Where the first lecture ended

• The first lecture produced the estimator and stopped there: \hat\beta = (X^\top X)^{-1} X^\top Y = \left(\sum_{i=1}^n X_i X_i^\top\right)^{-1}\sum_{i=1}^n X_i Y_i, the unique minimiser of \sum_{i=1}^n (Y_i - X_i^\top b)^2.

• Nothing so far says whether it is a good estimator.

• \hat\beta is a function of the sample, so it is random, and in general it will not equal \beta.

• Three questions, taken in this order: whether \hat\beta is right on average, how far it moves from sample to sample, and whether a better estimator of the same kind exists.

The model in matrix form

• Stack the n observations, as in the first lecture: \underset{n\times1}{Y} = \begin{pmatrix} Y_1 \\ \vdots \\ Y_n \end{pmatrix},\qquad \underset{n\times k}{X} = \begin{pmatrix} X_1^\top \\ \vdots \\ X_n^\top \end{pmatrix},\qquad \underset{n\times1}{U} = \begin{pmatrix} U_1 \\ \vdots \\ U_n \end{pmatrix}.

• The n scalar equations Y_i = X_i^\top\beta + U_i become one: \underset{n\times1}{Y} = \underset{n\times k}{X}\ \underset{k\times1}{\beta} + \underset{n\times1}{U}.

• \beta is fixed and unknown. X, Y and U are random, and U is not observed.

The four assumptions

• Assumption 1. Y = X\beta + U for some \beta \in \mathbb{R}^k.

• Assumption 2. \mathrm{E}\left[U \mid X\right] = 0.

• Assumption 3. \mathrm{Var}\left(U \mid X\right) = \sigma^2 I_n for some \sigma^2 > 0.

• Assumption 4. \operatorname{rank}(X) = k, so n \ge k and no regressor is an exact linear combination of the others.

• The numbering is the typed notes’. Assumption 4 is the sample rank condition, the sample counterpart of \operatorname{rank}\big(\mathrm{E}\left[X_i X_i^\top\right]\big) = k from the first lecture.

• Assumption 3 is new, and nothing before the variance of \hat\beta uses it.

What unbiased means

• \hat\beta is a function of the sample, hence a random vector: \hat\beta(\omega) = \left(\sum_{i=1}^n X_i(\omega) X_i(\omega)^\top\right)^{-1}\sum_{i=1}^n X_i(\omega) Y_i(\omega).

• With a continuously distributed error, \mathrm{P}(\hat\beta = \beta) = 0. Hitting \beta exactly is not the standard.

• Definition. An estimator \hat\theta of \theta is unbiased if \mathrm{E}\left[\hat\theta\right] = \theta, and conditionally unbiased given X if \mathrm{E}\left[\hat\theta \mid X\right] = \theta, whatever the true value of \theta: the estimator is a rule that does not know \theta, and the equality must hold for every value \theta could take.

• The conditional statement is the stronger one, and it is the one this lecture proves.

The sampling error

• Substitute Assumption 1 into the estimator; Assumption 4 is what makes (X^\top X)^{-1} exist: \begin{aligned} \hat\beta &= (X^\top X)^{-1} X^\top Y \\ &= (X^\top X)^{-1} X^\top (X\beta + U) \\ &= \underbrace{(X^\top X)^{-1} X^\top X}_{=\, I_k}\, \beta + (X^\top X)^{-1} X^\top U. \end{aligned}

• The first term collapses, and what is left is the sampling error: \boxed{\ \hat\beta = \underbrace{\beta}_{\text{signal}} + \underbrace{{\color{blue}(X^\top X)^{-1} X^\top U}}_{\text{noise}}\ }

• Everything in this section and the next is a statement about the noise term.

Taking expectations, first attempt

• Take expectations of the sampling error: \mathrm{E}\left[\hat\beta\right] = \beta + \mathrm{E}\left[{\color{blue}(X^\top X)^{-1} X^\top U}\right].

• Suppose only that \mathrm{E}\left[X_i U_i\right] = 0 for every i. Then \mathrm{E}\left[X^\top U\right] = \mathrm{E}\left[\sum_{i=1}^n X_i U_i\right] = \sum_{i=1}^n \mathrm{E}\left[X_i U_i\right] = 0.

• That is not enough, because in general \mathrm{E}\left[(X^\top X)^{-1} X^\top U\right] \;\text{ need not equal }\; \mathrm{E}\left[(X^\top X)^{-1}\right]\,\mathrm{E}\left[X^\top U\right].

• The two factors are built from the same X, and once \mathrm{E}\left[U \mid X\right] is not zero, the expectation of their product need not factor.

The expectation of a ratio

• The one-regressor case makes the failure visible: \mathrm{E}\left[\frac{\sum_i X_i U_i}{\sum_i X_i^2}\right] \;\text{ need not equal }\; \frac{\mathrm{E}\left[\sum_i X_i U_i\right]}{\mathrm{E}\left[\sum_i X_i^2\right]} = 0.

