cusumtest

Conduct a cusum test to assess whether there is a structural break in the equation for food demand. Input the predictor series as a matrix and input the response series as a vector.

Load the US food consumption data set Data_Consumption.mat, which contains annual measurements from 1927 through 1962 with missing data due to World War II in the matrix Data.

load Data_Consumption

Suppose that you want to develop a model for consumption as determined by food prices and disposable income, and assess its stability through the economic shock through the war.

Plot the series.

P = Data(:,1); % Food price index
I = Data(:,2); % Disposable income index
Q = Data(:,3); % Food consumption index

figure;
plot(dates,[P I Q])
axis tight
grid on
xlabel("Year")
ylabel("Index")
legend(["Price" "Income" "Consumption"],Location="southeast")

Measurements are missing from 1942 through 1947, which correspond to WWII.

Stabilize each series by applying the log transformation.

LP = log(P);
LI = log(I);
LQ = log(Q);

Assume that log consumption is a linear function of the logs of food price and income.

$${\epsilon }_t$$ is a Gaussian random variable with mean 0 and standard deviation $${\sigma }^2$$.

Identify the indices before WWII. Plot log consumption with respect to the logs of food price and income.

preWarIdx = (dates <= 1941);

figure
scatter3(LP(preWarIdx),LI(preWarIdx),LQ(preWarIdx),[],"ro");
hold on
scatter3(LP(~preWarIdx),LI(~preWarIdx),LQ(~preWarIdx),[],"b*");
legend(["Pre-war observations" "Post-war observations"], ...
    Location="best")
xlabel("Log Price")
ylabel("Log Income")
zlabel("Log Consumption")

% Obtain better view
h = gca;
h.CameraPosition = [4.3 -12.2 5.3];

Data relationships appear to be affected by the war.

Conduct a cusum test to assess whether there is a significant structural change. Use default values.

X = [LP LI];
y = LQ;
h = cusumtest(X,y)

h = logical
   0

h = 0 indicates that there is not enough evidence to reject the null hypothesis that the coefficients are equal across subsamples.

Conduct a cusum test to assess whether there is a structural change in the equation for food demand, where the time series are variables in a table.

Load the US food consumption data set Data_Consumption.mat, which contains annual measurements from 1927 through 1962 with missing data due to World War II in the table DataTable. Convert the table to a timetable, and remove rows containing missing values.

load Data_Consumption
dates = datetime(dates,12,31);
TT = table2timetable(DataTable,RowTimes=dates);
TT.Row = [];
TT = rmmissing(TT);

Apply the log transform to all variables in the table.

LogTT = varfun(@log,TT);
LogTT.Properties.VariableNames

ans = 1x3 cell
    {'log_P'}    {'log_I'}    {'log_Q'}

Conduct a cusum test to assess whether there is a structural change in the regression model of log food consumption log_Q on log price log_P and log income log_I.

h = cusumtest(LogTT)

h = logical
   0

By default, chowtest selects the last table variable as the response, and selects all other variables as predictors. You can select a different variable by using the ResponseVariable name-value argument, and you can choose a different set of predictor variables by using the PredictorVariables name-value argument.

Load the US food consumption data set Data_Consumption.mat. Consider a model for log food consumption as determined by log food prices and log disposable income.

load Data_Consumption
LogDT = varfun(@log,DataTable);
numObs = height(LogDT) - sum(any(ismissing(LogDT),2))

numObs = 
30

numPreds = width(LogDT) - 1

numPreds = 
2

Conduct a cusum test using default values. Return all test decision statistics.

