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In what follows $y = (y_1,\dots,y_n)$ is a $n\times 1$ vector of random variables and $X = (x_{ij})$ is a $n\times d$ random matrix ($n>d$ tipically) with $\text{rank}(X)=d$ with probability 1.

Write $\beta := E(X'X)^{-1}E(X'y)$ and $u := y - X\beta$, and let $\hat{\beta} := (X'X)^{-1}X'y$ denote the OLS estimator. Also let $\tilde{\beta}=A'y$ denote some linear estimator of the form $A = \varphi\circ X$, for some $\varphi:M(n\times d)\to M(n\times d)$ (measurable), where $M(n\times d)$ is the space of $n\times d$ matrices.

The Gauss-Markov theorem states that, if

  1. $E(u |X) =0$ almost surely;
  2. $E(uu'|X) = \sigma^2 \mathrm{Id} $ for some $\sigma>0$ (where $\mathrm{Id}$ is the identity matrix);

then, whenever $\tilde{\beta}$ is unbiased for $\beta$, it holds that the matrix $$ E[(\tilde{\beta}-\beta)(\tilde{\beta}-\beta)'] - E[(\hat{\beta}-\beta)(\hat{\beta}-\beta)'] $$ is positive semi-definite. Now, in every textbook that I've come across, the unbiasedness assumption is invoked to conclude that $E(\tilde{\beta} | X) = \beta$ (almost surely) but this conclusion is strictly stronger than unbiasedness. Indeed, since $\tilde{\beta} = A'y = A'X\beta + A'u$, and since $A$ is $X$-measurable, we have by the assumption in item 1 above that $E(\tilde{\beta}|X) = A'X\beta$. At this points the canonical argument concludes that $\beta = A'X\beta$ and so on.

In my understanding, however, the definition of unbiasedness only allows me to conclude, using iterated expectations, that $\beta = E(A'X)\beta$, that is, $E(A'X) = \mathrm{Id}$. Am I missing something or is it implicit that the estimator $\tilde\beta$ is conditionally unbiased?

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  • $\begingroup$ If I understand correctly, then, in the least square framework, $\tilde\beta$ being unbiased and $\tilde\beta$ being conditionally unbiased given $X$ is the same thing because $\tilde\beta$ can only exist when $X$ is given because it's a function of $X$. But I might not be understanding your question. $\endgroup$
    – mlofton
    Oct 25, 2020 at 7:08
  • $\begingroup$ I think this is not generally the case (maybe it's something particular to least squares) as the following (artificial) example shows: let $y_1,\dots,y_n$ be a random sample with $E(y_1) =: \beta$. Let, for the sake of argument, $x_i := y_i$, $i=1,\dots,n$. Put $\hat{\beta} = n^{-1}\sum_{i=1}^n x_i$ and $\tilde{\beta} = x_1$. Clearly $E(\hat{\beta}) = E(\tilde{\beta}) = \beta$ but $E(\tilde{\beta}\,|\, x_1,\dots, x_n) = x_1$. $\endgroup$
    – user127022
    Oct 25, 2020 at 12:22
  • $\begingroup$ I like your example and I think you're right. Forget my first comment. It was el-wrongo. I'll try to read-follow your answer later when I have more time. Hopefully, someone else can chime in because I think this is a very subtle topic-question. You're making me realize that maybe I don't understand Gauss Markov as well as I thought that I did !!!! $\endgroup$
    – mlofton
    Oct 26, 2020 at 14:24

1 Answer 1

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In the paper The Gauss-Markov Theorem and Random Regressors, Juliet Popper Shaffer writes:

If attention is restricted to linear estimators ... that are conditionally unbiased, given $X$, the Gauss-Markov theorem applies. If, however, the estimator is required only to be unconditionally unbiased, the Gauss-Markov theorem may or may not hold, depending on what is known about the distribution of $X$.

Therefore, in the assumptions of the Gauss-Markov theorem with random $X$, it should be stated explicitly that $E(\tilde{\beta}\,|\,X) =\beta$.


There is an additional passage found in the “canonical proofs” that also bothers me, namely, that the equality $E(\tilde{\beta}\,|\,X) =\beta$ should hold for all $\beta\in\mathbb{R}^d$, as usually unbiasedness (conditional or unconditional) is introduced with a given fixed probability measure in mind. Since this post refers to the method of proof, I've written a statement which explicitly asserts every assumption that is used in these proofs:

Theorem Fix a measurable space $(\Omega,\mathscr{A})$, a random $n\times d$ matrix ${X}$ and a $n\times 1$ random vector $v$. Let $\mathfrak{M}$ denote the set of all probability measures $P$ satisfying the following

  1. $P(\text{rank}(X) = d) = 1$
  2. $E_P\big(v'v\big)<\infty$
  3. $E_P(v| X) = 0$
  4. $E_P( v v'| X) = \mathbf{Id}$

Let moreover $\psi:M(n\times d)\to M(n\times d)$ be measurable (where $M(n\times d)$ denotes the vector space of $n\times d$ matrices), and put $ X_\psi = \psi\circ X$.

Suppose that, for all $P \in\mathfrak{M}$, for all $\beta\in\mathbb R^{d}$ and all $\sigma>0$, it holds that $$ E _P ( X_\psi'( X\beta + \sigma v)\,|\, X) = \beta,\qquad\text{$P $-a.s.} $$ Then, for any $P \in\mathfrak M$, any $\beta\in\mathbb R^{d}$ and any $\sigma>0$ it holds that the $d\times d$ matrix $$ \text{var}_P ( X_\psi'( X\beta + \sigma v)\,|\, X) - \text{var}_P (( X' X)^{-1} X'( X \beta + \sigma v)\,|\, X) $$ is positive semidefinite, where $\text{var}_P $ is defined through $$ \text{var}_P ( z) := E _P ( z z') - E _P ( z)E _P ( z') $$ for all random vectors $ z$ such that $E _P ( z' z)<\infty$.

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