• In general the expectation of a ratio is not the ratio of expectations. For any two scalar random variables Z and S, \mathrm{E}\left[\frac{Z}{S}\right] = \iint \frac{z}{s}\, f_{Z,S}(z,s)\, dz\, ds \;\text{ need not equal }\; \frac{\int z f_Z(z)\, dz}{\int s f_S(s)\, ds} = \frac{\mathrm{E}\left[Z\right]}{\mathrm{E}\left[S\right]}.

• The way out is to condition on X, which turns (X^\top X)^{-1}X^\top into a known matrix.

Conditional unbiasedness: the claim

• Claim. Under Assumptions 1, 2 and 4, \mathrm{E}\left[\hat\beta \mid X\right] = \beta.

• Assumption 3 is not used. Unbiasedness does not depend on the variance of the errors.

Proof of conditional unbiasedness

• Condition the sampling error on X: \begin{aligned} \mathrm{E}\left[\hat\beta \mid X\right] &= \mathrm{E}\left[\beta + (X^\top X)^{-1}X^\top U \mid X\right] \\ &= \beta + \mathrm{E}\left[(X^\top X)^{-1}X^\top U \mid X\right] \\ &= \beta + (X^\top X)^{-1}X^\top \underbrace{\mathrm{E}\left[U \mid X\right]}_{=\,0 \text{ by Assumption 2}} \\ &= \beta. \end{aligned}

• The third line takes out what is known, as in the first lecture. Given X, the matrix (X^\top X)^{-1}X^\top is known and comes out of the expectation.

• This is exactly the step the unconditional version could not take.

From conditional to unconditional

• The law of iterated expectations, from the first lecture, finishes the proof whenever \mathrm{E}\left[\hat\beta\right] exists: \mathrm{E}\left[\hat\beta\right] = \mathrm{E}\left[\mathrm{E}\left[\hat\beta \mid X\right]\right] = \mathrm{E}\left[\beta\right] = \beta.

• Conditional unbiasedness implies unbiasedness whenever the unconditional mean exists. The converse does not hold.

• Existence is not automatic: with n = k = 1, a regressor uniform on (0, 1) and a mean-zero error independent of it, the noise term U / X has no mean, although its conditional mean is zero.

The scalar case

• Take k = 1, so X is n \times 1 and X^\top X = \sum_{i=1}^n X_i^2, \qquad X^\top Y = \sum_{i=1}^n X_i Y_i.

• The estimator and its sampling error: \hat\beta = \frac{\sum_i X_i Y_i}{\sum_i X_i^2} = \beta + \frac{\sum_i X_i U_i}{\sum_i X_i^2}.

• Conditioning on X_1, \ldots, X_n makes the denominator and every X_i in the numerator known: \mathrm{E}\left[\hat\beta \mid X\right] = \beta + \frac{\sum_i X_i \overbrace{\mathrm{E}\left[U_i \mid X\right]}^{=\,0}}{\sum_i X_i^2} = \beta.

A weaker condition

• Replace Assumption 2 by \mathrm{E}\left[X_i U_i\right] = 0 \quad \text{for all } i, which, together with \mathrm{E}\left[U_i\right] = 0, says that each regressor is uncorrelated with the error.

• The first lecture showed that \beta is still identified under this condition.

• It is not enough for unbiasedness: \mathrm{E}\left[\hat\beta\right] = \beta + \underbrace{\mathrm{E}\left[(X^\top X)^{-1}X^\top U\right]}_{\text{need not be } 0}.

• So \hat\beta need not be unbiased. The noise term is a nonlinear function of X and U, and once \mathrm{E}\left[U \mid X\right] is not zero, nothing forces its mean to vanish.

Zero correlation does not pin down the conditional mean

• Take one regressor, so X_i is a scalar with density f_X, and write g(x) = \mathrm{E}\left[U_i \mid X_i = x\right]. By iterated expectations, \mathrm{E}\left[X_i U_i\right] = \mathrm{E}\left[\mathrm{E}\left[X_i U_i \mid X_i\right]\right] = \mathrm{E}\left[X_i\, g(X_i)\right] = \int x\, g(x) f_X(x)\, dx.

• Assumption 2 says g(x) = 0 at every x, which makes the integral zero. The weaker condition says only that the integral is zero.

• An integral vanishes when the positive and negative parts cancel, so \int x\, g(x) f_X(x)\, dx = 0 \;\text{ does not imply }\; g(x) = 0 \ \text{ for all } x.

The parabola

• Let the conditional mean of the error be a parabola in the regressor, centred at zero and shifted down so that \mathrm{E}\left[U_i\right] = \mathrm{E}\left[g(X_i)\right] = 0, and let the regressor be symmetric about zero. Then g is even, so x\, g(x) is odd and \mathrm{E}\left[X_i U_i\right] = 0, while \mathrm{E}\left[U_i \mid X_i\right] \neq 0 at every value of X_i except the two zeros of g.

• The parabola shows that the conditional mean is not zero. It does not by itself bias \hat\beta: with g even, the conditional mean of the noise term, \sum_i X_i g(X_i) / \sum_i X_i^2, changes sign when every X_i does. The regressors are symmetric about zero, so the noise term has mean zero. Bias needs one of those two symmetries broken, as on the next slide but one.

