Test of Heteroscedasticity

Different available test of heteroscedasticty, Detection of Heteroscedasticity using Graphical techniques will be presented in this category.

The Breusch-Pagan Test (Numerical Example)

To perform the Breusch-Pagan test for the detection of heteroscedasticity, use the data from the following file Table_11.3.

Step 1:

The estimated regression is $\hat{Y}_i = 9.2903 + 0.6378X_i$

Step 2:

The residuals obtained from this regression are:

$\hat{u}_i$$\hat{u}_i^2$$p_i$
-5.3130728.228730.358665
-8.0687665.104940.827201
6.4980142.224070.536485
0.553390.306240.003891
-6.8244546.573180.591743
1.364471.861770.023655
5.7977033.613330.427079
-3.5801512.817440.162854
0.986620.973420.012368
8.3090869.040850.877209
-2.257695.097150.064763
-1.335841.784460.022673
8.0420164.673910.821724
10.47524109.730661.3942
6.2309338.824510.493291
-9.0915382.655881.050197
-12.79183163.630992.079039
-16.84722283.828793.606231
-17.35860301.321043.828481
2.719557.395950.09397
2.397095.746040.073007
0.774940.600520.00763
9.4524889.349301.135241
4.8857123.870140.303286
4.5306320.526580.260804
-0.036140.001311.66E-05
-0.303220.091940.001168
9.5078690.399441.148584
-18.98076360.269094.577455
20.26355410.611595.217089

The estimated $\tilde{\sigma}^2$ is $\frac{\sum u_i^2}{n} = \frac{2361.15325}{30} = 78.7051$.

Compute a new variable $p_i = \frac{\hat{u}_i^2}{\hat{\sigma^2}}$

Step 3:

Assuming $p_i$ is linearly related to $X_i(=Z_i)$ and run the regression of $p_i=\alpha_1+\alpha_2Z_{2i}+v_i$.

The regression Results are: $\hat{p}_i=-0.74261 + 0.010063X_i$

Step 4:

Obtain the Explained Sum of Squares (ESS) = 10.42802.

Step 5:

Compute: $\Theta = \frac{1}{2} ESS = \frac{10.42802}{2}= 5.2140$.

The Breusch-Pagan test follows Chi-Square Distribution. The $\chi^2_{tab}$ value at a 5% level of significance and with ($k-1$) one degree of freedom is 3.8414. The $\chi_{cal}^2$ is greater than $\chi_{tab}^2$, therefore, results are statistically significant. There is evidence of heteroscedasticity at a 5% level of significance.

See More about Breusch-Pagan Test

Bruesch-Pagan-Test-of-Heteroscedasticity

White General Heteroscedasticity Test (Numerical Example)

One important assumption of Regression is that the variance of the Error Term is constant across observations. If the error has a constant variance, then the errors are called homoscedastic, otherwise heteroscedastic. In the case of heteroscedastic errors (non-constant variance), the standard estimation methods become inefficient. Typically, to assess the assumption of homoscedasticity, residuals are plotted.

White general Heteroscedasticity test

We will consider the following data, to test the presence of heteroscedasticity using White General Heteroscedasticity test.

IncomeEducationJob Experience
529
9.7418
28.4821
8.8812
21814
26.61016
25.41216
23.1129
22.51218
19.5125
21.7127
24.8139
30.11412
24.81417
28.51519
26156
38.91617
22.1161
33.11710
48.32117

White General Heteroscedasticity Test

To perform the White General Heteroscedasticity test, the general procedure is

Step 1: Run a regression and obtain $\hat{u}_i$ of this regression equation.

The regression model is: $income = \beta_1+\beta_2\, educ + \beta_3\, jobexp + u_i$

The Regression results are: $Income_i=-7.09686 + 1.93339 educ_{i} + 0.649365 jobexp_{i}$

Step 2: Run the following auxiliary regression

$$\hat{u}_i^2=\alpha_1+\alpha_2X_{2i}+\alpha_3 X_{3i}+\alpha_4 X_{2i}^2+\alpha_5X_{3i}^2+\alpha_6X_{2i}X_{3i}+vi $$

that is, regress the squared residuals on a constant, all the explanatory variables, the squared explanatory variables, and their respective cross-product.

Here in auxiliary regression education, $Y$ is income, $X_2$ is educ, and $X_3$ is jobexp.

The results from auxiliary regression are:

$$Y=42.6145  -0.10872\,X_{2i} – 5.8402\, X_{3i} -0.15273\, X_{2i}^2 + 0.200715\, X_{3i}^2 + 0.226517\,X_{2i}X_{3i}$$

Step 3: Formulate the null and alternative hypotheses

$H_0: \alpha_1=\alpha_2=\cdots=\alpha_p=0$

$H_1$: at least one of the $\alpha$s is different from zero

Step 4: Reject the null and conclude that there is significant evidence of heteroscedasticity when the statistic is bigger than the critical value.

The statistic with computed value is:

$$n \cdot R^2 \, \Rightarrow = 20\times 0.4488 = 8.977$$

The statistics follow asymptotically $\chi^2_{df}$, where $df=k-1$. The Critical value is $\chi^2_5$ at a 5% level of significance is  11.07. 

Since the calculated value is smaller than the tabulated value, therefore, the null hypothesis is accepted. Therefore, based on the White general heteroscedasticity test, there is no heteroscedasticity.

Download the data file: White’s test Related Data

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Park Glejser Test: Numerical Example

To detect the presence of heteroscedasticity using the Park Glejser test, consider the following data.

Year1992199319941995199619971998
Yt37484536255563
Xt4.56.53.532.58.57.5

The step-by-step procedure for conducting the Park Glejser test:

Step 1: Obtain an estimate of the regression equation

$$\hat{Y}_i = 19.8822 + 4.7173X_i$$

Obtain the residuals from this estimated regression equation:

Residuals-4.1103-2.54508.60711.9657-6.6756-4.97977.7377

Take the absolute values of these residuals and consider it as your dependent variables to perform the different functional forms suggested by Glejser.

Step 2: Regress the absolute values of $\hat{u}_i$ on the $X$ variable that is thought to be closely associated with $\sigma_i^2$. We will use the following function forms.

Sr. No.Functional FormResults
1)$|\hat{u}_t| = \beta_1 + \beta_2 X_i +v_i$ $|\hat{u}_i| = 5.2666-0.00681X_i,\quad R^2=0.00004$

$t_{cal} = -0.014$

   
2)$|\hat{u}_t| = \beta_1 + \beta_2 \sqrt{X_i} +v_i$$|\hat{u}_i| = 5.445-0.0962X_i,\quad R^2=0.000389$

$t_{cal} = -0.04414$

   
3)$|\hat{u}_t| = \beta_1 + \beta_2 \frac{1}{X_i} +v_i$$||\hat{u}_i| = 4.9124+1.3571X_i,\quad R^2=0.00332$

$t_{cal} = -0.12914$

   
4)$|\hat{u}_t| = \beta_1 + \beta_2 \frac{1}{\sqrt{X_i}} +v_i$

$\hat{u}_i| = 4.7375+1.0428X_i,\quad R^2=0.00209$ $t_{cal} = 0.10252$

Since none of the residual regression is significant, therefore, the hypothesis of heteroscedasticity is rejected. Therefore, we can say that there is no relationship between the absolute value of the residuals ($u_i$) and the explanatory variable $X$.

Error Variance is Proportional to Xi: Park Glejser Test

How to perform White General Heteroscedasticity?

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