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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Introduction to Dynamic Programming
We have studied the theory of dynamic programming in discrete time under certainty. Let's review what we know so far, so that we can start thinking about how to take to the computer.
The Problem
We want to find a sequence ${x_t}_{t=0}^\infty$ and a function $V^*
Step3:  Clearly the more points we have the better our approximation. But, more points means more computations and more time to get those approximations. Since we will be iterating over approximations, we might not want to use too many points, but be smart about the choice of points or we might want to use less points for a start and then increase the number of points once we have a good candidate solution to our fixed point problem.
In order to make it easy to define interpolated functions, we define a new class of Python object
Step4:  We can now define our interpolated sinus function as follows
Step5:  Optimal Growth
Let's start by computing the solution to an optimal growth problem, in which a social planner seeks to find paths ${c_t,k_t}$ such that
\begin{align}
\max_{{c_t,k_t}}&\sum_{t=0}^{\infty}\beta^{t}u(c_{t})\[.2cm]
\text{s.t. }&k_{t+1}\leq f(k_{t})+(1-\delta)k_{t}-c_{t}\[.2cm]
c_{t}\geq0,&\ k_{t}\geq0,\ k_{0}\text{ is given}.
\end{align}
As usual we assume that our utility function $u(\cdot)$ and production function $f(\cdot)$ are Neoclassical. Under these conditions we have seen that our problem satisfies the conditions of our previous theorem and thus we know a unique solution exists.
An example with analytical solution
Let's assume that $u(c)=\ln(c)$, $f(k)=k^\alpha$, and $\delta=1$. For this case we have seen that the solution implies
\begin{align}
&\text{Value Function
Step6:  Let's fix the value of the fundamental parameters so we can realize computations
Step7:  Now let's focus on the Value function iteration
Step8:  Here we have created a grid on $[gridmin,gridmax]$ that has a number of points given by gridsize. Since we know that the Value functions is stricly concave, our grid has more points closer to zero than farther away
Step9:  Now we need a function, which for given $V_0$ solves 
$$\sup\limits _{y\in G(x)}\left{ U(x,y)+\beta V(y)\right}.$$
Let's use one of Scipy's optimizing routines
Step10:  Since fminbound returns 
$$\arg\min\limits _{y\in [\underline x,\bar x]}\left{ U(x,y)+\beta V(y)\right}$$
we have to either replace our objective function for its negative or, better yet, define a function that uses fminbound and returns the maximum and the maximizer
Step13:  Note
We could have included other parameters to pass to our maximizer and maximum functions, e.g. to allow us to manipulate the options of fminbound
The Bellman Operator
Step14:  Given a linear interpolation of our guess for the Value function, $V_0=w$, the first function returns a LinInterp object, which is the linear interpolation of the function generated by the Bellman Operator on the finite set of points on the grid. The second function returns what Stachurski (2009) calls a w-greedy policy, i.e. the function that maximizes the RHS of the Bellman Operator.
Now we are ready to work on the iteration
Step15:  Our initial guess $V_0$
Step16:  After one interation
Step17:  Doing it by hand is too slow..let's automate this process
Step18:  Does it look like we converged? Let's compare our estimated Value function V1 and the actual function Va and compute the error at each point.
Step19:  Exercise
Use the policy function to compute the optimal policy. Compare it to the actual one
Do the same for the consumption function. Find the savings rate and plot it.
Construct the paths of consumption and capital starting from $k_0=.1$. Show the time series and the paths in the consumption-capital space
Estimate the level of steady state capital and consumption. Show graphically that it is lower than the Golden Rule Level.
Repeat the exercise with other values of $\alpha,\beta,\delta,\sigma,k_0$. Can you write a function or class such that it will generate the whole analysis for given values of the parameters and functions. Can you generalize it in order to analyze the effects of changing the utility or production functions?
Solution
Since we already have V1, we can just apply policy(V1) to get the result
Step20:  Since $c = f(k) + (1-\delta) k - k'$
Step21:  Since $s = f(k) - c = k' - (1-\delta) k = $ investment,
Step22:  Start with $k_0=0.1$ and follow the economy for 20 periods.