• Read the horizontal axis as education and the vertical axis as ability: ability is low at middling education and high at both ends.

What survives under the weaker condition

• Under \mathrm{E}\left[X_i U_i\right] = 0 and \mathrm{E}\left[U_i\right] = 0 in place of Assumption 2: \beta is identified, and \hat\beta is still the method-of-moments and least-squares estimator.

• What is lost: X_i^\top\beta need not be the conditional mean any more, and \hat\beta is no longer guaranteed to be unbiased at any sample size.

The definition

• For a random vector V that is n \times 1 with finite second moments, and with \mathrm{E}\left[V\right] = (\mathrm{E}\left[V_1\right], \ldots, \mathrm{E}\left[V_n\right])^\top, \mathrm{Var}\left(V\right) = \mathrm{E}\left[\underset{n\times 1}{(V - \mathrm{E}\left[V\right])}\ \underset{1\times n}{(V - \mathrm{E}\left[V\right])^\top}\right] \qquad (n \times n).

• The product inside is an outer product, not an inner product, so the result is a matrix rather than a number.

The matrix entry by entry

• Entry (i,j) is \mathrm{E}\left[(V_i - \mathrm{E}\left[V_i\right])(V_j - \mathrm{E}\left[V_j\right])\right], so \mathrm{Var}\left(V\right) = \begin{pmatrix} \mathrm{Var}\left(V_1\right) & \mathrm{Cov}\left(V_1, V_2\right) & \cdots & \mathrm{Cov}\left(V_1, V_n\right) \\ \mathrm{Cov}\left(V_2, V_1\right) & \mathrm{Var}\left(V_2\right) & & \vdots \\ \vdots & & \ddots & \\ \mathrm{Cov}\left(V_n, V_1\right) & \cdots & & \mathrm{Var}\left(V_n\right) \end{pmatrix}.

• The variances sit on the diagonal and the covariances off it.

Symmetry

• \mathrm{Var}\left(V\right)^\top = \mathrm{Var}\left(V\right), because \mathrm{Cov}\left(V_i, V_j\right) = \mathrm{Cov}\left(V_j, V_i\right).

• The same statement from the definition, using that the transpose passes through the expectation entry by entry, and then (AB)^\top = B^\top A^\top: \mathrm{Var}\left(V\right)^\top = \mathrm{E}\left[\big((V - \mathrm{E}\left[V\right])(V - \mathrm{E}\left[V\right])^\top\big)^\top\right] = \mathrm{E}\left[(V - \mathrm{E}\left[V\right])(V - \mathrm{E}\left[V\right])^\top\right] = \mathrm{Var}\left(V\right).

Positive semidefiniteness

• Claim. \mathrm{Var}\left(V\right) is positive semidefinite.

• Proof. Take any non-random a \in \mathbb{R}^n and set W = a^\top (V - \mathrm{E}\left[V\right]), a scalar. Then \begin{aligned} a^\top \mathrm{Var}\left(V\right) a &= a^\top \mathrm{E}\left[(V - \mathrm{E}\left[V\right])(V - \mathrm{E}\left[V\right])^\top\right] a \\ &= \mathrm{E}\left[\underbrace{a^\top (V - \mathrm{E}\left[V\right])}_{W}\ \underbrace{(V - \mathrm{E}\left[V\right])^\top a}_{W}\right] \\ &= \mathrm{E}\left[W^2\right] \;\ge\; 0. \end{aligned}

• The constant a moves inside the expectation because expectation is linear.

• This is the argument the first lecture used for \mathrm{E}\left[X_i X_i^\top\right], with W_i = X_i^\top c.

When it is positive definite

• For a \neq 0, with W = a^\top (V - \mathrm{E}\left[V\right]) as above, track the equality case: a^\top \mathrm{Var}\left(V\right) a = 0 \iff \mathrm{E}\left[W^2\right] = 0 \iff a^\top (V - \mathrm{E}\left[V\right]) = 0, that is, a_1 (V_1 - \mathrm{E}\left[V_1\right]) + \cdots + a_n (V_n - \mathrm{E}\left[V_n\right]) = 0.

• So \mathrm{Var}\left(V\right) is positive definite exactly when no non-zero linear combination of V_1, \ldots, V_n is constant, that is, when no exact linear relation holds among the centred variables V_i - \mathrm{E}\left[V_i\right].

Variance under a linear map

• Claim. For a non-random m \times n matrix \Gamma, \boxed{\ \mathrm{Var}\left(\Gamma V\right) = \Gamma \mathrm{Var}\left(V\right) \Gamma^\top \ } \qquad (m \times m).

