LECT: Add lecture on lake model

Author: mmcky

lectures/_toc.yml

@@ -23,6 +23,7 @@ parts:
   - file: schelling
   - file: solow
   - file: cobweb
+  - file: lake_model
 - caption: Other
   numbered: true
   chapters:

lectures/lake_model.md

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+---
+jupytext:
+  text_representation:
+    extension: .md
+    format_name: myst
+    format_version: 0.13
+    jupytext_version: 1.14.4
+kernelspec:
+  display_name: Python 3 (ipykernel)
+  language: python
+  name: python3
+---
+
+# A Lake Model of Employment
+
+\
+The importance of the Perron-Frobenius theorem stems from the fact that
+firstly in the real world most matrices we encounter are nonnegative matrices.
+
+Secondly, a lot of important models are simply linear iterative models that
+begin with an initial condition $x_0$ and then evolve recursively by the rule
+$x_{t+1} = Ax_t$ or in short $x_t = A^tx_0$.
+
+This theorem helps characterise the dominant eigenvalue $r(A)$ which
+determines the behavior of this iterative process. 
+
+We now illustrate the power of the Perron-Frobenius theorem by showing how it
+helps us to analyze a model of employment and unemployment flows in a large
+population.
+
+This model is sometimes called the **lake model** because there are two pools of workers:
+1. those who are currently employed.
+2. those who are currently unemployed but are seeking employment.
+
+The "flows" between the two lakes are as follows:
+1. workers exit the labor market at rate $d$.
+2. new workers enter the labor market at rate $b$.
+3. employed workers separate from their jobs at rate $\alpha$.
+4. unemployed workers find jobs at rate $\lambda$.
+
+Let $e_t$ and $u_t$ be the number of employed and unemployed workers at time $t$ respectively.
+
+The total population of workers is $n_t = e_t + u_t$.
+
+The number of unemployed and employed workers thus evolve according to:
+
+```{math}
+:label: lake_model
+\begin{aligned}
+    u_{t+1} = (1-d)(1-\lambda)u_t + \alpha(1-d)e_t + bn_t = ((1-d)(1-\lambda) +& b)u_t + (\alpha(1-d) + b)e_t \\
+    e_{t+1} = (1-d)\lambda u_t + (1 - \alpha)(1-d)e_t&
+\end{aligned}
+```
+
+We can arrange {eq}`lake_model` as a linear system of equations in matrix form $x_{t+1} = Ax_t$ such that:
+
+$$
+x_{t+1} =
+\begin{bmatrix}
+    u_{t+1} \\
+    e_{t+1}
+\end{bmatrix}
+, \; A =
+\begin{bmatrix}
+    (1-d)(1-\lambda) + b & \alpha(1-d) + b \\
+    (1-d)\lambda & (1 - \alpha)(1-d)
+\end{bmatrix}
+\; \text{and} \;
+x_t =
+\begin{bmatrix}
+    u_t \\
+    e_t
+\end{bmatrix}
+$$
+
+Suppose at $t=0$ we have $x_0 = \begin{bmatrix} u_0 \\ e_0 \end{bmatrix}$.
+
+Then, $x_1=Ax_0$, $x_2=Ax_1=A^2x_0$ and thus $x_t = A^tx_0$.
+
+Thus the long run outcomes of this system depend on the initial condition $x_0$ and the matrix $A$.
+
+$A$ is a nonnegative and irreducible matrix and thus we can use the Perron-Frobenius theorem to obtain some useful results about A.
+
+Note that $colsum_j(A) = 1 + b - d$ for $j=1,2$ and by {eq}`PF_bounds` we can thus conclude that the dominant eigenvalue
+$r(A) = 1 + b - d$.
+
+If we consider $g = b - d$ as the overall growth rate of the total labor force, then $r(A) = 1 + g$.
+
+We can thus find a unique positive vector $\bar{x} = \begin{bmatrix} \bar{u} \\ \bar{e} \end{bmatrix}$
+such that $A\bar{x} = r(A)\bar{x}$ and $\begin{bmatrix} 1 & 1 \end{bmatrix} \bar{x} = 1$.
+
+Since $\bar{x}$ is the eigenvector corresponding to the dominant eigenvalue $r(A)$ we can also call this the dominant eigenvector.
+
+This eigenvector plays an important role in determining long run outcomes as is illustrated below.
+
+```{code-cell} ipython3
+:tags: []
+