[h,H,Stat,W] = cusumtest(LogDT)

h = logical
   0

H=1×28 table
               H1       H2       H3       H4       H5       H6       H7       H8       H9       H10      H11      H12      H13      H14      H15      H16      H17      H18      H19      H20      H21      H22      H23      H24      H25      H26      H27      H28 
              _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____    _____

    Test 1    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false    false

Stat=1×28 table
              Stat1      Stat2       Stat3       Stat4       Stat5       Stat6       Stat7      Stat8      Stat9     Stat10     Stat11     Stat12      Stat13     Stat14    Stat15    Stat16    Stat17    Stat18    Stat19    Stat20    Stat21    Stat22    Stat23    Stat24    Stat25    Stat26      Stat27      Stat28 
              _____    _________    ________    ________    ________    ________    _______    _______    _______    _______    ______    ________    ________    ______    ______    ______    ______    ______    ______    ______    ______    ______    ______    ______    ______    _______    ________    ________

    Test 1     NaN     -0.012438    0.064511    -0.50784    -0.55747    -0.42687    -2.7881    -3.0973    -3.7625    -3.5417    -1.913    -0.65794    -0.35743    2.3762    3.3104    3.7509    2.8851    3.7395    4.2295    4.641     4.2412    4.496     3.2467    2.0001    1.5324    0.81729    0.053352    -0.98812

W=1×28 table
              W1         W2             W3            W4            W5            W6           W7            W8            W9           W10         W11         W12         W13         W14          W15          W16          W17           W18          W19          W20          W21           W22          W23          W24          W25           W26           W27           W28   
              ___    ___________    __________    __________    ___________    _________    _________    __________    __________    _________    ________    ________    ________    ________    _________    _________    __________    _________    _________    _________    __________    _________    _________    _________    __________    __________    __________    _________

    Test 1    NaN    -0.00012823    0.00079327    -0.0059004    -0.00051169    0.0013464    -0.024342    -0.0031882    -0.0068572    0.0022762    0.016791    0.012938    0.003098    0.028181    0.0096305    0.0045417    -0.0089262    0.0088086    0.0050515    0.0042415    -0.0041209    0.0026269    -0.012879    -0.012851    -0.0048219    -0.0073721    -0.0078755    -0.010737

cusumtest returns the overall rejection decision h, and tables containing the sequence of rejection decision from the forward recursions of the cusum test H, the corresponding sequence of test statistics Stat, and the corresponding recursive residuals W. Each table contains numObs - numPreds + Intercept = 28 variables corresponding to results of each recursion in the cusum test. Stat1 = W1 = NaN indicates the presence of a model intercept.

Determine whether an explanatory model of real gross national product (RGNP) is stable by plotting recursive residuals.

Load the Nelson-Plosser data set Data_NelsonPlosser.mat, which contains the table of data DataTable.

load Data_NelsonPlosser

The time series in the data set contain annual, macroeconomic measurements from 1860 to 1970. For more details, a list of variables, and descriptions, enter Description in the command line.

Convert the table to a timetable. Focus the sample to measurements from the end of 1915 through the end of 1970.

dates = datetime(dates,12,31);
span = isbetween(dates,datetime(1915,12,31),datetime(1970,12,31),"closed");
TT = table2timetable(DataTable,RowTimes=dates);
TT.Dates = [];
TT = TT(span,:);

Consider a predictive model of the US RGNP GNPR given measurements of the industrial production index IPI, total employment E, and real wages WR.

Plot the series in the model.

prednames = ["IPI" "E" "WR"];

tiledlayout(2,2)
for j = ["GNPR" prednames]
    nexttile
    plot(TT.Time,TT{:,j})
    ylabel(j)
end

To address exponential growth, apply the log transform to the series.

LogTT = varfun(@log,TT);

LogTT is a timetable containing the transformed variables in TT, but with names prepended with log_.

Assume that an appropriate multiple regression model to describe real GNP is

Conduct a cusum test to assess whether all regression coefficients are stable. Print a test summary to the command line. Plot the test statistics.

lprednames = "log_" + prednames;
cusumtest(LogTT,ResponseVariable="log_GNPR", ...
    PredictorVariables=lprednames,Display="summary")

RESULTS SUMMARY

***************
Test 1

Test type: cusum
Test direction: forward
Intercept: yes
Number of iterations: 52

Decision: Fail to reject coefficient stability
Significance level: 0.0500

ans = logical
   0

The cusum series does not cross the critical lines, which indicates model stability.