• Proof. \mathrm{E}\left[\Gamma V\right] = \Gamma \mathrm{E}\left[V\right], so \begin{aligned} \mathrm{Var}\left(\Gamma V\right) &= \mathrm{E}\left[\big(\Gamma V - \Gamma\mathrm{E}\left[V\right]\big)\big(\Gamma V - \Gamma\mathrm{E}\left[V\right]\big)^\top\right] \\ &= \mathrm{E}\left[\Gamma (V - \mathrm{E}\left[V\right]) (V - \mathrm{E}\left[V\right])^\top \Gamma^\top\right] \\ &= \Gamma \mathrm{E}\left[(V - \mathrm{E}\left[V\right])(V - \mathrm{E}\left[V\right])^\top\right] \Gamma^\top = \Gamma \mathrm{Var}\left(V\right) \Gamma^\top, \end{aligned} using (AB)^\top = B^\top A^\top on the second factor, and linearity to move the non-random \Gamma and \Gamma^\top outside the expectation.

• Check against the scalar case: with m = n = 1 this is \mathrm{Var}\left(\gamma V\right) = \gamma^2 \mathrm{Var}\left(V\right).

Adding a constant

• For a non-random \alpha that is m \times 1, \mathrm{Var}\left(\alpha + \Gamma V\right) = \mathrm{Var}\left(\Gamma V\right) = \Gamma \mathrm{Var}\left(V\right) \Gamma^\top.

• The constant shifts the mean, so the deviations from the mean are unchanged and it drops out.

• This is why \beta will drop out of the variance of the sampling error.

The conditional version

• The same definition with conditional expectations throughout: \mathrm{Var}\left(V \mid X\right) = \mathrm{E}\left[\big(V - \mathrm{E}\left[V \mid X\right]\big)\big(V - \mathrm{E}\left[V \mid X\right]\big)^\top \,\Big|\, X\right].

• Everything above holds conditionally, with \Gamma allowed to be a function of X. Given X, such a \Gamma is known and comes out of the conditional expectation, by taking out what is known, as in the proof of unbiasedness.

• When the conditional mean is zero, the definition simplifies. Under Assumption 2, \mathrm{Var}\left(U \mid X\right) = \mathrm{E}\left[U U^\top \mid X\right].

The variance matrix of \hat\beta

• \hat\beta is k \times 1, so \mathrm{Var}\left(\hat\beta \mid X\right) is k \times k: the conditional variance of each estimated coefficient on the diagonal, the conditional covariances between them off it.

• Conditioning on X again, for the same reason it worked for unbiasedness: it makes (X^\top X)^{-1}X^\top a known matrix.

The sandwich formula

• Start from the sampling error and name the matrix that multiplies U: \hat\beta = \beta + \underbrace{(X^\top X)^{-1} X^\top}_{\Gamma_X}\, U.

• \beta is a constant and drops out, and \Gamma_X is known given X, so the rule for a linear map applies: \begin{aligned} \mathrm{Var}\left(\hat\beta \mid X\right) &= \mathrm{Var}\left(\beta + \Gamma_X U \mid X\right) \\ &= \mathrm{Var}\left(\Gamma_X U \mid X\right) \\ &= \Gamma_X \mathrm{Var}\left(U \mid X\right) \Gamma_X^\top. \end{aligned}

The sandwich formula written out

• Substituting \Gamma_X = (X^\top X)^{-1}X^\top, with \Gamma_X^\top = X (X^\top X)^{-1} because X^\top X and its inverse are symmetric: \boxed{\ \mathrm{Var}\left(\hat\beta \mid X\right) = (X^\top X)^{-1} X^\top\ \mathrm{Var}\left(U \mid X\right)\ X (X^\top X)^{-1}\ }

• The box uses Assumptions 1 and 4, and needs the errors to have finite conditional second moments, so that \mathrm{Var}\left(U \mid X\right) exists. Assumption 2 enters only to write the middle factor as \mathrm{E}\left[U U^\top \mid X\right].

• Apart from requiring \mathrm{Var}\left(U \mid X\right) to exist, nothing yet has been assumed about the errors’ conditional variances or covariances.

Assumption 3: homoskedasticity and no autocorrelation

• Assumption 3. \mathrm{Var}\left(U \mid X\right) = \sigma^2 I_n for some \sigma^2 > 0, that is, \mathrm{Var}\left(U \mid X\right) = \begin{pmatrix} \sigma^2 & & 0 \\ & \ddots & \\ 0 & & \sigma^2 \end{pmatrix}.

• Read the diagonal: every error has the same conditional variance, \mathrm{Var}\left(U_1 \mid X\right) = \cdots = \mathrm{Var}\left(U_n \mid X\right) = \sigma^2.

• This is homoskedasticity. It rules out errors that are more variable at some values of the regressors than at others.

• Read the off-diagonal entries: for i \neq j, with the first equality by Assumption 2, \mathrm{Cov}\left(U_i, U_j \mid X\right) = \mathrm{E}\left[U_i U_j \mid X\right] = 0.

• The errors are conditionally uncorrelated across observations. This is no autocorrelation, also called no serial correlation.

• Assumption 3 is these two statements together, and nothing more. It does not say the errors are independent, and it does not say they are normal.