Step24:  Steady state level of capital and consumption are given by these last elements of these lists
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Python Code:
from __future__ import division
%pylab --no-import-all
%matplotlib inline
from numpy import interp
interp?
x = np.linspace(0, np.pi, 100)
plt.figure(1)
plt.plot(x, np.sin(x), label='Actual Function')
for  i in np.arange(3,11,2):
    fig1 = plt.figure(1)
    xp = np.linspace(0, np.pi, i)
    yp = np.sin(xp)
    y = interp(x, xp, yp)
    plt.plot(x, y, label='Interpolation ' + str(i))
    fig2 = plt.figure(2)
    plt.title('Error with up to ' + str(i) + ' points in interpolation')
    plt.ylabel('Error')
    plt.plot(y - np.sin(x), label=str(i))
    plt.legend(loc=8)
plt.figure(1)
plt.legend(loc=8)
plt.show()
fig1
fig2
Explanation: Introduction to Dynamic Programming
We have studied the theory of dynamic programming in discrete time under certainty. Let's review what we know so far, so that we can start thinking about how to take to the computer.
The Problem
We want to find a sequence ${x_t}_{t=0}^\infty$ and a function $V^*:X\to\mathbb{R}$ such that 
$$V^{\ast}\left(x_{0}\right)=\sup\limits {\left{ x{t}\right} {t=0}^{\infty}}\sum\limits {t=0}^{\infty}\beta^{t}U(x_{t},x_{t+1})$$ 
subject to $x_{t+1}\in G(x_{t})\subseteq X\subseteq\mathbb{R}^K$, for all $t\geq0$ and $x_0\in\mathbb{R}$ given. We assume $\beta\in(0,1)$.
We have seen that we can analyze this problem by solving instead the related problem 
$$V(x)=\sup\limits _{y\in G(x)}\left{ U(x,y)+\beta V(y)\right} ,\text{ for all }x\in X.$$
Basic Results
Assumptions
$G\left(x\right)$  is nonempty for all $x\in X$ ; and for all $x_{0}\in X$  and $\mathbf{x}\in \Phi (x_{0})$, $\lim\nolimits_{n\rightarrow\infty}\sum_{t=0}^{n}\beta^{t}U(x_{t},x_{t+1})$ exists and is finite.
$X$  is a compact subset of $\mathbb{R}^{K}$, $G$ is nonempty, compact-valued and continuous. Moreover, 
$U:\mathbf{X}{G}\rightarrow\mathbb{R}$  is continuous, where $\mathbf{X}{G}=\left{ (x,y)\in X\times X:y\in G(x)\right}$
$U$ is strictly concave and $G$  is convex
For each $y\in X$, $U(\cdot,y)$  is strictly increasing in each of its first $K$  arguments, and $G$  is monotone in the sense that $x\leq x^{\prime}$  implies $G(x)\subset G(x^{\prime})$.
$U$  is continuously differentiable on the interior of its domain $\mathbf{X}_{G}$.
Let $\Phi (x_{t})={{x_{s}}{s=t}^{\infty}:x{s+1}\in G(x_{s})\text{, for }s=t,t+1,...}$ and assume that $\lim_{t\rightarrow\infty}\beta^{t}V\left(x_{t}\right)=0$  for all $\left(x,x_{1},x_{2},...\right)\in \Phi (x)$.
If all of these conditions are satisfied, then we have the following
Theorem
There exists a unique (value) function 
$$V^(x_0)=V(x_0),$$ 
which is continuous, strictly increasing, strictly concave, and differentiable. Also, there exists a unique path ${x^t}{t=0}^\infty$, which starting from the given $x_0$ attains the value $V^(x_0)$. The path can be found through a unique continuous policy function $\pi: X\to X$ such that $x^_{t+1}=\pi(x^_t)$.*
Taking it to the computer
Ok. Now that we know the conditions for the existence and uniqueness (plus other characteristics) of our problem, how do we go about solving it?
The idea is going to be simple and is based on what we saw when we proved the contraction mapping theorem and the proof of the previous theorem (Yes I know...we split this in various steps and intermediate results, which might have confused you). 
