\%\text{.}\)

Subsection 11.4.3 Conceptual Review Questions

What is a recurrence relation and how it is translated into a matrix?
What is a dynamical system? What does it mean for such a system to be discrete and linear?
How can matrices be used as models?
Why are eigenvalues and eigenvectors important in matrix models?
What is a Leslie matrix and what does it measure?
What is a Markov chain and how does it model probability?

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