Smaller variance of estimators in matrix sense When comparing asymptotic variances, you usually conclude that a covariance matrix A is more efficient whenever the difference B - A ≥ 0 (where ≥ 0 means that the difference is a positive semidefinite matrix).

However, this comparison involves all elements in the matrices - both the diagonal elements (asymptotic variances of each beta estimator) and the off-diagonal ones (covariances between estimators) -, while we are usually interested in estimators for a particular beta coefficient having small standard errors.

Could anyone explain why is this method preferred to a comparison that entails only the diagonal elements? Could it be the case that some diagonal element is smaller in B while B - A ≥ 0?

Thank you!
Relative efficiency between two unbiased estimators $\widehat{\theta }_{A}$
and $\widehat{\theta }_{B}$ of an unknown parameter vector $\theta _{0}\in 
\mathbb{R}^{K}$ is usually defined as follow (see for instance Ruud, 2001).
The estimator $\widehat{\theta }_{A}$ is said to be efficient relative to $
\widehat{\theta }_{B}$ if we have:
$$ \mathrm{V}\left[ \widehat{\theta }_{A}\right] \equiv \Omega _{A}&lt;&lt;\Omega
_{B}\equiv \mathrm{V}\left[ \widehat{\theta }_{B}\right] .$$

If $\Omega _{B}-\Omega _{A}$ is positive definite, then the diagonal terms
of $\Omega _{B}$ and $\Omega _{A}$ are necessarily such that $\sigma
_{Bii}^{2}&gt;\sigma _{Aii}^{2}.$ Indeed if $v^{T}\left( \Omega _{B}-\Omega
_{A}\right) v&gt;0$ for any $v\neq 0$ then for $v=e_{i}$ we catch the diagonal
terms out of $\Omega _{B}-\Omega _{A}$ and find that $\sigma
_{Bii}^{2}&gt;\sigma _{Aii}^{2}.$ The converse, however, it is not true: the
mere conditions $\sigma _{Bii}^{2}&gt;\sigma _{Aii}^{2}$ do not garantee that $
\Omega _{B}-\Omega _{A}$ be positive definite. Covariances matter.  


The answer to your second question is: no it never happens that $\sigma _{Bii}^{2}&lt;\sigma _{Aii}^{2}$ when $\Omega _{B}&gt;&gt;\Omega _{A}$.  
The answer to your first question is: it is important to consider all covariance terms. If we want that the confidence ellipse (at any threshold) of $\widehat{\theta }_{A}$ be nested within the confidence ellipse of $\widehat{\theta }_{B}$ then we need $\Omega _{A}&lt;&lt;\Omega _{B}$ and not just $\sigma _{Aii}^{2}&lt;\sigma _{Bii}^{2}$.   


See Ruud, (2001, chapter 9) for a proof and detailed explanations. An example is provided here, illustrating that the confidence ellipses are nested when $\Omega _{B}-\Omega _{A}$ positive definite, and not nested if $\Omega _{B}-\Omega _{A}$ is not positive definite.  

Example: 
\begin{eqnarray*}
\widehat{\theta }_{A}\left( a\right) &amp;\sim &amp;\mathcal{N}\left( 0,\Omega
_{A}\left( a\right) \right) ,\qquad \widehat{\theta }_{B}\sim \mathcal{N}%
\left( 0,\Omega _{B}\right) \\
\Omega _{A}\left( a\right) &amp;=&amp;\left( 
\begin{array}{cc}
4 &amp; a \\ 
a &amp; 9%
\end{array}%
\right) ,\qquad \Omega _{B}=\left( 
\begin{array}{cc}
5 &amp; 0 \\ 
0 &amp; 10%
\end{array}%
\right).
\end{eqnarray*}

For $a=0$ the matrix $\Omega _{B}-\Omega _{A}\left( 0\right) $ is positive
definite and the $95\%$ threshold confidence ellipses are nested. The Figure
below (left panel) represents the iso-curves centered on $\theta _{0}=0$ and
whose equations are given by $v^{T}\left( \Omega _{A}\left( 0\right) \right)
^{-1}v=5.99$ and $x^{T}\Omega _{B}^{-1}x=5.99$ and illustrates that for $a=0$
the later is nested within the former. This is no longer true for $a=-5$
(right panel) in which case $\Omega _{B}-\Omega _{A}\left( -5\right) $ is no
longer positive definite. In this case the probability that $\widehat{\theta 
}_{A}$ is further away than $\widehat{\theta }_{B}$ from the true value $%
\theta _{0}=0$ is positive, and $\widehat{\theta }_{A}$ is no longer
efficient (case with $a=-5$) relatively to $\widehat{\theta }_{B}$. Note
that the variances always satisfy $\sigma _{Bii}^{2}&gt;\sigma _{Aii}^{2}$ but
this is not sufficient for the ellipses to be nested, the covariances must
also satisfy $\left( \sigma _{B11}^{2}-\sigma _{A11}^{2}\right) \left(
\sigma _{B22}^{2}-\sigma _{A22}^{2}\right) -\left( \sigma _{B12}-\sigma
_{A12}\right) ^{2}&gt;0$, which is not the case in this example for $a=-5$.