Remember that our Bellman Operator $T: C(X)\to C(X)$ defined as 
$$T(V(x))\equiv\sup\limits _{y\in G(x)}\left{ U(x,y)+\beta V(y)\right}$$
assigns a continuous, strictly increasing, strictly concave function $T(V)$ to each continuous, increasing, and concave function $V$ defined on $X$. Since $T(V)$ is a contraction mapping, we know that if $V_0$ is any initial continuous, increasing, and concave function defined on $X$, then $T^n(V_0)\to V^*$. This is precisely what we are going to do using the computer (well we will also do it by hand for a couple of examples).
Value function iteration
So, now that we have a strategy to tackle the problem, and you have learned some basic Python at Code Academy and IPython in our other notebook, we are ready to write some code and do some dynamic economic analysis.
But before we start, there is one issue I want to highlight. Notice that our state space $X$ is not assumed to be finite, and clearly the fact that our functions are continuous imply that we cannot be in a finite problem. So how do we represent such an infinite object in a computer, which only has finite memory? The solution is to take an approximation to the function, what Stachurski (2009) calls a fitted function. There are various methods to approximate functions (see Judd (1998) for an excellent presentation). The simplest method is a linear interpolation, which is what we will use here. 
The idea behind linear interpolation is quite simple. Assume we want to approximate the function $V: X\to X$, $X\subseteq\mathbb{R}$. The only thing we need is a finite set ${x_i}{i=0}^N\subseteq X$ for which we compute the value under $V$, i.e. we create the finite set of values ${V_i=V(x_i)}{i=0}^N$. Then our approximation to the function $V$, $\hat V$, will be defined as
$$\hat V(x)=V_{i-1}+\frac{V_i-V_{i-1}}{x_i-x_{i-1}}(x-x_{i-1}) \quad\text{ if } x_{i-1}\le x < x_i.$$
In principle we could construct our own interpolation function, but Scipy has already optimized approximation algorithms, so let's use that package instead. Let's see what a linear interpolation of $\sin(x)$ would look like.
End of explanation
class LinInterp:
    "Provides linear interpolation in one dimension."
    def __init__(self, X, Y):
        Parameters: X and Y are sequences or arrays
        containing the (x,y) interpolation points.
        
        self.X, self.Y = X, Y
    def __call__(self, z):
        Parameters: z is a number, sequence or array.
        This method makes an instance f of LinInterp callable,
        so f(z) returns the interpolation value(s) at z.
        
        if isinstance(z, int) or isinstance(z, float):
            return interp ([z], self.X, self.Y)[0]  
        else:
            return interp(z, self.X, self.Y)
Explanation: Clearly the more points we have the better our approximation. But, more points means more computations and more time to get those approximations. Since we will be iterating over approximations, we might not want to use too many points, but be smart about the choice of points or we might want to use less points for a start and then increase the number of points once we have a good candidate solution to our fixed point problem.
In order to make it easy to define interpolated functions, we define a new class of Python object
End of explanation
xp = np.linspace(0, np.pi, 10)
yp = np.sin(xp)
oursin = LinInterp(xp, yp)
plt.plot(oursin(x));
Explanation: We can now define our interpolated sinus function as follows
End of explanation
%%file optgrowthfuncs.py
def U(c, sigma=1):
    '''This function returns the value of utility when the CRRA
    coefficient is sigma. I.e. 
    u(c,sigma)=(c**(1-sigma)-1)/(1-sigma) if sigma!=1 
    and 
    u(c,sigma)=ln(c) if sigma==1
    Usage: u(c,sigma)
    '''
    if sigma!=1:
        u = (c**(1-sigma)-1) / (1-sigma)
    else:
        u = np.log(c)
    return u
def F(K, L=1, alpha=.3, A=1):
    '''
    Cobb-Douglas production function
    F(K,L)=K^alpha L^(1-alpha)
    '''
    return A * K**alpha * L**(1-alpha)
def Va(k, alpha=.3, beta=.9):
    ab = alpha*beta
    return np.log(1-ab) / (1-beta) + ab * np.log(ab) / ((1-beta) * (1-ab)) + alpha * np.log(k) / (1-ab)
def opk(k, alpha=.3, beta=.9):
    return alpha * beta * k**alpha
def opc(k, alpha=.3, beta=.9):
    return (1-alpha*beta)*k**alpha
# %load optgrowthfuncs.py
def U(c, sigma=1):
    '''This function returns the value of utility when the CRRA
    coefficient is sigma. I.e. 