The variance of OLS under Assumption 3

• Put \mathrm{Var}\left(U \mid X\right) = \sigma^2 I_n into the sandwich: \begin{aligned} \mathrm{Var}\left(\hat\beta \mid X\right) &= (X^\top X)^{-1} X^\top\ \sigma^2 I_n\ X (X^\top X)^{-1} \\ &= \sigma^2 (X^\top X)^{-1} \underbrace{X^\top X (X^\top X)^{-1}}_{=\, I_k} \\ &= \sigma^2 (X^\top X)^{-1}. \end{aligned}

• Under Assumptions 1 to 4: \boxed{\ \mathrm{E}\left[\hat\beta \mid X\right] = \beta, \qquad \mathrm{Var}\left(\hat\beta \mid X\right) = \sigma^2 (X^\top X)^{-1}\ }

• This pair is the central result of the lecture.

What \sigma^2 is

• From Assumptions 2 and 3 together, \sigma^2 = \mathrm{Var}\left(U_i \mid X\right) = \mathrm{E}\left[U_i^2 \mid X\right], and the law of iterated expectations gives \sigma^2 = \mathrm{E}\left[U_i^2\right] = \mathrm{Var}\left(U_i\right), the last step because \mathrm{E}\left[U_i\right] = \mathrm{E}\left[\mathrm{E}\left[U_i \mid X\right]\right] = 0 by Assumption 2.

• \sigma^2 is one unknown number, the same for every observation. Estimating it is a question for the next lecture.

One regressor

• Take k = 1 and no intercept, so Y_i = \beta X_i + U_i and X is n \times 1. Then X^\top X = \sum_i X_i^2 and \mathrm{Var}\left(\hat\beta \mid X\right) = \frac{\sigma^2}{\sum_{i=1}^n X_i^2}.

• Two things move the variance. Noisier errors raise it. A larger \sum_i X_i^2 lowers it: with no intercept this is the sum of squares of the regressor about zero. With an intercept the same formula gives the variance of the slope, with \sum_i (X_i - \bar X)^2, the spread of the regressor about its mean, in place of \sum_i X_i^2.

Why spread in the regressor helps

• With the regressor bunched together, many quite different slopes fit the data about equally well, so the estimate moves a lot from sample to sample.

• With the regressor spread out, the slope is pinned down.

Without homoskedasticity

• If the errors are heteroskedastic but still conditionally uncorrelated across observations, so that \mathrm{Var}\left(U \mid X\right) is diagonal with entries that may depend on X, \mathrm{Var}\left(U \mid X\right) = \Omega = \begin{pmatrix} \sigma_1^2 & & 0 \\ & \ddots & \\ 0 & & \sigma_n^2 \end{pmatrix}, the sandwich does not collapse. The middle factor becomes X^\top \Omega X = \begin{pmatrix} X_1 & \cdots & X_n \end{pmatrix} \Omega \begin{pmatrix} X_1^\top \\ \vdots \\ X_n^\top \end{pmatrix} = \sum_{i=1}^n \sigma_i^2 X_i X_i^\top, so \mathrm{Var}\left(\hat\beta \mid X\right) = (X^\top X)^{-1} \left(\sum_{i=1}^n \sigma_i^2 X_i X_i^\top\right) (X^\top X)^{-1}.

• Unbiasedness is untouched. Only the variance formula changes.

Where each assumption was used

Linear estimators

• Definition. An estimator of \beta is linear if it can be written \tilde\beta = A Y for a matrix A that is k \times n and depends on X alone, not on Y.

• OLS is linear, with A = (X^\top X)^{-1} X^\top.

• Given X, each entry of a linear estimator is a fixed linear combination of the observations on the dependent variable.

Two more linear estimators

• The two-point slope. With k = 1, Y_i = \beta X_i + U_i and X_1 \neq X_n, \tilde\beta = \frac{Y_n - Y_1}{X_n - X_1} = -\frac{1}{X_n - X_1} Y_1 + \frac{1}{X_n - X_1} Y_n. It uses two observations and throws the rest away.

• Weighted least squares. For a known n \times n matrix W that is symmetric and positive definite, and that may depend on X but not on Y, \tilde\beta = (X^\top W X)^{-1} X^\top W\, Y. OLS is the case W = I_n. With W diagonal, this estimator weights observation i by the ith diagonal entry w_i, as in the first lecture’s slide on other linear estimators. The inverse (X^\top W X)^{-1} exists: for c \neq 0, Xc \neq 0 by Assumption 4, so c^\top X^\top W X c = (Xc)^\top W (Xc) > 0.

Both are unbiased

• The two-point slope, substituting Y_i = \beta X_i + U_i and using Assumption 2: \tilde\beta = \beta + \frac{U_n - U_1}{X_n - X_1}, \qquad \mathrm{E}\left[\tilde\beta \mid X\right] = \beta.

• Weighted least squares, substituting Y = X\beta + U and using Assumption 2: \tilde\beta = \beta + (X^\top W X)^{-1} X^\top W\, U, \qquad \mathrm{E}\left[\tilde\beta \mid X\right] = \beta.

