Steady State
İKT379 / Lecture 4
Lecture 4 · Week 4 · Gujarati ch. 3 · about 50 min

Classical Linear Regression Model Assumptions

The seven assumptions of the classical linear regression model and what each one looks like on a graph, the variances and standard errors of the OLS estimators, the Gauss–Markov theorem (OLS is BLUE), and r² with its properties.

By the end you can
  • List the seven CLRM assumptions with their formulas and draw homoscedasticity, heteroscedasticity and autocorrelation
  • Compute the estimated error variance, Var(b2), Var(b1) and the standard errors by hand
  • State the Gauss–Markov theorem and explain linear, unbiased and best with sampling-distribution graphs
  • Compute r² and r and list the properties of r²

How the exam asks it: three core exam templates

Exam questions on CLRM assumptions evaluate: (1) calculating standard errors and precision formulas by hand, (2) applying and proving the Gauss–Markov theorem (BLUE), and (3) computing r2r^2 and correlation rr directly from summary deviation sums.

Exam template 1: Hand-calculating error variance and standard errors

Question: Five students report study hours X=(1,2,3,4,5)X = (1, 2, 3, 4, 5) and exam scores Y=(4,6,7,8,10)Y = (4, 6, 7, 8, 10). The fitted line is Y^i=2.8+1.4Xi\hat{Y}_i = 2.8 + 1.4X_i, with residual sum of squares RSS=0.40RSS = 0.40, ∑xi2=10\sum x_i^2 = 10, and ∑Xi2=55\sum X_i^2 = 55. a) Calculate the estimated error variance σ^2\hat{\sigma}^2 and standard error of regression σ^\hat{\sigma}.
b) Calculate the variance and standard error of the slope estimator, Var(b2)Var(b_2) and SE(b2)SE(b_2).
c) Calculate the variance and standard error of the intercept estimator, Var(b1)Var(b_1) and SE(b1)SE(b_1), and report the estimated regression line with standard errors in parentheses.

Worked example · Full step-by-step solution for Template 1 (Variances and standard errors)0/9

n=5n = 5; RSS=0.40RSS = 0.40; ∑xi2=10\sum x_i^2 = 10; ∑Xi2=55\sum X_i^2 = 55; Y^i=2.8+1.4Xi\hat{Y}_i = 2.8 + 1.4X_i.

Exam template 2: Gauss–Markov theorem and proving unbiasedness

Question: Under the classical linear regression assumptions: a) State what each letter in BLUE stands for.
b) Write the OLS slope as a linear combination b2=∑kiYib_2 = \sum k_i Y_i and state the properties of the weights kik_i.
c) Prove that E(b2)=β2E(b_2) = \beta_2 (unbiasedness), and state which specific assumption is required for this result.

Worked example · Full step-by-step solution for Template 2 (Gauss–Markov & unbiasedness)0/9

Slope estimator b2=∑xiyi∑xi2b_2 = \dfrac{\sum x_i y_i}{\sum x_i^2} with true PRF Yi=β1+β2Xi+uiY_i = \beta_1 + \beta_2 X_i + u_i.

Exam template 3: Computing r-squared and r from deviation sums

Question: An exam provides summary deviation sums from a consumption regression: ∑xiyi=16,800\sum x_i y_i = 16{,}800, ∑xi2=33,000\sum x_i^2 = 33{,}000, and ∑yi2=8,890\sum y_i^2 = 8{,}890. a) Compute the OLS slope coefficient b2b_2.
b) Calculate the coefficient of determination r2r^2 using the deviation formula r2=b22∑xi2∑yi2r^2 = b_2^2 \dfrac{\sum x_i^2}{\sum y_i^2}.
c) Calculate the Pearson correlation coefficient rr and determine its sign.

Worked example · Full step-by-step solution for Template 3 (r-squared and correlation r)0/9

∑xiyi=16,800\sum x_i y_i = 16{,}800; ∑xi2=33,000\sum x_i^2 = 33{,}000; ∑yi2=8,890\sum y_i^2 = 8{,}890.

