% Problem Set 13
% Wed. Nov. 28

clc; 
clear; 
close all;

%2. a.
%Input
x = [1:1:15]';
%T = number of observations
T = size(x,1);
%Output
y = [0.58, 1.10, 1.20, 1.30, 1.95, 2.55, 2.60, 2.90, 3.45, 3.50, 3.60, 4.10, 4.35, 4.40, 4.50]';
%Add constant to Input
x = [ones(15,1), x];
%K = number of regressors
K = size(x,2);
xprimx = x'*x;
xprimy = x'*y;
b = inv(xprimx)*xprimy;
b1 = b(1,1);
b2 = b(2,1);

%2.b Written.

%2. c.
e = y - b1*x(:,1) - b2*x(:,2);
sig2 = e'*e/(T-K);

%2.d.
W = sig2*inv(xprimx);

%2.e.
lny = log(y);
%When we include the first row of constants and then take the log it
%becomes zero. Re-estimate with ln(b1) = 1 to get an estimate, say a, we will be 
%able to recover b1 via exp(a) = b1. 
lnx = [ones(15,1), log(x(:,2))];
lnb = inv(lnx'*lnx)*(lnx'*lny);
b12e = exp(lnb(1,1));
b22e = lnb(2,1);

%2.f Written.

%2.g 
y2a = x*b;
y2e = b12e*x(:,2).^(b22e);

plot(y2a,'r-');
hold on;
plot(y2e,'k:');
plot(y,'b-+');
legend('linear','log linear','actual','location','NorthWest');

%matrix size
m = 15;
n = 1000;
mu = 0;
sig2 = (y-b12e*x(:,2).^(b22e))'*(y-b12e*x(:,2).^(b22e))/(T-K);
%fill a 15x1000 matrix with the log linear estimates of output
temp1 = ones(m,n);
temp2 = b12e*x(:,2).^(b22e);
temp3 = zeros(m,n);
for i = 1:1:15
temp3(i,:) = temp1(i,:)*temp2(i,1);
end

rand('seed',1234);
egen = mu + sqrt(sig2)*randn(m,n);
y2ematrix = temp3 + egen;

lny2ematrix = log(y2ematrix);
lnbmatrix = 9999*ones(2,1000);
for i = 1:1:1000;
lnbmatrix(:,i) = inv(lnx'*lnx)*(lnx'*lny2ematrix(:,1));
end