• Being unbiased does not single OLS out. When n > k, there are many conditionally unbiased linear estimators.

The variance of the weighted estimator

• The route that gave the sandwich formula applies, with A = (X^\top W X)^{-1} X^\top W in place of \Gamma_X. Since W^\top = W, the matrix X^\top W X and its inverse are symmetric too, so A^\top = W X (X^\top W X)^{-1}. Under Assumptions 1, 3 and 4, \mathrm{Var}\left(\tilde\beta \mid X\right) = \sigma^2 (X^\top W X)^{-1} X^\top W\, W X (X^\top W X)^{-1}, which reduces to \sigma^2 (X^\top X)^{-1} when W = I_n.

• Which of these matrices is smallest, and what smallest means for matrices, are the next questions.

The comparison to settle

• For every linear conditionally unbiased \tilde\beta, compare the two conditional variance matrices: \underset{k\times k}{\mathrm{Var}\left(\tilde\beta \mid X\right)} - \underset{k\times k}{\mathrm{Var}\left(\hat\beta \mid X\right)} \ \overset{?}{\ge}\ 0.

• Here \ge 0 means positive semidefinite.

Reading the matrix ordering

• If a k \times k matrix M is positive semidefinite, then c^\top M c \ge 0 for every c \in \mathbb{R}^k.

• Take c to be the jth coordinate vector e_j, the vector with a 1 in position j and a 0 everywhere else: e_j = (0, \ldots, 0, 1, 0, \ldots, 0)^\top, \qquad \text{the } 1 \text{ sitting in position } j.

• Writing the quadratic form out entry by entry, and using (e_j)_l = 0 for every l \neq j and (e_j)_j = 1, e_j^\top M e_j = \sum_{l=1}^k \sum_{m=1}^k (e_j)_l\, M_{lm}\, (e_j)_m = M_{jj}.

• So every diagonal entry of a positive semidefinite matrix is non-negative.

• Apply this to M = \mathrm{Var}\left(\tilde\beta \mid X\right) - \mathrm{Var}\left(\hat\beta \mid X\right), whose jth diagonal entry is the difference of the two variances of the jth coefficient: \mathrm{Var}\left(\tilde\beta_j \mid X\right) \ \ge\ \mathrm{Var}\left(\hat\beta_j \mid X\right), \qquad j = 1, \ldots, k.

• A general c says the same about every linear combination c^\top\beta of the coefficients, since \mathrm{Var}\left(c^\top V \mid X\right) = c^\top \mathrm{Var}\left(V \mid X\right) c: \mathrm{Var}\left(c^\top \tilde\beta \mid X\right) \ \ge\ \mathrm{Var}\left(c^\top \hat\beta \mid X\right) \qquad \text{for every } c \in \mathbb{R}^k.

Proof, step 1: linearity and unbiasedness force A X = I_k

• Take conditional expectations of \tilde\beta = A Y: \begin{aligned} \beta = \mathrm{E}\left[\tilde\beta \mid X\right] &= \mathrm{E}\left[A(X\beta + U) \mid X\right] \\ &= A X \beta + A \underbrace{\mathrm{E}\left[U \mid X\right]}_{=\,0 \text{ by Assumption 2}} \\ &= A X \beta. \end{aligned}

• This must hold for every value of \beta, which is what unbiasedness means, so \boxed{\ A X = I_k\ }

• Linearity plus conditional unbiasedness is exactly this one condition.

Step 1 checked on the examples

• OLS: A = (X^\top X)^{-1} X^\top, so A X = (X^\top X)^{-1} X^\top X = I_k.

• Weighted least squares: A = (X^\top W X)^{-1} X^\top W, so A X = (X^\top W X)^{-1} X^\top W X = I_k.

• The two-point slope, with k = 1: A has -1/(X_n - X_1) in position 1 and 1/(X_n - X_1) in position n, so A X = \frac{X_n - X_1}{X_n - X_1} = 1.

Proof, step 2: the two sampling errors

• Using A X = I_k for the general estimator, and the sampling error for OLS, \hat\beta = \beta + (X^\top X)^{-1} X^\top U, \qquad \tilde\beta = A X \beta + A U = \beta + A U.

• Both estimators differ from \beta only through a known matrix times U. Everything that follows is an expectation of U U^\top.

The transpose rule for covariance matrices

• For a k \times 1 random vector V and an \ell \times 1 random vector Z, \mathrm{Cov}\left(V, Z\right) = \mathrm{E}\left[(V - \mathrm{E}\left[V\right])(Z - \mathrm{E}\left[Z\right])^\top\right] \qquad (k \times \ell), and transposing gives \mathrm{Cov}\left(V, Z\right)^\top = \mathrm{E}\left[(Z - \mathrm{E}\left[Z\right])(V - \mathrm{E}\left[V\right])^\top\right] = \mathrm{Cov}\left(Z, V\right).

• The same definition and the same rule hold conditionally, with \mathrm{E}\left[\cdot\right] replaced by \mathrm{E}\left[\cdot \mid X\right] throughout. Step 4 needs this, because both orders appear there.