Cheat-sheet & the 4-step method

Why assumptions?

OLS always gives numbers: b1b_1 and b2b_2 can be computed from any data. But we want more than numbers. We want to know how close b1b_1 and b2b_2 are to the true β1\beta_1 and β2\beta_2. Since the PRF is Yi=β1+β2Xi+uiY_i = \beta_1 + \beta_2 X_i + u_i, YiY_i depends on uiu_i. So unless we say how the uiu_i are generated, we cannot say anything about how good b1b_1 and b2b_2 are. The classical linear regression model (CLRM) is the PRF plus seven assumptions.

Notation as on the slides: b1,b2b_1, b_2 are the OLS estimators of β1,β2\beta_1, \beta_2, and small letters are deviations from the mean: xi=Xi−Xˉx_i = X_i - \bar X, yi=Yi−Yˉy_i = Y_i - \bar Y.

The seven assumptions

#AssumptionIn symbols
1Linear regression model: linear in the parametersYi=β1+β2Xi+uiY_i = \beta_1 + \beta_2 X_i + u_i
2X values fixed in repeated samples (or independent of the error)cov(Xi,ui)=0cov(X_i, u_i) = 0
3Zero mean value of the disturbanceE(ui∣Xi)=0E(u_i \mid X_i) = 0
4Homoscedasticity (equal variance of uiu_i)Var(ui∣Xi)=E(ui2)=σ2Var(u_i \mid X_i) = E(u_i^2) = \sigma^2
5No autocorrelation between the disturbancescov(ui,uj)=0cov(u_i, u_j) = 0 for i≠ji \neq j
6More observations than parametersn>2n \gt 2 here
7Variability in the X values∑xi2>0\sum x_i^2 \gt 0

What each one means in words:

  1. Linear in parameters. The β\beta‘s enter to the first power. X2X^2 or ln⁡X\ln X is fine; β22\beta_2^2 is not.
  2. Fixed X. We choose the X values (income 80, 100, …, 260) and draw Y for them again and again. The analysis is conditional on these X values.
  3. Zero mean. The factors left in uu do not push Y up or down on average at any X. Then E(Y∣Xi)=β1+β2XiE(Y \mid X_i) = \beta_1 + \beta_2 X_i: the PRF is the line of conditional means.
  4. Homoscedasticity. “Equal spread”: the variance of uu is the same number σ2\sigma^2 at every X. The violation is heteroscedasticity: Var(ui∣Xi)=σi2Var(u_i \mid X_i) = \sigma_i^2, changing with ii.
  5. No autocorrelation. Knowing one disturbance tells you nothing about another. Typical violation: in time series, a shock this quarter carries over to the next.
  6. n greater than the number of parameters. Two parameters need at least three observations; with two points the line goes through both and n−2=0n - 2 = 0 leaves nothing to estimate σ2\sigma^2.
  7. Variability in X. If every XiX_i were the same, ∑xi2=0\sum x_i^2 = 0 and b2=∑xiyi/∑xi2b_2 = \sum x_i y_i / \sum x_i^2 cannot be computed. You cannot learn how Y responds to X if X never changes.
Diagram · What the CLRM assumes about uᵢdraw these on the exam
XYPRFX₁X₂X₃mean of each curvelies on the PRF
PRF E(Y | Xᵢ) = β₁ + β₂Xᵢf(u | Xᵢ)

Assumption: zero mean of the disturbance, E(uᵢ | Xᵢ) = 0. At every X the distribution of Y is centred on the PRF: positive and negative u cancel on average, so E(Y | Xᵢ) = β₁ + β₂Xᵢ.

Check your understanding

In a wage regression the spread of the disturbance grows with years of experience. Which assumption is violated?

Check your understanding

A sample has X=5X = 5 for every one of its 20 observations. Which assumption fails, and what happens?

Check your understanding

Which model violates assumption 1 (linear in the parameters)?

