Posted by: admin 3 years, 1 month ago

(Comments)

So my goal today is pretty simple, be confident with the methodology of statistic in Maatlab! 

The video today that I will watch is 

Couple of impoortant pooint is

The code can be found here

couple of interesting element

• what does y(:,1) means if y is a matrix ? the answer is you want to access all row and only the first column. 

The full code

Demo 1: Cointegration

Demo from the April 14, 2011 webinar titled "Cointegration and Pairs Trading with Econometrics Toolbox."

See also Demo 2

Contents

 Copyright 2011, The MathWorks, Inc.
 All rights reserved.

clear; close all; clc

Interest Rate Data

load Data_Canada
Y = Data(:,3:end); %meaning all row but only from column 3 to end

figure
plot(dates,Y,'LineWidth',2)
xlabel('Year')
ylabel('Percent')
names = series(3:end);
legend(names,'location','NW')
title('{\bf Canadian Interest Rates, 1954-1994}')
axis tight
grid on

Pretest for the Order of Integration

y1 = Y(:,1); % Short-term rate

% Levels data:
fprintf('=== Test y1 for a unit root ===\n\n')
[h1,pVal1] = adftest(y1,'model','ARD') % Left-tail probability

fprintf('\n=== Test y1 for stationarity ===\n\n')
[h0,pVal0] = kpsstest(y1,'trend',false) % Right-tail probability

% Differenced data:
fprintf('\n=== Test (1-L)y1 for a unit root ===\n\n')
[h1D,pVal1D] = adftest(diff(y1),'model','ARD') % Left-tail probability

fprintf('\n=== Test (1-L)y1 for stationarity ===\n\n')
[h0D,pVal0D] = kpsstest(diff(y1),'trend',false) % Right-tail probability

figure
plot(dates(2:end),diff(Y),'LineWidth',2)
names = series(3:end);
legend(names,'location','NW')
title('{\bf Differenced Data}')
axis tight
grid on

=== Test y1 for a unit root ===


h1 =

     0


pVal1 =

    0.2867


=== Test y1 for stationarity ===

Warning: Test statistic #1 above tabulated critical values:
minimum p-value = 0.010 reported. 

h0 =

     1


pVal0 =

    0.0100


=== Test (1-L)y1 for a unit root ===

Warning: Test statistic #1 below tabulated critical values:
minimum p-value = 0.001 reported. 

h1D =

     1


pVal1D =

  1.0000e-003


=== Test (1-L)y1 for stationarity ===

Warning: Test statistic #1 below tabulated critical values:
maximum p-value = 0.100 reported. 

h0D =

     0


pVal0D =

    0.1000

Engle-Granger Test for Cointegration

% Run the test with both "tau" (t1) and "z" (t2) statistics:
fprintf('\n=== Engle-Granger tests for cointegration ===\n\n')
[hEG,pValEG] = egcitest(Y,'test',{'t1','t2'})

=== Engle-Granger tests for cointegration ===


hEG =

     0     1


pValEG =

    0.0526    0.0202

Identify the Cointegrating Relation

% Return the results of the cointegrating regression:
[~,~,~,~,reg] = egcitest(Y,'test','t2');

c0 = reg.coeff(1);
b = reg.coeff(2:3);
figure
C = get(gca,'ColorOrder');
set(gca,'NextPlot','ReplaceChildren','ColorOrder',circshift(C,3))
plot(dates,Y*[1;-b]-c0,'LineWidth',2)
title('{\bf Cointegrating Relation}')
axis tight
grid on

VEC Model Estimation, Simulation, Forecasting

% See Documentation:
%
%   Econometrics Toolbox\User's Guide
%   \Mulivariate Time Series Models
%   \Cointegration and Error Correction
%   \Identifying Single Cointegrating Relations
%   \Estimating VEC Model Parameters

Multiple Cointegrating Relations

% Permutations of the data variables:
P0 = perms([1 2 3]);
[~,idx] = unique(P0(:,1)); % Rows of P0 with unique regressand y1
P = P0(idx,:); % Unique regressions
numPerms = size(P,1);

% Preallocate:
T0 = size(Y,1);
HEG = zeros(1,numPerms);
PValEG = zeros(1,numPerms);
CIR = zeros(T0,numPerms);

