Using the unbiasedness assumption in the proof of the Gauss-Markov Theorem 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

*

*$E(u |X) =0$ almost surely;

*$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?
 A: 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

*

*$P(\text{rank}(X) = d) = 1$

*$E_P\big(v'v\big)<\infty$

*$E_P(v| X) = 0$

*$E_P( v v'| X) = \mathbf{Id}$
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 $E _P (  z'  z)<\infty$.
