Heteroskedasticity

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We wrapped up our review.

Today

Heteroskedasticity

Schedule

This week

First assignment!

Submit one file (HTML, PDF, or Word) that includes

1. Your written answers
2. Figures and regression output
3. The R code that generated your answers.

This file should be a rendered RMarkdown or Quarto file (.rmd or .qmd).

Important

• We should be able to easily find your answers for each question.
• Do not copy. (You will receive a zero.)

Prerequisite material

The \(\chi^2\) distribution

Some test statistics are distributed as \(\chi^2\) random variables.

The \(\chi^2\) distribution is just another example of a common (named) distribution (like the Normal distribution, the \(t\) distribution, and the \(F\)).

The shape of the \(\chi^2\) distribution depends on a single parameter: - We will call this parameter \(k\) - Our test statistics will refer to \(k\) as degrees of freedom.

Prerequisite material

The \(\chi^2\) distribution

Three examples of \(\chi_k^2\): \(\textcolor{#314f4f}{k = 1}\), \(\textcolor{#e64173}{k = 2}\), and \(\textcolor{orange}{k = 9}\)

Prerequisite material

The \(\chi^2\) distribution

Probability of observing a more extreme test statistic \(\widehat{\text{LM}}\) under H0

Heteroskedasticity

Let’s write down our current assumptions

1. Our sample (the \(x_k\)’s and \(y_i\)) was randomly drawn from the population.

2. \(y\) is a linear function of the \(\beta_k\)’s and \(u_i\).

3. There is no perfect multicollinearity in our sample.

4. The explanatory variables are exogenous: \(\mathop{\boldsymbol{E}}\left[ u \middle| X \right] = 0 \left(\implies \mathop{\boldsymbol{E}}\left[ u \right] = 0\right)\).

5. The disurbances have constant variance \(\sigma^2\) and zero covariance, i.e.,

   • \(\mathop{\boldsymbol{E}}\left[ u_i^2 \middle| X \right] = \mathop{\text{Var}} \left( u_i \middle| X \right) = \sigma^2 \implies \mathop{\text{Var}} \left( u_i \right) = \sigma^2\)
   • \(\mathop{\text{Cov}} \left( u_i, \, u_j \middle| X \right) = \mathop{\boldsymbol{E}}\left[ u_i u_j \middle| X \right] = 0\) for \(i\neq j\)

6. The disturbances come from a Normal distribution, i.e., \(u_i \overset{\text{iid}}{\sim} \mathop{\text{N}}\left( 0, \sigma^2 \right)\).

Heteroskedasticity

Today we’re focusing on assumption #5:

5. The disurbances have constant variance \(\sigma^2\) and zero covariance, i.e.,

• \(\mathop{\boldsymbol{E}}\left[ u_i^2 \middle| X \right] = \mathop{\text{Var}} \left( u_i \middle| X \right) = \sigma^2 \implies \mathop{\text{Var}} \left( u_i \right) = \sigma^2\)
• \(\mathop{\text{Cov}} \left( u_i, \, u_j \middle| X \right) = \mathop{\boldsymbol{E}}\left[ u_i u_j \middle| X \right] = 0\) for \(i\neq j\)

Heteroskedasticity

Classic example of heteroskedasticity: The funnel

Variance of \(u\) increases with \(x\)

Heteroskedasticity

Another example of heteroskedasticity: (double funnel?)

Variance of \(u\) increasing at the extremes of \(x\)

Heteroskedasticity

Another example of heteroskedasticity:

Differing variances of \(u\) by group

Heteroskedasticity

What and why?

Heteroskedasticity is present when the variance of \(u\) changes with any combination of our explanatory variables \(x_1\), through \(x_k\) (henceforth: \(X\)).

Heteroskedasticity

Consequences

So what are the consquences of heteroskedasticity? Bias? Inefficiency?

First, let’s check if it has consquences for the the unbiasedness of OLS.