    u(c,sigma)=(c**(1-sigma)-1)/(1-sigma) if sigma!=1 
    and 
    u(c,sigma)=ln(c) if sigma==1
    Usage: u(c,sigma)
    '''
    if sigma!=1:
        u = (c**(1-sigma)-1) / (1-sigma)
    else:
        u = np.log(c)
    return u
def F(K, L=1, alpha=.3, A=1):
    '''
    Cobb-Douglas production function
    F(K,L)=K^alpha L^(1-alpha)
    '''
    return A * K**alpha * L**(1-alpha)
def Va(k, alpha=.3, beta=.9):
    ab = alpha*beta
    return np.log(1-ab) / (1-beta) + ab * np.log(ab) / ((1-beta) * (1-ab)) + alpha * np.log(k) / (1-ab)
def opk(k, alpha=.3, beta=.9):
    return alpha * beta * k**alpha
def opc(k, alpha=.3, beta=.9):
    return (1-alpha*beta)*k**alpha
Explanation: Optimal Growth
Let's start by computing the solution to an optimal growth problem, in which a social planner seeks to find paths ${c_t,k_t}$ such that
\begin{align}
\max_{{c_t,k_t}}&\sum_{t=0}^{\infty}\beta^{t}u(c_{t})\[.2cm]
\text{s.t. }&k_{t+1}\leq f(k_{t})+(1-\delta)k_{t}-c_{t}\[.2cm]
c_{t}\geq0,&\ k_{t}\geq0,\ k_{0}\text{ is given}.
\end{align}
As usual we assume that our utility function $u(\cdot)$ and production function $f(\cdot)$ are Neoclassical. Under these conditions we have seen that our problem satisfies the conditions of our previous theorem and thus we know a unique solution exists.
An example with analytical solution
Let's assume that $u(c)=\ln(c)$, $f(k)=k^\alpha$, and $\delta=1$. For this case we have seen that the solution implies
\begin{align}
&\text{Value Function: } &  V(k)=&\frac{\ln(1-\alpha\beta)}{1-\beta}+\frac{\alpha\beta\ln(\alpha\beta)}{(1-\alpha\beta)(1-\beta)}+\frac{\alpha}{1-\alpha\beta}\ln(k)\[.2cm]
&\text{Optimal Policy: } & \pi\left(k\right)=&\beta\alpha k^{\alpha} \[.2cm]
&\text{Optimal Consumption Function: } & c=&\left(1-\beta\alpha\right)k^{\alpha}\[.2cm]
\end{align}
We will use these to compare the solution found by iteration of the Value function described above. Copy the Python functions you had defined in the previous notebook into the cell below and define Python functions for the actual optimal solutions given above.
End of explanation
alpha = .3
beta = .9
sigma = 1
delta = 1
Explanation: Let's fix the value of the fundamental parameters so we can realize computations
End of explanation
# Grid of values for state variable over which function will be approximated
gridmin, gridmax, gridsize = 0.1, 5, 300
grid = np.linspace(gridmin, gridmax**1e-1, gridsize)**10
Explanation: Now let's focus on the Value function iteration:
End of explanation
plt.hist(grid, bins=50);
plt.xlabel('State Space');
plt.ylabel('Number of Points');
plt.plot(grid, grid,'r.');
plt.title('State Space Grid');
Explanation: Here we have created a grid on $[gridmin,gridmax]$ that has a number of points given by gridsize. Since we know that the Value functions is stricly concave, our grid has more points closer to zero than farther away
End of explanation
from scipy.optimize import fminbound
fminbound?