Precision: variances and standard errors

b1b_1 and b2b_2 change from sample to sample. Under the assumptions, their variances are:

Variance and standard error of the slopeKey formulaFormula sheet →
Var(b2)=σ2∑xi2SE(b2)=σ∑xi2Var(b_2) = \frac{\sigma^2}{\sum x_i^2} \qquad SE(b_2) = \frac{\sigma}{\sqrt{\sum x_i^2}}
σ2\sigma^2
variance of the disturbance u
xix_i
X_i - \bar X
∑xi2\sum x_i^2
total variation of X around its mean

Use it when you need the precision of the slope estimate, or the SE for a t test or a confidence interval.

Variance and standard error of the interceptKey formulaFormula sheet →
Var(b1)=∑Xi2n∑xi2 σ2SE(b1)=∑Xi2n∑xi2 σVar(b_1) = \frac{\sum X_i^2}{n\sum x_i^2}\,\sigma^2 \qquad SE(b_1) = \sqrt{\frac{\sum X_i^2}{n\sum x_i^2}}\,\sigma
∑Xi2\sum X_i^2
sum of the squared raw X values (capital X)
nn
sample size

Use it when you need the precision of the intercept. Careful: the numerator has capital X, the denominator small x.

σ2\sigma^2 itself is unknown, so we estimate it from the residuals:

Estimated error varianceKey formulaFormula sheet →
σ^2=∑ei2n−2=RSSn−2σ^=σ^2\hat\sigma^2 = \frac{\sum e_i^2}{n-2} = \frac{RSS}{n-2} \qquad \hat\sigma = \sqrt{\hat\sigma^2}
RSSRSS
\sum e_i^2 = residual sum of squares
n−2n - 2
degrees of freedom (two parameters estimated)
σ^\hat\sigma
standard error of the regression

Use it when you need a number for sigma-squared before computing any variance or standard error. EViews prints sigma-hat as 'S.E. of regression'.

Covariance of the two estimatorsFormula sheet →
cov(b1,b2)=−Xˉ Var(b2)cov(b_1, b_2) = -\bar X\, Var(b_2)
Xˉ\bar X
sample mean of X

Use it when you are asked how b1 and b2 move together: with a positive mean of X they are negatively related (overestimate the slope, underestimate the intercept).

Check your understanding

Holding σ2\sigma^2 fixed, sample A has X values bunched near the mean; sample B has widely spread X values. Which gives the more precise slope?

Worked example · Standard errors by hand (five students)0/6

Study hours X=1,2,3,4,5X = 1, 2, 3, 4, 5, scores Y=4,6,7,8,10Y = 4, 6, 7, 8, 10. From Lesson 3: Y^=2.8+1.4X\hat Y = 2.8 + 1.4X, residuals −0.2,0.4,0,−0.4,0.2-0.2, 0.4, 0, -0.4, 0.2. Find σ^2\hat\sigma^2, SE(b2)SE(b_2) and SE(b1)SE(b_1).

Worked example · The consumption example: reproduce the EViews standard errors0/6

Gujarati Table 2.4 (the slides' EViews example): n=10n = 10, Xˉ=170\bar X = 170, ∑xi2=33,000\sum x_i^2 = 33{,}000, ∑Xi2=322,000\sum X_i^2 = 322{,}000, RSS=337.2727RSS = 337.2727, b2=0.509091b_2 = 0.509091.

Your turn

A regression with n=12n = 12 has RSS=50RSS = 50 and ∑xi2=125\sum x_i^2 = 125. Find SE(b2)SE(b_2).

The Gauss–Markov theorem: OLS is BLUE

For the slope b2b_2:

Worked example · Show that b₂ is linear and unbiased (standard derivation)0/6

Start from b2=∑xiyi∑xi2b_2 = \dfrac{\sum x_i y_i}{\sum x_i^2} and the PRF Yi=β1+β2Xi+uiY_i = \beta_1 + \beta_2 X_i + u_i, with X fixed and E(ui)=0E(u_i) = 0.