Heteroskedasticity

Proof: Assuming \(y_i = \beta_0 + \beta_1 x_i + u_i\)

\[ \begin{aligned} \hat{\beta}_1 &= \dfrac{\sum_i\left(y_i-\overline{y}\right)\left(x_i-\overline{x}\right)}{\sum_i\left(x_i -\overline{x}\right)^2} \\ &= \dfrac{\sum_i\left(\left[ \beta_0 + \beta_1 x_i + u_i \right]- \left[ \beta_0 + \beta_1 \overline{x} + \overline{u} \right] \right)\left(x_i-\overline{x}\right)}{\sum_i\left(x_i -\overline{x}\right)^2} \\ &= \dfrac{\sum_i\left(\beta_1 \left[ x_i - \overline{x} \right] + \left[u_i - \overline{u}\right] \right)\left(x_i-\overline{x}\right)}{\sum_i\left(x_i -\overline{x}\right)^2} \\ &= \dfrac{\sum_i\left(\beta_1 \left[ x_i - \overline{x} \right]^2 + \left[ x_i - \overline{x} \right] \left[u_i - \overline{u}\right]\right)}{\sum_i\left(x_i -\overline{x}\right)^2} \\ &= \beta_1 + \dfrac{\sum_i\left(x_i - \overline{x}\right) \left(u_i - \overline{u}\right)}{\sum_i\left(x_i -\overline{x}\right)^2} \end{aligned} \]

Heteroskedasticity

\[ \begin{aligned} \hat{\beta}_1 &= \cdots = \beta_1 + \dfrac{\sum_i\left(x_i - \overline{x}\right) \left(u_i - \overline{u}\right)}{\sum_i\left(x_i -\overline{x}\right)^2} \\ &= \beta_1 + \dfrac{\sum_i\left(x_i - \overline{x}\right) u_i - \overline{u} \sum_i\left(x_i - \overline{x}\right)}{\sum_i\left(x_i -\overline{x}\right)^2} \\ &= \beta_1 + \dfrac{\sum_i\left(x_i - \overline{x}\right) u_i - \overline{u} \left(\sum_i x_i - \sum_i \overline{x}\right)}{\sum_i\left(x_i -\overline{x}\right)^2} \\ &= \beta_1 + \dfrac{\sum_i\left(x_i - \overline{x}\right) u_i - \overline{u} \left(\sum_i x_i - n \overline{x}\right)}{\sum_i\left(x_i -\overline{x}\right)^2} \\ &= \beta_1 + \dfrac{\sum_i\left(x_i - \overline{x}\right) u_i - \overline{u} \textcolor{#e64173}{\left(\sum_i x_i - \sum_i x_i\right)}}{\sum_i\left(x_i -\overline{x}\right)^2} \\ &= \beta_1 + \dfrac{\sum_i\left(x_i - \overline{x}\right) u_i}{\sum_i\left(x_i -\overline{x}\right)^2} \quad \text{😅} \end{aligned} \]

Heteroskedasticity

Consequences: Bias

We now want to see if heteroskedasticity biases the OLS estimator for \(\beta_1\).

Heteroskedasticity

Consequences: Efficiency

In the presence of heteroskedasticity, OLS is no longer the most efficient (best) linear unbiased estimator.

Heteroskedasticity

Consequences: Inference

OLS standard errors are biased in the presence of heteroskedasticity.

• Wrong confidence intervals

• Problems for hypothesis testing (both \(t\) and \(F\) tests)

• It’s hard to learn much without sound inference.

Heteroskedasticity

Solutions

1. Tests to determine whether heteroskedasticity is present.

2. Remedies for (1) efficiency and (2) inference

Testing for heteroskedasticity

While we might have solutions for heteroskedasticity, the efficiency of our estimators depends upon whether or not heteroskedasticity is present.

1. The Goldfeld-Quandt test

2. The White test

Testing for heteroskedasticity

The Goldfeld-Quandt test

Focuses on a specific type of heteroskedasticity: whether the variance of \(u_i\) differs between two groups.¹

Remember how we used our residuals to estimate the \(\sigma^2\)?

\[ s^2 = \dfrac{\text{SSE}}{n-1} = \dfrac{\sum_i e_i^2}{n-1} \]

We will use this same idea to determine whether there is evidence that our two groups differ in the variances of their disturbances, effectively comparing \(s^2_1\) and \(s^2_2\) from our two groups.