Explanation: Now we need a function, which for given $V_0$ solves 
$$\sup\limits _{y\in G(x)}\left{ U(x,y)+\beta V(y)\right}.$$
Let's use one of Scipy's optimizing routines
End of explanation
# Maximize function V on interval [a,b]
def maximum(V, a, b, **kwargs):
    return float(V(fminbound(lambda x: -V(x), a, b, **kwargs)))
# Return Maximizer of function V on interval [a,b]
def maximizer(V, a, b, **kwargs):
    return float(fminbound(lambda x: -V(x), a, b, **kwargs))
Explanation: Since fminbound returns 
$$\arg\min\limits _{y\in [\underline x,\bar x]}\left{ U(x,y)+\beta V(y)\right}$$
we have to either replace our objective function for its negative or, better yet, define a function that uses fminbound and returns the maximum and the maximizer
End of explanation
# The following two functions are used to find the optimal policy and value functions using value function iteration
# Bellman Operator
def bellman(w):
    The approximate Bellman operator.
    Parameters: w is a LinInterp object (i.e., a 
    callable object which acts pointwise on arrays).
    Returns: An instance of LinInterp that represents the optimal operator.
    w is a function defined on the state space.
    
    vals = []
    for k in grid:
        kmax = F(k, alpha=alpha) + (1-delta) * k
        h = lambda kp: U(kmax - kp, sigma) + beta * w(kp)
        vals.append(maximum(h, 0, kmax))
    return LinInterp(grid, vals)
# Optimal policy
def policy(w):
    
    For each function w, policy(w) returns the function that maximizes the 
    RHS of the Bellman operator.
    Replace w for the Value function to get optimal policy.
    The approximate optimal policy operator w-greedy (See Stachurski (2009)). 
    Parameters: w is a LinInterp object (i.e., a 
    callable object which acts pointwise on arrays).
    Returns: An instance of LinInterp that captures the optimal policy.
    
    vals = []
    for k in grid:
        kmax = F(k,alpha=alpha) + (1-delta) * k
        h = lambda kp: U(kmax - kp,sigma) + beta * w(kp)
        vals.append(maximizer(h, 0, kmax))
    return LinInterp(grid, vals)
Explanation: Note
We could have included other parameters to pass to our maximizer and maximum functions, e.g. to allow us to manipulate the options of fminbound
The Bellman Operator
End of explanation
# Parameters for the optimization procedures
count = 0
maxiter = 1000
tol = 1e-6
print('tol=%f' % tol)
Explanation: Given a linear interpolation of our guess for the Value function, $V_0=w$, the first function returns a LinInterp object, which is the linear interpolation of the function generated by the Bellman Operator on the finite set of points on the grid. The second function returns what Stachurski (2009) calls a w-greedy policy, i.e. the function that maximizes the RHS of the Bellman Operator.
Now we are ready to work on the iteration
End of explanation
V0 = LinInterp(grid,U(grid))
plt.figure(1)
plt.plot(grid, V0(grid), label='V'+str(count));
plt.plot(grid, Va(grid), label='Actual');
plt.legend(loc=0);
Explanation: Our initial guess $V_0$
End of explanation
plt.plot(grid, V0(grid), label='V'+str(count));
count += 1
V0 = bellman(V0)
plt.figure(1)
plt.plot(grid, V0(grid), label='V'+str(count));
plt.plot(grid, Va(grid), label='Actual');
plt.legend(loc=0);
plt.show();
Explanation: After one interation
End of explanation
fig, ax = plt.subplots()
ax.set_xlim(grid.min(), grid.max())
ax.plot(grid, Va(grid), label='Actual', color='k', lw=2, alpha=0.6);
count=0
maxiter=200
tol=1e-6
while count<maxiter:
    V1 = bellman(V0)
    err = np.max(np.abs(np.array(V1(grid))-np.array(V0(grid))))
    if np.mod(count,10)==0:
        ax.plot(grid, V1(grid), color=plt.cm.jet(count / maxiter), lw=2, alpha=0.6);
        #print('%d %2.10f ' % (count,err))
    V0 = V1
    count += 1
    if err<tol:
        print(count)
        break
ax.plot(grid, V1(grid), label='Estimated', color='r', lw=2, alpha=0.6);
ax.legend(loc='lower right')
plt.draw();
fig
Explanation: Doing it by hand is too slow..let's automate this process
End of explanation
print(err)
err = Va(grid)-V1(grid)
plt.plot(grid,err);
print(err.max()-err.min())
fig, ax = plt.subplots()
ax.set_ylim(-10, -7)
ax.set_xlim(grid.min(), grid.max())
ax.plot(grid, Va(grid), label='Actual')
ax.plot(grid, V1(grid)+err.mean(), label='Estimated')
ax.legend(loc='lower right')
plt.show();
Explanation: Does it look like we converged? Let's compare our estimated Value function V1 and the actual function Va and compute the error at each point.