Proof, step 3: the covariance equals the variance of OLS

• Both estimators are conditionally unbiased, so their deviations from their conditional means are the two noise terms of step 2: \begin{aligned} \mathrm{Cov}\left(\tilde\beta, \hat\beta \mid X\right) &= \mathrm{E}\left[\big(\tilde\beta - \beta\big)\big(\hat\beta - \beta\big)^\top \,\Big|\, X\right] \\ &= \mathrm{E}\left[A U\, U^\top X (X^\top X)^{-1} \,\big|\, X\right] \\ &= A\, \underbrace{\mathrm{E}\left[U U^\top \mid X\right]}_{=\,\sigma^2 I_n \text{ by Assumptions 2 and 3}}\, X (X^\top X)^{-1} \\ &= \sigma^2 \underbrace{A X}_{=\, I_k} (X^\top X)^{-1} \\ &= \sigma^2 (X^\top X)^{-1}. \end{aligned}

• The right-hand side is \mathrm{Var}\left(\hat\beta \mid X\right), so \boxed{\ \mathrm{Cov}\left(\tilde\beta, \hat\beta \mid X\right) = \mathrm{Var}\left(\hat\beta \mid X\right)\ }

• Every linear conditionally unbiased estimator has the same covariance with OLS, namely the variance of OLS itself. This is the one surprising step in the proof.

Proof, step 4: the difference of the variances

• As for a scalar, expand the variance of the difference using the definition; the cross terms are the two covariance matrices: \begin{aligned} \mathrm{Var}\left(\tilde\beta - \hat\beta \mid X\right) = \mathrm{Var}\left(\tilde\beta \mid X\right) &+ {\color{blue}\mathrm{Var}\left(\hat\beta \mid X\right)} \\ &- {\color{blue}\mathrm{Cov}\left(\tilde\beta, \hat\beta \mid X\right)} - {\color{blue}\mathrm{Cov}\left(\hat\beta, \tilde\beta \mid X\right)}. \end{aligned}

• Step 3 gives the first covariance term, the second is its transpose by the rule above, and \mathrm{Var}\left(\hat\beta \mid X\right) is symmetric. All three terms in blue equal \mathrm{Var}\left(\hat\beta \mid X\right), and they combine into one with a minus sign: \mathrm{Var}\left(\tilde\beta - \hat\beta \mid X\right) = \mathrm{Var}\left(\tilde\beta \mid X\right) - \mathrm{Var}\left(\hat\beta \mid X\right).

• The left-hand side is a conditional variance matrix, hence positive semidefinite. Therefore \mathrm{Var}\left(\tilde\beta \mid X\right) - \mathrm{Var}\left(\hat\beta \mid X\right) \ \ge\ 0, which is the theorem.

Uniqueness

• Suppose some linear conditionally unbiased \tilde\beta attains the same variance matrix. By the identity of step 4, \mathrm{Var}\left(\tilde\beta \mid X\right) = \mathrm{Var}\left(\hat\beta \mid X\right) \quad\Longrightarrow\quad \mathrm{Var}\left(\tilde\beta - \hat\beta \mid X\right) = 0.

• A random vector with zero conditional variance equals its conditional mean (the equality case of positive semidefiniteness, one coordinate at a time), so \tilde\beta - \hat\beta = d(X) for some function of X alone.

• Both are conditionally unbiased, so taking conditional expectations, \underbrace{\mathrm{E}\left[\tilde\beta \mid X\right]}_{=\,\beta} = \underbrace{\mathrm{E}\left[\hat\beta \mid X\right]}_{=\,\beta} + d(X) \quad\Longrightarrow\quad d(X) = 0.

• Therefore \tilde\beta = \hat\beta. A linear conditionally unbiased competitor that is also best attains the same variance matrix: both M = \mathrm{Var}\left(\tilde\beta \mid X\right) - \mathrm{Var}\left(\hat\beta \mid X\right) and -M are then positive semidefinite, so c^\top M c = 0 for every c, and a symmetric matrix with that property is zero. So OLS is not just a best linear unbiased estimator; it is the only one.

Where Assumption 3 was used

• Assumption 3 entered in step 3, where \mathrm{E}\left[U U^\top \mid X\right] was replaced by \sigma^2 I_n, and again when \sigma^2 (X^\top X)^{-1} was recognised as \mathrm{Var}\left(\hat\beta \mid X\right). Without it the cancellation that produced \sigma^2 A X fails, and the covariance need not equal the variance of OLS.

• Assumptions 1, 2 and 4 give unbiasedness, and Assumptions 1 and 4 with finite conditional second moments give the sandwich formula. Assumption 3 is what makes OLS efficient.

• If Assumption 3 fails, OLS is still unbiased but need not be best: Assumption 3 is sufficient for efficiency, not necessary.

Adding normality

• Assumption 5. U \mid X \sim N(0, \sigma^2 I_n). Assumptions 1 to 5 define the classical normal regression model.