Testing for heteroskedasticity

The Goldfeld-Quandt test

Operationally,

Testing for heteroskedasticity

The Goldfeld-Quandt test

The G-Q test statistic

\[ F_{\left(n^{\star}-k,\, n^{\star}-k\right)} = \dfrac{\text{SSE}_2/(n^\star-k)}{\text{SSE}_1/(n^\star-k)} = \dfrac{\text{SSE}_2}{\text{SSE}_1} \]

follows an \(F\) distribution (under the null hypothesis) with \(n^{\star}-k\) and \(n^{\star}-k\) degrees of freedom.¹

Testing for heteroskedasticity

The Goldfeld-Quandt test

Testing for heteroskedasticity

The Goldfeld-Quandt test

\(F_{375,\,375} = \dfrac{\textcolor{#e64173}{\text{SSE}_2 = 18,203.4}}{\textcolor{#314f4f}{\text{SSE}_1 = 1,039.5}} \approx 17.5 \implies\) p-value \(< 0.001\)

\(\therefore\) We reject H0: \(\sigma^2_1 = \sigma^2_2\) and conclude there is statistically significant evidence of heteroskedasticity.

Testing for heteroskedasticity

The Goldfeld-Quandt test

The problem…

Testing for heteroskedasticity

The Goldfeld-Quandt test

Testing for heteroskedasticity

The Goldfeld-Quandt test

\(F_{375,\,375} = \dfrac{\textcolor{#e64173}{\text{SSE}_2 = 14,516.8}}{\textcolor{#314f4f}{\text{SSE}_1 = 14,937.1}} \approx 1 \implies\) p-value \(\approx 0.609\)

\(\therefore\) We fail to reject H0: \(\sigma^2_1 = \sigma^2_2\) while heteroskedasticity is present.

Testing for heteroskedasticity

The White test

Breusch and Pagan (1981) attempted to solve this issue of being too specific with the functional form of the heteroskedasticity.

• Regress \(e_i^2\) on \(X = \left[ 1,\, x_1,\, x_2,\, \ldots,\, x_k \right]\) and test for joint significance.
• Allows the data to show if/how the variance of \(u_i\) correlates with \(X\).
• If \(\sigma_i^2\) correlates with \(X\), then we have heteroskedasticity.

Testing for heteroskedasticity

The White test

Toward this goal, Hal White took advantage of the fact that we can replace the homoskedasticity requirement with a weaker assumption:

• Old: \(\mathop{\text{Var}} \left( u_i \middle| X \right) = \sigma^2\)

• New: \(u^2\) is uncorrelated with the explanatory variables (i.e., \(x_j\) for all \(j\)), their squares (i.e., \(x_j^2\)), and the first-degree interactions (i.e., \(x_j x_h\)).

Testing for heteroskedasticity

The White test

An outline of White’s test for heteroskedasticity:

Testing for heteroskedasticity

The White test

White’s test statistic is

\[ \text{LM} = n \times R_e^2 \qquad \text{Under H}_0,\, \text{LM} \overset{\text{d}}{\sim} \chi_k^2 \]

where \(R^2_e\) comes from the regression of \(e^2\) on the explanatory variables, their squares, and their interactions.

\[ e^2 = \alpha_0 + \underbrace{\sum_{h=1}^k \alpha_h x_h}_{\text{Expl. variables}} + \underbrace{\sum_{j=1}^k \alpha_{k+j} x_j^2}_{\text{Squared terms}} + \underbrace{\sum_{\ell = 1}^{k-1} \sum_{m = \ell + 1}^k \alpha_{\ell,m} x_\ell x_m}_{\text{Interactions}} + v_i \]

Note: The \(k\) (for our \(\chi_k^2\)) equals the number of estimated parameters in the regression above (the \(\alpha_j\)), excluding the intercept \(\left( \alpha_0 \right)\).

Testing for heteroskedasticity

The White test

Practical note: If a variable is equal to its square (e.g., binary variables), then you don’t (can’t) include it. The same rule applies for interactions.

