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Identify when OLS qualifies as the best linear unbiased estimator by stating the required conditions.

FRM® Part I · Quantitative Analysis · Regression Diagnostics · FRM1.QA.9.h

The idea

The Gauss-Markov theorem says OLS is the best linear unbiased estimator, BLUE, when five conditions hold. The model is linear in the parameters, though it may still use logs or squares of the variables. The errors have zero mean conditional on the regressors, E(ε|X) = 0, which is called exogeneity. It means no regressor is correlated with the error, and it is a different condition from independent observations. No regressor is an exact linear combination of the others, so the coefficients are identified. The errors have constant variance (homoskedasticity). The errors are uncorrelated with one another (no serial correlation). Best means smallest variance among all linear unbiased estimators. Normality of errors is not required for BLUE. It is needed only for exact small-sample t and F tests. Violating exogeneity breaks unbiasedness, while heteroskedasticity or autocorrelation keeps OLS unbiased but no longer best. Exams expect the five conditions stated precisely.

A practice question

From the question bank, with the reason each wrong option is wrong.

Which set of conditions makes OLS the best linear unbiased estimator (BLUE) under the Gauss-Markov theorem?

  1. Linear in parameters, normally distributed errors, no perfect collinearity, and a large sample.Why not: Normality and a large sample are not Gauss-Markov conditions; the theorem needs homoskedastic, uncorrelated errors.
  2. Zero conditional mean of errors, no perfect collinearity, normally distributed errors, and R² above 0.5.Why not: R² and normality are not required, and the linearity, homoskedasticity and no-autocorrelation conditions are missing.
  3. Linear, zero-mean errors given X, no perfect collinearity, constant variance, no autocorrelation. - correct
  4. Linear in parameters, homoskedastic errors, uncorrelated regressors, and a large sample.Why not: Regressors may be correlated with each other (only perfect collinearity is excluded); zero conditional error mean is missing.

Gauss-Markov requires linearity in parameters, zero-mean errors conditional on the regressors, no perfect multicollinearity, constant error variance and uncorrelated errors. Normality is not among them.

A harder one

A risk analyst finds that the regression errors are strongly skewed, but linearity, exogeneity, no perfect collinearity, constant variance and no serial correlation all hold. Which statement is correct?

  1. OLS is no longer BLUE, because BLUE requires normally distributed errors.Why not: Normality is not a Gauss-Markov condition.
  2. OLS is biased because skewed errors have a non-zero mean.Why not: Skewness concerns the shape of the distribution; the errors can have zero mean and remain skewed, so unbiasedness is retained.
  3. OLS remains unbiased but is no longer efficient, because skewness is a form of heteroskedasticity.Why not: Skewness is not heteroskedasticity; the variance can be constant while the distribution is asymmetric.
  4. OLS is still BLUE; normality is needed only for exact small-sample t and F tests. - correct

Gauss-Markov does not require normality. Skewed errors with the five conditions in place leave OLS BLUE; skewness only affects the exactness of small-sample tests, and non-linear estimators might beat OLS.

The hardest tier

An analyst estimates y_t = a + b y_(t-1) + e_t by OLS, where the errors e_t are serially correlated. Which statement is correct?

  1. OLS is unbiased but no longer best, as with any serially correlated errors.Why not: This holds when the regressors are exogenous; here the lagged dependent variable is correlated with the error.
  2. OLS remains BLUE, because the model is linear in the parameters.Why not: Linearity is only one condition; exogeneity and uncorrelated errors are violated.
  3. OLS cannot be computed, since y_(t-1) and y_t are perfectly collinear.Why not: The two variables are highly related, not perfectly collinear; OLS can be computed.
  4. Exogeneity fails because y_(t-1) is correlated with e_t, so OLS is biased and inconsistent, not merely inefficient. - correct

If e_t is correlated with e_(t-1), and y_(t-1) contains e_(t-1), then the regressor is correlated with the error. The zero conditional mean condition fails, so OLS is biased and inconsistent. Autocorrelation with exogenous regressors would have caused only inefficiency.

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