End of explanation
optcapital = policy(V1)
plt.figure(1)
plt.plot(grid, optcapital(grid), label='Estimated Policy Function');
plt.plot(grid, opk(grid), label='Actual Policy Function');
plt.legend(loc='lower right');
err = opk(grid)-optcapital(grid)
plt.plot(grid,err);
print(err.max()-err.min())
print(err.mean())
Explanation: Exercise
Use the policy function to compute the optimal policy. Compare it to the actual one
Do the same for the consumption function. Find the savings rate and plot it.
Construct the paths of consumption and capital starting from $k_0=.1$. Show the time series and the paths in the consumption-capital space
Estimate the level of steady state capital and consumption. Show graphically that it is lower than the Golden Rule Level.
Repeat the exercise with other values of $\alpha,\beta,\delta,\sigma,k_0$. Can you write a function or class such that it will generate the whole analysis for given values of the parameters and functions. Can you generalize it in order to analyze the effects of changing the utility or production functions?
Solution
Since we already have V1, we can just apply policy(V1) to get the result
End of explanation
def optcons(k, k1=None):
    '''
    Return value of consumption when capitakl today is k
    '''
    if k1 is None:
        c = F(k) + (1-delta)*k - optcapital(k)
    else:
        c = F(k) + (1-delta)*k - k1
    return c
plt.figure(1)
plt.plot(grid, optcons(grid), label='Estimated Consumption Function');
plt.plot(grid, opc(grid), label='Actual Consumption Function');
plt.legend(loc='lower right');
err = opc(grid)-optcons(grid)
plt.plot(grid,err);
print(err.max()-err.min())
print(err.mean())
Explanation: Since $c = f(k) + (1-\delta) k - k'$
End of explanation
def optsav(k):
    '''
    Estimated savings function
    '''
    s = F(k) - optcons(k)
    return s
def ops(k):
    s = F(k) - opc(k)
    return s
plt.figure(1)
plt.plot(grid, optsav(grid), label='Estimated Savings Function');
plt.plot(grid, ops(grid), label='Actual Savings Function');
plt.legend(loc='lower right');
err = ops(grid)-optsav(grid)
plt.plot(grid,err);
print(err.max()-err.min())
print(err.mean())
Explanation: Since $s = f(k) - c = k' - (1-\delta) k = $ investment,
End of explanation
T = 20
kt = []
ct = []
k0 = 0.1
kt.append(k0)
for t in range(0,T):
    k1 = optcapital(k0)
    c0 = optcons(k0, k1)
    k0 = k1
    ct.append(c0)
    kt.append(k0)
plt.plot(kt)
plt.xlabel(r'$t$')
plt.ylabel(r'$k_t$')
plt.plot(ct)
plt.xlabel(r'$t$')
plt.ylabel(r'$c_t$')
print(kt)
print(ct)
Explanation: Start with $k_0=0.1$ and follow the economy for 20 periods.
End of explanation
css = ct[-1]
kss = kt[-1]
print('Steady state capital is ', kss)
print('Steady state consumption is ', css)
k0 = 0.1
maxerr = 1e-14
maxiter = 1000
count = 0
while count<maxiter:
    k1 = optcapital(k0)
    err = np.abs(k1-k0)
    if err<maxerr:
        kss_true = k1
        break
    else:
        k0 = k1
        count += 1
css_true = optcons(kss, kss)
from IPython.display import display, Markdown
display(Markdown(
   rf
       True theoretical steady state value of capital {kss_true:1.5}.
       Approximated numerical steady state value of capital {kss:1.5}.
       True theoretical steady state value of consumption {css_true:1.5}.
       Approximated numerical steady state value of consumption {css:1.5}.
       Convergence to steady state in {count} iterations.
   ))
print(kss_true)
print(kss)
print(count)
print(css_true)
print(css)
Explanation: Steady state level of capital and consumption are given by these last elements of these lists
End of explanation
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