Testing for heteroskedasticity

The White test

Example: Consider the model¹ \(y = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + u\)

Step 1: Estimate the model; obtain residuals \((e)\).

Step 2: Regress \(e^2\) on explanatory variables, squares, and interactions. \[ \begin{aligned} e^2 = &\alpha_0 + \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 + \alpha_4 x_1^2 + \alpha_5 x_2^2 + \alpha_6 x_3^2 \\ &+ \alpha_7 x_1 x_2 + \alpha_8 x_1 x_3 + \alpha_9 x_2 x_3 + v \end{aligned} \]

Record the R2 from this equation (call it \(R_e^2\)).

Step 3: Test H0: \(\alpha_1 = \alpha_2 = \cdots = \alpha_9 = 0\) using \(\text{LM} = n R^2_e \overset{\text{d}}{\sim} \chi_9^2\).

Testing for heteroskedasticity

The White test

The White test for this simple linear regression.

\[ \begin{aligned} e_i^2 &= \hat{\alpha}_0 + \hat{\alpha}_1 x_{1i} \textcolor{#e64173}{+ \hat{\alpha}_2 x^2_{1i}} & \widehat{\text{LM}} &= 185.8 &\mathit{p}\text{-value} < 0.001 \end{aligned} \]

Testing for heteroskedasticity

Examples

Goal: Estimate the relationship between standardized test scores (outcome variable) and (1) student-teacher ratio and (2) income, i.e.,

\[ \left(\text{Test score}\right)_i = \beta_0 + \beta_1 \text{Ratio}_{i} + \beta_2 \text{Income}_{i} + u_i \tag{1} \]

Potential issue: Heteroskedasticity… and we do not observe \(u_i\).

Solution:

1. Estimate the relationship in \((1)\) using OLS.
2. Test for heteroskedasticity.

• Goldfeld-Quandt
• White

Testing for heteroskedasticity

Examples

We will use testing data from the dataset Caschool in the Ecdat R package.

# Load packages
library(pacman)
p_load(tidyverse, Ecdat)
# Select and rename desired variables; assign to new dataset
test_df = select(Caschool, test_score = testscr, ratio = str, income = avginc)
# Format as tibble
test_df = as_tibble(test_df)
# View first 2 rows of the dataset
head(test_df, 2)

#> # A tibble: 2 × 3
#>   test_score ratio income
#>        <dbl> <dbl>  <dbl>
#> 1       691.  17.9  22.7 
#> 2       661.  21.5   9.82

Testing for heteroskedasticity

Examples

Let’s begin by estimating our model

\[ \left(\text{Test score}\right)_i = \beta_0 + \beta_1 \text{Ratio}_{i} + \beta_2 \text{Income}_{i} + u_i \]

# Estimate the model
est_model = lm(test_score ~ ratio + income, data = test_df)
# Summary of the estimate
tidy(est_model)

#> # A tibble: 3 × 5
#>   term        estimate std.error statistic   p.value
#>   <chr>          <dbl>     <dbl>     <dbl>     <dbl>
#> 1 (Intercept)  639.       7.45       85.7  5.70e-267
#> 2 ratio         -0.649    0.354      -1.83 6.79e-  2
#> 3 income         1.84     0.0928     19.8  4.38e- 62

Testing for heteroskedasticity

Examples

Now, let’s see what the residuals suggest about heteroskedasticity

# Add the residuals to our dataset
test_df$e = residuals(est_model)

Testing for heteroskedasticity

Example: Goldfeld-Quandt

Income looks potentially heteroskedastic; let’s test via Goldfeld-Quandt.

# Arrange the data by income
test_df = arrange(test_df, income)

Testing for heteroskedasticity

Example: Goldfeld-Quandt

Income looks potentially heteroskedastic; let’s test via Goldfeld-Quandt.

# Arrange the data by income
test_df = arrange(test_df, income)
# Re-estimate the model for the last and first 158 observations
est_model1 = lm(test_score ~ ratio + income, data = tail(test_df, 158))
est_model2 = lm(test_score ~ ratio + income, data = head(test_df, 158))

Testing for heteroskedasticity

Example: Goldfeld-Quandt

Income looks potentially heteroskedastic; let’s test via Goldfeld-Quandt.

