How do weak instruments violate full rank condition? So I know that if the instruments $(Z_i)$ in 2SLS regression are weak i.e. $X_i$ and $Z_i$ are uncorrelated, then full rank condition for matrix $\Sigma_{ZX}=E[Z_iX_i]$ will fail to hold, but how can one prove this formally? 
 A: If they are indeed uncorrelated, then we do not talk about weak instrument asymptotics. Then, the rank condition for identification that $\Sigma_{ZX}$ have full column rank will fail.
There is lots of literature about weak instruments, but one way to formalize weak instruments is to postulate the following relationship between regressors and instruments (based on an exercise in Hayashi, Econometrics):
$$
x_i=\frac{1}{\sqrt{n}}z_i+v_i
$$
This formalizes the notion that we do wish to have asymptotics in which a (possibly small) correlation between endogenous regressor and IV accumulates without bound, so that we ultimately get consistency. Instead, we formalize the relationship between the two so that the knowledge about the relationship between regressor and IV remains bounded even as $n\to\infty$, as the coefficient relating the two shrinks to zero as $n\to\infty$.
It is of course a bit unusual to think of a parameter as sample-size dependent, but then again, all asymptotics are fiction, and ultimately the test as to whether they are useful is the extent to which they are capable of reflecting actual finite-sample situations. And such weak-IV asymptotics have been widely found to be useful in the literature.
Indeed, one can show that IV is not consistent under such asymptotics:
Consider
$$
y_i=x_i\delta+\epsilon_i\;\;(i=1,\ldots,n),
$$
where $y_i, x_i, \epsilon_i$ are scalar. We further have a scalar instrument
$z_i$ related to $x_i$ via
$$x_i=z_i\beta+v_i$$
Let $g_i:=(z_i\epsilon_i,\; z_iv_i)'$. Make the following assumptions, which are standard except for condition 4:

*

*$(z_i,\epsilon_i,v_i)$ is an i.i.d. sequence.


*Likewise, $g_i$ is i.i.d. with mean zero and $E(g_ig_i')=S$ positive definite.


*$E(z_i^2)=\sigma_z^2>0$ (this is a variance as the notation suggests if $E(z_i)=0$, but the latter is not necessary).


*$\beta= 1/\sqrt{n}$.
Let $\hat{\delta}:=s_{zx}^{-1}s_{zy}$, where
$$
s_{zx}=\frac{1}{n}\sum_iz_ix_i, \;\;s_{zy}=\frac{1}{n}\sum_iz_iy_i.
$$
We first show that $\sqrt{n}s_{zx}\xrightarrow{d}\sigma_z^2+a$, where $a\sim N(0,S_{22})$ ($S_{ij}$ is the $(i,j)$-element of $S$).
\begin{eqnarray*}
\sqrt{n}s_{zx}&=&\sqrt{n}\frac{1}{n}\sum_iz_ix_i\\
&=&\sqrt{n}\frac{1}{n}\sum_iz_i\left(\frac{z_i}{\sqrt{n}}+v_i\right)\\
&=&\underbrace{\frac{1}{n}\sum_iz_i^2}_{\xrightarrow{p}\sigma^2_z \text{ by LLN}}+\underbrace{\frac{1}{\sqrt{n}}\sum_iz_iv_i}_{\xrightarrow{d} a\text{ by CLT}}\\
\end{eqnarray*}
(A sum of expressions converging in probability and distribution converges in distribution together.)
Next, we show that $\hat{\delta}-\delta\xrightarrow{d} (\sigma_z^2+a)^{-1}b$, where $b\sim N(0,S_{11})$ and $(a,b)\sim N(0,S)$ are jointly normal.
Using the standard sampling error representation we can write
\begin{eqnarray*}
\hat{\delta}-\delta&=&\frac{\frac{1}{n}\sum_iz_i\epsilon_i}{\frac{1}{n}\sum_iz_ix_i}\\
&=&\frac{\frac{1}{\sqrt{n}}\sum_iz_i\epsilon_i}{\frac{1}{\sqrt{n}}\sum_iz_ix_i}\\
&\xrightarrow{d}&\frac{b}{\sigma_z^2+a},
\end{eqnarray*}
where we are allowed to look at the distribution of the ratio due to joint normality.
As $\hat{\delta}-\delta$ converges in distribution to some r.v., and hence not in probability to the number zero. The estimator is therefore inconsistent.