+def lake_model(α, λ, d, b):
+    g = b-d
+    A = np.array([[(1-d)*(1-λ) + b,   α*(1-d) + b],
+              [(1-d)*λ,          (1-α)*(1-d)]])
+
+    ū = (1 + g - (1 - d) * (1 - α)) / (1 + g - (1 - d) * (1 - α) + (1 - d) * λ)
+    ē = 1 - ū
+    x̄ = np.array([ū, ē])
+    x̄.shape = (2,1)
+    
+    ts_length = 1000
+    x_ts_1 = np.zeros((2, ts_length))
+    x_ts_2 = np.zeros((2, ts_length))
+    x_0_1 = (5.0, 0.1)
+    x_0_2 = (0.1, 4.0)
+    x_ts_1[0, 0] = x_0_1[0]
+    x_ts_1[1, 0] = x_0_1[1]
+    x_ts_2[0, 0] = x_0_2[0]
+    x_ts_2[1, 0] = x_0_2[1]
+    
+    for t in range(1, ts_length):
+        x_ts_1[:, t] = A @ x_ts_1[:, t-1]
+        x_ts_2[:, t] = A @ x_ts_2[:, t-1]
+        
+    fig, ax = plt.subplots()
+    
+    # Set the axes through the origin
+    for spine in ["left", "bottom"]:
+        ax.spines[spine].set_position("zero")
+    for spine in ["right", "top"]:
+        ax.spines[spine].set_color("none")
+        
+    ax.set_xlim(-2, 6)
+    ax.set_ylim(-2, 6)
+    ax.set_xlabel("unemployed workforce")
+    ax.set_ylabel("employed workforce")
+    ax.set_xticks((0, 6))
+    ax.set_yticks((0, 6))
+    s = 10
+    ax.plot([0, s * ū], [0, s * ē], "k--", lw=1)
+    ax.scatter(x_ts_1[0, :], x_ts_1[1, :], s=4, c="blue")
+    ax.scatter(x_ts_2[0, :], x_ts_2[1, :], s=4, c="green")
+    
+    ax.plot([ū], [ē], "ko", ms=4, alpha=0.6)
+    ax.annotate(r'$\bar{x}$', 
+            xy=(ū, ē),
+            xycoords="data",
+            xytext=(20, -20),
+            textcoords="offset points",
+            arrowprops=dict(arrowstyle = "->"))
+
+    x, y = x_0_1[0], x_0_1[1]
+    ax.plot([x], [y], "ko", ms=2, alpha=0.6)
+    ax.annotate(f'$x_0 = ({x},{y})$', 
+            xy=(x, y),
+            xycoords="data",
+            xytext=(0, 20),
+            textcoords="offset points",
+            arrowprops=dict(arrowstyle = "->"))
+    
+    x, y = x_0_2[0], x_0_2[1]
+    ax.plot([x], [y], "ko", ms=2, alpha=0.6)
+    ax.annotate(f'$x_0 = ({x},{y})$', 
+            xy=(x, y),
+            xycoords="data",
+            xytext=(0, 20),
+            textcoords="offset points",
+            arrowprops=dict(arrowstyle = "->"))
+    
+    plt.show()
+```
+
+```{code-cell} ipython3
+lake_model(α=0.01, λ=0.1, d=0.02, b=0.025)
+```
+
+If $\bar{x}$ is an eigenvector corresponding to the eigenvalue $r(A)$ then all the vectors in the set
+$D := \{ x \in \mathbb{R}^2 : x = \alpha \bar{x} \; \text{for some} \; \alpha >0 \}$ are also eigenvectors corresponding
+to $r(A)$.
+
+This set is represented by a dashed line in the above figure.
+
+We can observe that for two distinct initial conditions $x_0$ the sequence of iterates $(A^t x_0)_{t \geq 0}$ move towards D over time.
+
+This suggests that all such sequences share strong similarities in the long run, determined by the dominant eigenvector $\bar{x}$.
+
+In the example illustrated above we considered parameter such that overall growth rate of the labor force $g>0$.
+
+Suppose now we are faced with a situation where the $g<0$, i.e, negative growth in the labor force.
+
+This means that $b-d<0$, i.e., workers exit the market faster than they enter.
+
+What would the behavior of the iterative sequence $x_{t+1} = Ax_t$ be now?
+
+This is visualised below.
+
+```{code-cell} ipython3
+lake_model(α=0.01, λ=0.1, d=0.025, b=0.02)
+```
+
+Thus, while the sequence of iterates still move towards the dominant eigenvector $\bar{x}$ however in this case
+they converge to the origin.
+
+This is a result of the fact that $r(A)<1$, which ensures that the iterative sequence  $(A^t x_0)_{t \geq 0}$ will converge
+to some point, in this case to $(0,0)$.
+
+This leads us into the next result.