• Given X, \hat\beta = \beta + (X^\top X)^{-1}X^\top U is a constant plus a linear function of U, and Assumption 5 fixes the conditional distribution of U. For non-random \alpha and \Gamma, such a function of a normal vector is normal: V \sim N(\mu, \Sigma) \quad\Longrightarrow\quad \alpha + \Gamma V \sim N\big(\alpha + \Gamma \mu,\ \Gamma \Sigma \Gamma^\top\big).

• Under Assumptions 1, 4 and 5, since Assumption 5 implies Assumptions 2 and 3, the conditional mean and variance are the ones already computed: \boxed{\ \hat\beta \mid X \ \sim\ N\big(\beta,\ \sigma^2 (X^\top X)^{-1}\big)\ }

• Nothing before this slide needed normality. It is what confidence intervals and tests will be built on.

If Assumption 3 fails: generalised least squares

• Suppose now that \mathrm{Var}\left(U \mid X\right) = \Omega for a known positive definite \Omega, not necessarily diagonal, with \Omega \neq \sigma^2 I_n for every \sigma^2 > 0. Let \Omega^{-1/2} be the symmetric positive definite matrix with \Omega^{-1/2}\Omega^{-1/2} = \Omega^{-1}, built from the eigenvalue decomposition \Omega = C \Lambda C^\top, and write \Omega^{1/2} for the inverse of \Omega^{-1/2}.

• Multiply the model through by \Omega^{-1/2}: \underbrace{\Omega^{-1/2} Y}_{Y^*} = \underbrace{\Omega^{-1/2} X}_{X^*} \beta + \underbrace{\Omega^{-1/2} U}_{U^*}, \qquad \mathrm{E}\left[U^* \mid X\right] = 0, \qquad \mathrm{Var}\left(U^* \mid X\right) = \Omega^{-1/2} \Omega \Omega^{-1/2} = I_n, by taking out what is known and by the rule for a linear map.

• The transformed model satisfies all four assumptions (X^* has rank k because \Omega^{-1/2} is nonsingular), so Gauss-Markov applies to it. Since A \mapsto A\,\Omega^{1/2} matches the linear conditionally unbiased estimators of the two models one to one, OLS on the transformed model is also best among estimators linear in Y and conditionally unbiased, and it is \hat\beta^* = (X^{*\top} X^*)^{-1} X^{*\top} Y^* = (X^\top \Omega^{-1} X)^{-1} X^\top \Omega^{-1} Y, which is weighted least squares with W = \Omega^{-1}.

• In practice \Omega is not known. Later lectures return to this.

Summary: unbiasedness and variance

• Substituting the model into the estimator gives the sampling error \hat\beta = \beta + (X^\top X)^{-1} X^\top U. Everything follows from it.

• Unbiasedness. Under Assumptions 1, 2 and 4, conditioning on X makes (X^\top X)^{-1}X^\top known, so \mathrm{E}\left[\hat\beta \mid X\right] = \beta, and the law of iterated expectations gives \mathrm{E}\left[\hat\beta\right] = \beta whenever that mean exists. The weaker condition \mathrm{E}\left[X_i U_i\right] = 0 is not enough, because once \mathrm{E}\left[U \mid X\right] is not zero, the expectation of the noise term need not vanish, and the parabola shows that zero correlation does not make the conditional mean zero.

• Variance. Under Assumptions 1 and 4, given finite conditional second moments, \mathrm{Var}\left(\hat\beta \mid X\right) = (X^\top X)^{-1} X^\top \mathrm{Var}\left(U \mid X\right) X (X^\top X)^{-1}, and adding Assumption 3, that \mathrm{Var}\left(U \mid X\right) = \sigma^2 I_n, gives \mathrm{Var}\left(\hat\beta \mid X\right) = \sigma^2 (X^\top X)^{-1}. With one regressor and no intercept this is \sigma^2 / \sum_i X_i^2: noisier errors raise it; a larger \sum_i X_i^2 lowers it.

Summary: efficiency

• Gauss-Markov. Under Assumptions 1 to 4, OLS has the smallest conditional variance among all estimators that are linear in Y and conditionally unbiased, whatever the value of \beta. For every such \tilde\beta, the difference of the two conditional variance matrices is positive semidefinite: \mathrm{Var}\left(\tilde\beta \mid X\right) - \mathrm{Var}\left(\hat\beta \mid X\right) \ \ge\ 0, so in particular \mathrm{Var}\left(\tilde\beta_j \mid X\right) \ge \mathrm{Var}\left(\hat\beta_j \mid X\right) for every coefficient.

• The proof. Linearity and conditional unbiasedness force A X = I_k. Assumption 3 then makes \mathrm{Cov}\left(\tilde\beta, \hat\beta \mid X\right) = \mathrm{Var}\left(\hat\beta \mid X\right). Expanding \mathrm{Var}\left(\tilde\beta - \hat\beta \mid X\right), which is positive semidefinite, leaves exactly the difference of the two variances.

• Uniqueness. Equality of the variances forces \tilde\beta = \hat\beta.

• Assumption 3 is what buys efficiency. Assumptions 1, 2 and 4 give unbiasedness on their own.