% Run all tests:
for i = 1:numPerms

    YPerm = Y(:,P(i,:));
    [h,pVal,~,~,reg] = egcitest(YPerm,'test','t2');
    HEG(i) = h;
    PValEG(i) = pVal;
    c0i = reg.coeff(1);
    bi = reg.coeff(2:3);
    CIR(:,i) = YPerm*[1;-bi]-c0i;

end

fprintf('\n=== Different Engle-Granger tests, same data ===\n\n')
HEG,PValEG

% Plot the cointegrating relations:
figure
C = get(gca,'ColorOrder');
set(gca,'NextPlot','ReplaceChildren','ColorOrder',circshift(C,3))
plot(dates,CIR,'LineWidth',2)
title('{\bf Multiple Cointegrating Relations}')
legend(strcat({'Cointegrating relation  '}, ...
     num2str((1:numPerms)')),'location','NW');
axis tight
grid on

=== Different Engle-Granger tests, same data ===


HEG =

     1     1     0


PValEG =

    0.0202    0.0290    0.0625

Johansen Test for Cointegration

fprintf('\n=== Johansen tests for cointegration ===\n')
[hJ,pValJ] = jcitest(Y,'model','H1','lags',1:2);

=== Johansen tests for cointegration ===

************************
Results Summary (Test 1)

Data: Y
Effective sample size: 39
Model: H1
Lags: 1
Statistic: trace
Significance level: 0.05

r  h  stat      cValue   pValue   eigVal   
========================================
0  1  35.3442   29.7976  0.0104   0.3979  
1  1  15.5568   15.4948  0.0490   0.2757  
2  0  2.9796    3.8415   0.0843   0.0736  

************************
Results Summary (Test 2)

Data: Y
Effective sample size: 38
Model: H1
Lags: 2
Statistic: trace
Significance level: 0.05

r  h  stat      cValue   pValue   eigVal   
========================================
0  0  25.8188   29.7976  0.1346   0.2839  
1  0  13.1267   15.4948  0.1109   0.2377  
2  0  2.8108    3.8415   0.0937   0.0713  

VEC Model Estimation

[~,~,~,~,mles] = jcitest(Y,'model','H1','lags',2,'display','params');

B = mles.r2.paramVals.B % Cointegrating relations with rank = 2 restriction

****************************
Parameter Estimates (Test 1)

r = 0
------
B1 =
   -0.1848    0.5704   -0.3273
    0.0305    0.3143   -0.3448
    0.0964    0.1485   -0.1406

B2 =
   -0.6046    1.6615   -1.3922
   -0.1729    0.4501   -0.4796
   -0.1631    0.5759   -0.5231

c1 =
    0.1420
    0.1517
    0.1508


r = 1
------
A =
   -0.6259
   -0.2261
   -0.0222

B =
    0.7081
    1.6282
   -2.4581

B1 =
    0.0579    1.0824   -0.8718
    0.1182    0.4993   -0.5415
    0.1050    0.1667   -0.1600

B2 =
   -0.5462    2.2436   -1.7723
   -0.1518    0.6605   -0.6169
   -0.1610    0.5966   -0.5366

c0 =
    2.2351

c1 =
   -0.0366
    0.0872
    0.1444


r = 2
------
A =
   -0.6259    0.1379
   -0.2261   -0.0480
   -0.0222    0.0137

B =
    0.7081   -2.4407
    1.6282    6.2883
   -2.4581   -3.5321

B1 =
    0.2438    0.6395   -0.6729
    0.0535    0.6533   -0.6107
    0.1234    0.1228   -0.1403

B2 =
   -0.3857    1.7970   -1.4915
   -0.2076    0.8158   -0.7146
   -0.1451    0.5524   -0.5089

c0 =
    2.0901
   -3.0289

c1 =
   -0.0104
    0.0137
    0.1528


B =

    0.7081   -2.4407
    1.6282    6.2883
   -2.4581   -3.5321

Testing Cointegrating Vectors

fprintf('\n=== Test y1, y2,y3 for stationarity ===\n\n')
[h0J,pVal0J] = jcontest(Y,1,'BVec',{[1 0 0]',[0 1 0]',[0 0 1]'})

=== Test y1, y2,y3 for stationarity ===


h0J =

     1     1     1


pVal0J =

  1.0e-003 *

    0.3368    0.1758    0.1310

Step by step on video

1. ADF test whether the data is non stationary and have unit root

2. Cointegration test Engle Granger

• ← PhD Journey Day 21 - Finally the abstract is submitted
• PhD Journey Day 23 - Still continue in mastering VECM →