K_save_households.m

%% Utility Maximization and Intertemporal Consumption
% *back to* <https://fanwangecon.github.io *Fan*>*'s* <https://fanwangecon.github.io/Math4Econ/ 
% *Intro Math for Econ*>*,*  <https://fanwangecon.github.io/M4Econ/ *Matlab Examples*>*, 
% or* <https://fanwangecon.github.io/MEconTools/ *MEconTools*> *Repositories*
%% Model Components and Maximization Problem
% Suppose we have a household who will $z_2$ income tomorrow, and has $z_1$ 
% dollar income income today. He needs to determine how much to save/borrow. There 
% is no uncertainty in this problem, we solve the problem with uncertainty again 
% in: <https://fanwangecon.github.io/Math4Econ/nonlinear/RiskyAsset.html Protofolio 
% Choice: Investments in Risky (stocks) and Safe (bank) Assets, and Financing 
% Risky Investments with Bank Loans>.
% 
% We can write down the model where we maximize utility over choices $c_{today}, 
% c_{tomorrow}$:
%% 
% * *Utility*: $U(c_{today}, c_{tomrrow}) = \log(c_{today}) + \beta \cdot \log(c_{tomorrow})$
% * *Budget Today*: $c_{today} + b = z_1$
% * *Budget Tomorrow*: $c_{tomorrow} = b\cdot (1+ r ) + z_2$
%% 
% We can rewrite the problem as:
%% 
% * $\max_{b} \left\{ \log(z_1-b) + \beta \log(b \cdot (1 + r) + z_2 ) \right\}$
%% 
% _Note_: the only choice in this model is $b$, that will determine consumption 
% today and tomorrow.
% 
% _Note_: Does the interest rate have any effects when there are no inheritances 
% in the second period ($z_2 = 0$)? Change the second period inheritances in the 
% code below to analyze the effect of interest rate.
%% Open set for Choice set
% Even though the budget constraint seems to allow for $0$ consumption today 
% and tomorrow, but $\log$ utility is not defined at $0$, hence the maximization 
% problem is undefined at $c_{today}=0$ and $c_{tomorrow}=0$. Hence, the actual 
% choice set for $save$ is an open interval:
%% 
% * $b \in \left(-\frac{z_2}{1+r}, z_1\right)$, (_which means_ $0$ _and_ $b$ 
% _are not in the domain_)
%% 
% Our Maximization problem is hence:
%% 
% * $\max_{b \in \left(-\frac{z_2}{1+r}, z_1\right) } \left\{ \log(z_1-b) + 
% 0.95 \log(b \cdot (1 + 0.03) + z_2 ) \right\}$
%% 
% If you choose save $z_1$ or more, consumption today will be undefined (death 
% today). If you borrow more than endowment tomorrow divided by interest rate, 
% you will not be able to pay back your debts (we assume no default). Given this 
% choice set, we could view this as a constrained maximization problem, but the 
% constraints never bind.
%% Finding Optimal Choices--Brute Force Grid
% A brute-force way of solving for this problem is to generate a vectors of 
% values for saving between 0 and b, but not including them, evaluate the utility 
% function at these values, and then find the max. This method works when there 
% are more choices as well. Experiment with the following function by adjusting 
% the parameters, including the discount factor, interest rate, and wealth in 
% the first and second period. 

% Model Parameters
beta = 0.95;
r = 0.05;
wealth_1 = 10;
wealth_2 = 1;
% Generate a Vector of Points
choice_grid_count = 1000;
% This creates 100 equi-distance points, not at 0 and b, but between 0 and b
save_grid = linspace(-wealth_2/(1+r)+0.0001, wealth_1-0.0001, choice_grid_count);
% Evaluate utility
utility_at_savegrid = log(wealth_1 - save_grid) + beta*log(wealth_2 + save_grid*(1+r));
% Find Max
[max_utility, max_utility_index] = max(utility_at_savegrid);
% max_utility is the highest utility onthe choice grid
max_utility
% out of the choice grid points, which nth choice grid gives highest utility
max_utility_index
% we can find the savings level at the index
optimal_savings_choice = save_grid(max_utility_index)
% Plot
figure();
hold on;
plot(save_grid, utility_at_savegrid);
scatter(optimal_savings_choice, max_utility, 'filled');
xlabel('Borrow Save Choices along Grid');
ylabel('Utility at different savings choices');
title({'Optimal Borrow Save choice (red dot) on feasible choice grid';...
      ['optimal Borrow Save = ', num2str(optimal_savings_choice)]});
xlim([-wealth_2/(1+r), wealth_1])  
plot(ones(size(save_grid))*0, utility_at_savegrid, 'k--');
grid on;
%% Analytical Solution
% You can use the symbolic toolbox to take derivative and find root:

syms z beta r x
funcU = log(z - x) + beta*log((z/2) + x*(1+r))
dUdx = diff(funcU, x)
xOpti = solve(dUdx==0, x)
%% Supply Curve For Capital
% With the optimal capital choice as a function of interest rate, we can plot 
% out the supply for capital.

z=10;
beta=0.92;
grid_points = 21;
% Rate Vector
r = linspace(1.0,1.2,grid_points);
% Supply Curve
% use the . for division because it is a vector divided by another vector
s=(z*beta*(1+r)-(z/2))./((1+r)*(1+beta));
% Plot
figure();
plot(s,r);
xlabel('Savings (Saved at Bank, to be lent out to firms)');
ylabel('Interest Rate');
title({'Inverse Supply For Capital'});
grid on;
%% 
%