# Arrange the data by income
test_df = arrange(test_df, income)
# Re-estimate the model for the last and first 158 observations
est_model1 = lm(test_score ~ ratio + income, data = tail(test_df, 158))
est_model2 = lm(test_score ~ ratio + income, data = head(test_df, 158))
# Grab the residuals from each regression
e_model1 = residuals(est_model1)
e_model2 = residuals(est_model2)

Testing for heteroskedasticity

Example: Goldfeld-Quandt

Income looks potentially heteroskedastic; let’s test via Goldfeld-Quandt.

# Arrange the data by income
test_df = arrange(test_df, income)
# Re-estimate the model for the last and first 158 observations
est_model1 = lm(test_score ~ ratio + income, data = tail(test_df, 158))
est_model2 = lm(test_score ~ ratio + income, data = head(test_df, 158))
# Grab the residuals from each regression
e_model1 = residuals(est_model1)
e_model2 = residuals(est_model2)
# Calculate SSE for each regression
(sse_model1 = sum(e_model1^2))

#> [1] 19305.01

(sse_model2 = sum(e_model2^2))

#> [1] 29537.83

Testing for heteroskedasticity

Example: Goldfeld-Quandt

Remember the Goldfeld-Quandt test statistic?

\(F_{n^\star-k,\,n^\star-k} = \dfrac{\text{SSE}_2}{\text{SSE}_1}\) \(\approx\dfrac{29,537.83}{19,305.01}\) \(\approx1.53\)

# G-Q test statistic
(f_gq = sse_model2/sse_model1)

#> [1] 1.530061

# p-value
pf(q = f_gq, df1 = 158-3, df2 = 158-3, lower.tail = F)

#> [1] 0.004226666

Testing for heteroskedasticity

Example: Goldfeld-Quandt

The Goldfeld-Quandt test statistic and its null distribution

Testing for heteroskedasticity

Example: Goldfeld-Quandt

Putting it all together:

H0: \(\sigma_1^2 = \sigma_2^2\) vs. HA: \(\sigma_1^2 \neq \sigma_2^2\)

Goldfeld-Quandt test statistic: \(F \approx 1.53\)

p-value \(\approx 0.00423\)

∴ Reject H0 (p-value is less than 0.05).

Conclusion: We find statistically significant evidence that \(\sigma_1^2 \neq \sigma_2^2\).
There is statistically significant evidence of heteroskedasticity (at the 5% level).

Testing for heteroskedasticity

Example: Goldfeld-Quandt

What if we had chosen to focus on student-to-teacher ratio?

# Arrange the data by ratio
test_df = arrange(test_df, ratio)
# Re-estimate the model for the last and first 158 observations
est_model3 = lm(test_score ~ ratio + income, data = tail(test_df, 158))
est_model4 = lm(test_score ~ ratio + income, data = head(test_df, 158))
# Grab the residuals from each regression
e_model3 = residuals(est_model3)
e_model4 = residuals(est_model4)
# Calculate SSE for each regression
(sse_model3 = sum(e_model3^2))

#> [1] 26243.52

(sse_model4 = sum(e_model4^2))

#> [1] 29101.52

Testing for heteroskedasticity

Example: Goldfeld-Quandt

\(F_{n^\star-k,\,n^\star-k} = \dfrac{\text{SSE}_4}{\text{SSE}_3} \approx\dfrac{29,101.52}{26,243.52} \approx1.11\)

which has a p-value of approximately 0.2603.

Heteroskedasticity

Example: White

Let’s test the same model and data with the White test.