Conduct cusum tests to assess whether there are structural changes in the equation for food demand around World War II. Implement forward and backward recursive regressions to obtain the test statistics.

Load the US food consumption data set Data_Consumption.mat, which contains annual measurements from 1927 through 1962 with missing data due to the World War II in the table DataTable. Convert the table to a timetable, and remove rows containing missing values.

load Data_Consumption
dates = datetime(dates,12,31);
TT = table2timetable(DataTable,RowTimes=dates);
TT.Row = [];
TT = rmmissing(TT);

Consider a model for log food consumption as determined by log food prices and log disposable income, and assess its stability through the economic shock through the war.

Apply the log transform to all variables in the table.

LogTT = varfun(@log,TT);

Conduct forward and backward cusum tests using a 5% level of significance for each test. Plot the cusums. Return the recursive residuals.

[h,~,~,W] = cusumtest(LogTT,Direction=["forward" "backward"], ...
    Plot="on");

RESULTS SUMMARY

***************
Test 1

Test type: cusum
Test direction: forward
Intercept: yes
Number of iterations: 27

Decision: Fail to reject coefficient stability
Significance level: 0.0500

***************
Test 2

Test type: cusum
Test direction: backward
Intercept: yes
Number of iterations: 27

Decision: Fail to reject coefficient stability
Significance level: 0.0500

The plots and test results at the command line indicate that neither test rejects the null hypothesis that coefficients are stable.

Compare the results of the cusum tests with the results of a Chow test. Unlike cusum tests, Chow tests require a guess for the time point at which the structural break occurs. Specify that the break point is 1941.

bp = find(LogTT.Time >= datetime(1941,12,31),1);
chowtest(LogTT,bp,Display="summary");

RESULTS SUMMARY

***************
Test 1

Sample size: 30
Breakpoint: 15

Test type: breakpoint
Coefficients tested: All

Statistic: 5.5400
Critical value: 3.0088

P value: 0.0049
Significance level: 0.0500

Decision: Reject coefficient stability

The test results reject the null hypothesis that the coefficients are stable.

The Chow and cusum test results are not consistent. For details on cusum test limitations, see Limitations.

Check whether a cusum of squares test can detect a structural break in volatility in simulated data.

Simulate a series of data from this regression model

$$x_t$$ is a series of observations from three standard Gaussian predictor variables. $${\epsilon }_{1t}$$ and $${\epsilon }_{2t}$$ are series of Gaussian innovations both with mean 0 and standard deviation 0.1 and 0.2, respectively.

rng(1); % For reproducibility
T = 100;
X = randn(T,3);
sigma1 = 0.1;
sigma2 = 0.2;
e = [sigma1*randn(T/2,1); sigma2*randn(T/2,1)];
b = (1:3)';
y = X*b + e;

Conduct a cusum of squares test using a 5% level of significance. Plot the test statistics and critical region bands. Indicate that there is no model intercept. Request to return whether the test statistics cross into critical region at each iteration.

[~,H] = cusumtest(X,y,Test="cusumsq",Plot="on", ...
    Direction=["forward" "backward"],Display="off", ...
    Intercept=false);

Because the test statistics cross the critical lines at least once for both tests, the tests reject the null hypothesis of constant volatility at 5% level. The test statistics change direction around iteration 50, which is consistent with the simulated break in volatility in the data.

H is a 2-by-97 logical matrix containing the sequence of decisions for each iteration of each cusum of squares test. The first row corresponds to the forward cusum of squares test, and the second row corresponds to the backward cusum of squares test.

For the forward test, determine the iterations that result in the test statistics crossing the critical line.

bp = find(H(1,:) == 1)

bp = 1×35

    24    25    26    27    28    29    30    31    32    33    34    35    36    37    38    39    40    41    42    43    44    45    46    47    48    49    50    51    52    53    54    55    56    57    58