Diagram · Sampling distribution of b₂Gauss–Markov: OLS is BLUE
value of the estimateβ₂ (true)E(β̃₂)biasOLS: E(b₂) = β₂biased
OLS b₂biased β̃₂

Repeat the sampling many times and b₂ takes a different value each time: it has a sampling distribution. OLS is unbiased: its distribution is centred on the true β₂, E(b₂) = β₂. A biased estimator β̃₂ is centred somewhere else; the gap is the bias.

Check your understanding

What does the B ('best') in BLUE mean?

Check your understanding

The disturbances are heteroscedastic but all other assumptions hold. What happens to OLS?

r²: goodness of fit and its properties

From Lesson 3: TSS=ESS+RSSTSS = ESS + RSS, that is ∑yi2=∑y^i2+∑ei2\sum y_i^2 = \sum \hat y_i^2 + \sum e_i^2.

Coefficient of determinationKey formulaFormula sheet →
r2=ESSTSS=1−RSSTSS=b22 ∑xi2∑yi2=(∑xiyi)2∑xi2∑yi2r^2 = \frac{ESS}{TSS} = 1 - \frac{RSS}{TSS} = b_2^2\,\frac{\sum x_i^2}{\sum y_i^2} = \frac{\left(\sum x_i y_i\right)^2}{\sum x_i^2 \sum y_i^2}
r2r^2
share of the total variation in Y explained by the regression
yiy_i
Y_i - \bar Y

Use it when you need the fit of a two-variable regression; the last two forms work straight from the deviation sums without residuals.

Correlation coefficientFormula sheet →
r=±r2=∑xiyi∑xi2∑yi2r = \pm\sqrt{r^2} = \frac{\sum x_i y_i}{\sqrt{\sum x_i^2 \sum y_i^2}}
rr
sample correlation between X and Y, with the same sign as b_2

Use it when you are asked for the correlation between X and Y after a two-variable regression.

Diagram · The Ballentine: r² as overlap0 ≤ r² ≤ 1
YXESSRSSr² = ESS/TSS
Y (TSS)Xexplained (ESS)

Overlap (shaded) = the part of Y's variation explained by X (ESS). The rest of the Y circle is unexplained (RSS). r² = shaded area / whole Y circle = ESS/TSS.

Worked example · r² and r without residuals (consumption example)0/6

Table 2.4: ∑xiyi=16,800\sum x_i y_i = 16{,}800, ∑xi2=33,000\sum x_i^2 = 33{,}000, ∑yi2=8,890\sum y_i^2 = 8{,}890.

Your turn

A regression has b2=−2b_2 = -2, ∑xi2=20\sum x_i^2 = 20 and ∑yi2=100\sum y_i^2 = 100. Find the correlation coefficient rr.

Exam practice

Exam question 1

n=22n = 22, RSS=180RSS = 180. Find σ^2\hat\sigma^2, the estimated variance of the disturbance.

Exam question 2

Same regression (σ^2=9\hat\sigma^2 = 9) with ∑xi2=400\sum x_i^2 = 400. Find SE(b2)SE(b_2).

Exam question 3

Data: X=2,4,6,8,10X = 2, 4, 6, 8, 10 and the regression gives RSS=6RSS = 6. Find SE(b1)SE(b_1), to three decimals.

Exam question 4

∑xiyi=60\sum x_i y_i = 60, ∑xi2=40\sum x_i^2 = 40, ∑yi2=100\sum y_i^2 = 100. Find r2r^2.

Exam question 5

Which assumption is NOT needed to prove that b2b_2 is unbiased?

Exam question 6

In a regression of consumption on income, r2=0.81r^2 = 0.81 and b2=0.6b_2 = 0.6. Which statement is correct?

Exam question 7

Xˉ=50\bar X = 50 and Var(b2)=0.004Var(b_2) = 0.004. Find cov(b1,b2)cov(b_1, b_2).

Exam question 8

You draw three sideways bell curves on the PRF at X1<X2<X3X_1 \lt X_2 \lt X_3. They have the same width but their centres are above the PRF at every X. Which assumption is violated?

Summary and review

Review deck · 18 cards0/18 mastered