Recall: We saved our residuals as e in our dataset, i.e.,

# Estimate the model
est_model = lm(test_score ~ ratio + income, data = test_df)
# Add the residuals to our dataset
test_df$e = residuals(est_model)

Heteroskedasticity

Example: White

The White test adds squared terms and interactions to initial regression specification (the right-hand side) \[ \begin{align} \textcolor{#314f4f}{u_{i}^2} =& \textcolor{#314f4f}{\alpha_{0} + \alpha_{1} \text{Ratio}_{i} + \alpha_{2} \text{Income}_{i} } \\ &+ \textcolor{#e64173}{\alpha_{3} \text{Ratio}_{i}^2 + \alpha_{4} \text{Income}_{i}^2 + \alpha_{5} \text{Ratio}_{i}\times\text{Income}_{i}} \\ &+ \textcolor{#314f4f}{w_{i}} \end{align} \]

The White test tests the null hypothesis
H0: \(\textcolor{#e64173}{\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = \alpha_5 = 0}\)

We just need to write some R code to test H0.

Heteroskedasticity

Example: White

Aside: R has funky notation for squared terms and interactions in lm():

• Squared terms use I(), e.g., lm(y ~ I(x^2))

• Interactions use : between the variables, e.g., lm(y ~ x1:x2)

Example: Regress y on quadratic of x1 and x2:

# Pretend quadratic regression w/ interactions
lm(y ~ x1 + x2 + I(x1^2) + I(x2^2) + x1:x2, data = pretend_df)

Heteroskedasticity

Example: White

Step 1: Regress \(e_i^2\) on 1st degree, 2nd degree, and interactions

# Regress squared residuals on quadratic of explanatory variables
white_model = lm(
  I(e^2) ~ ratio + income + I(ratio^2) + I(income^2) + ratio:income,
  data = test_df
)
# Grab the R-squared
(white_r2 = summary(white_model)$r.squared)

Heteroskedasticity

Example: White

Step 2: Collect \(R_e^2\) from the regression.

# Regress squared residuals on quadratic of explanatory variables
white_model = lm(
  I(e^2) ~ ratio + income + I(ratio^2) + I(income^2) + ratio:income,
  data = test_df
)
# Grab the R-squared
(white_r2 = summary(white_model)$r.squared)

#> [1] 0.07332222

Heteroskedasticity

Example: White

Step 3: Calculate White test statistic \(\text{LM} = n \times R_e^2 \approx 420 \times 0.073\)

# Regress squared residuals on quadratic of explanatory variables
white_model = lm(
  I(e^2) ~ ratio + income + I(ratio^2) + I(income^2) + ratio:income,
  data = test_df
)
# Grab the R-squared
white_r2 = summary(white_model)$r.squared
# Calculate the White test statistic
(white_stat = 420 * white_r2)

#> [1] 30.79533

Heteroskedasticity

Example: White

Step 4: Calculate the associated p-value (where \(\text{LM} \overset{d}{\sim} \chi_k^2\)); here, \(k=5\)

# Regress squared residuals on quadratic of explanatory variables
white_model = lm(
  I(e^2) ~ ratio + income + I(ratio^2) + I(income^2) + ratio:income,
  data = test_df
)
# Grab the R-squared
white_r2 = summary(white_model)$r.squared
# Calculate the White test statistic
white_stat = 420 * white_r2
# Calculate the p-value
pchisq(q = white_stat, df = 5, lower.tail = F)

#> [1] 1.028039e-05

Heteroskedasticity

Example: White

Putting everything together…

H0: \(\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = \alpha_5 = 0\) vs. HA: \(\alpha_i \neq 0\) for some \(i \in \{1,\, 2,\,\ldots,\, 5\}\)

Heteroskedasticity

Example: White

The White test statistic and its null distribution

Heteroskedasticity

Review questions

Heteroskedasticity

Review questions

• Q: What is the definition of heteroskedasticity?

Heteroskedasticity

Review questions

• Q: Why are we concerned about heteroskedasticity?

Heteroskedasticity

Review questions

• Q: Does plotting \(y\) against \(x\), tell us anything about heteroskedasticity?

Heteroskedasticity

Review questions

• Q: Does plotting \(e\) against \(x\), tell us anything about heteroskedasticity?

Heteroskedasticity

Review questions

• Q: Since we cannot observe the \(u_i\)’s, what do we use to learn about heteroskedasticity?

Heteroskedasticity

Review questions

• Q: Which test do you recommend to test for heteroskedasticity? Why?

Testing for heteroskedasticity

The Breusch-Pagan test

Breusch and Pagan (1981) attempted to solve this issue of being too specific with the functional form of the heteroskedasticity.

• Allows the data to show if/how the variance of \(u_i\) correlates with \(X\).

• If \(\sigma_i^2\) correlates with \(X\), then we have heteroskedasticity.

• Regresses \(e_i^2\) on \(X = \left[ 1,\, x_1,\, x_2,\, \ldots,\, x_k \right]\) and tests for joint significance.

Testing for heteroskedasticity

The Breusch-Pagan test

How to implement:

Testing for heteroskedasticity

The Breusch-Pagan test

The B-P test statistic¹ is

\[ \text{LM} = n \times R^2_{e} \]

where \(R^2_e\) is the \(R^2\) from the regression

\[ e_i^2 = \alpha_0 + \alpha_1 x_{1i} + \alpha_2 x_{2i} + \cdots + \alpha_k x_{ki} + v_i \]

Under the null, \(\text{LM}\) is asymptotically distributed as \(\chi^2_k\).

Testing for heteroskedasticity

The Breusch-Pagan test

Problem: We’re still assuming a fairly restrictive functional form between our explanatory variables \(X\) and the variances of our disturbances \(\sigma^2_i\).

Testing for heteroskedasticity

The Breusch-Pagan test

Breusch-Pagan tests are still sensitive to functional form.

\[ \begin{aligned} e_i^2 &= \hat{\alpha}_0 + \hat{\alpha}_1 x_{1i} & \widehat{\text{LM}} &= 1.26 &\mathit{p}\text{-value} \approx 0.261 \\ e_i^2 &= \hat{\alpha}_0 + \hat{\alpha}_1 x_{1i} \textcolor{#e64173}{+ \hat{\alpha}_2 x^2_{1i}} & \widehat{\text{LM}} &= 185.8 &\mathit{p}\text{-value} < 0.001 \end{aligned} \]

Testing for heteroskedasticity

Example: Breusch-Pagan

Let’s test the same model with the Breusch Pagan.

Recall: We saved our residuals as e in our dataset, i.e.,

test_df$e = residuals(est_model)

Testing for heteroskedasticity

Example: Breusch-Pagan

In B-P, we first regress \(e_i^2\) on the explanatory variables,

# Regress squared residuals on explanatory variables
bp_model = lm(I(e^2) ~ ratio + income, data = test_df)

Testing for heteroskedasticity

Example: Breusch-Pagan

and use the resulting \(R^2\) to calculate a test statistic.

# Regress squared residuals on explanatory variables
bp_model = lm(I(e^2) ~ ratio + income, data = test_df)
# Grab the R-squared
(bp_r2 = summary(bp_model)$r.squared)

#> [1] 3.23205e-05

Testing for heteroskedasticity

Example: Breusch-Pagan

The Breusch-Pagan test statistic is

\(\text{LM} = n \times R^2_e\)

# B-P test statistic
bp_stat = 420 * bp_r2
# Calculate the p-value
pchisq(q = bp_stat, df = 2, lower.tail = F)

#> [1] 0.9932357

Testing for heteroskedasticity

Example: Breusch-Pagan

H0: \(\alpha_1 = \alpha_2 = 0\) vs. HA: \(\alpha_1 \neq 0\) and/or \(\alpha_2 \neq 0\) for the model \(u_i^2 = \alpha_0 + \alpha_1 \text{Ratio}_i + \alpha_2 \text{Income}_i + w_i\)

Breusch-Pagan test statistic: \(\widehat{\text{LM}} \approx 0.014\)

p-value \(\approx 0.993\)

∴ Fail to reject H0 (the p-value is greater than 0.05)

Conclusion: We do not find statistically significant evidence of heteroskedasticity at the 5-percent level.

(We find no evidence of a linear relationship between \(u_i^2\) and the explanatory variables.)

Testing for heteroskedasticity

Example: Breusch-Pagan

The Breusch-Pagan test statistic and its null distribution