Help computing asymptotic variance of a weird first difference estimator in a fixed effects model I'm working on an econometrics problem set, and I'm having some major problems computing asymptotic variance for this estimator. I'm considering a fixed-effects model 
$$
Y_{it} = \beta_1 X_{it} + \alpha_i + u_{it}
$$
With $t=1,2$. Letting $\Delta Y_i=Y_{i2}-Y_{i1},\Delta Y_i=X_{i2}-Y_{i1},\Delta u_i=u_{i2}-u_{i1}$, I am considering estimators 
$$
\hat{\beta}_1=\frac{\hat{Cov}(\Delta Y_{i},\Delta X_{i})}{\hat{Var}(\Delta X_{i})}
$$
and 
$$
\tilde{\beta}_1=\frac{\sum_{i=1}^{n}\Delta Y_{i}\Delta X_{i}}{\sum_{i=1}^{n}\Delta X_{i}}
$$
And I would like to know which, in general, will have a higher asymptotic variance. So for $\tilde{\beta}_1$, I've not had a problem I don't think. I won't copy out my whole derivation because I'm rather sure that it's correct, but I get that for large $n$, 
$$
\sqrt{n}(\hat{\beta}_{1,FD}-\beta_{1})\approx \mathcal N\left(0,\frac{Var(\Delta u_{i}\Delta X_{i})}{(\mathbf{E}[(\Delta X_{i})^{2}])^{2}}\right)
$$
What a mess. 
I'm having some major difficulty with $\hat{\beta}_1$ though. Here is what I have so far. We would like to compute $\sqrt{n}(\hat{\beta}_{1}-\beta)$. That's going to be equal to 
$$
\sqrt{n}\frac{\hat{Cov}(\Delta u_{i},\Delta X_{i})}{\hat{Var}(\Delta X_{i})}=\sqrt{n}\frac{\frac{1}{n}\sum_{i=1}^{n}\Delta u_{i}(\Delta X_{i}-\frac{1}{n}\sum_{i=1}^{n}\Delta X_{i})}{\hat{Var}(\Delta X_{i})}
$$
Since the numerator has mean 0 (exogeneity assumption) we can apply the central limit theorem to it to get
$$
\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^{n}\Delta u_{i}(\Delta X_{i}-\frac{1}{n}\sum_{i=1}^{n}\Delta X_{i})\right)\rightarrow_{d}\mathcal N(0,Var(\Delta u_{i}(\Delta X_{i}-\frac{1}{n}\sum_{i=1}^{n}\Delta X_{i})))
$$
, so I get the following for large $n$:
$$
\sqrt{n}\frac{\frac{1}{n}\sum_{i=1}^{n}\Delta u_{i}(\Delta X_{i}-\frac{1}{n}\sum_{i=1}^{n}\Delta X_{i})}{\hat{Var}(\Delta X_{i})}\approx \mathcal N\left(0,\frac{Var(\Delta u_{i}(\Delta X_{i}-\frac{1}{n}\sum_{i=1}^{n}\Delta X_{i}))}{Var^{2}(\Delta X_{i})}\right)
$$
But that seems wrong. I feel like there shouldn't be the sum over $i$ in there to begin with. Am I close? Anyone have any hints? 
 A: The first thing to note is that your notation is quite inconsistent, and you do not specify your exogeneity assumptions. 
The model you have is the following
$$
\begin{align}
Y_{it} &= \alpha_i + X_{it}\beta_1 + U_{it},\, t=1,2 \\
\mathbb{E}(U_{it}\mid \boldsymbol{X}_1, \ldots, \boldsymbol{X}_n) &= 0\\
%\mathbb{E}(U_{it}) &= 0\\
\mathbb{E}(U_{it}^2\mid \boldsymbol{X}_1, \ldots, \boldsymbol{X}_n) &= \sigma^2\, \forall i=1, \dots, n;\,t=1,2
\end{align}
$$
that is, we impose a strong exogeneity condition. Here $\boldsymbol{X}_i = [X_{i1}, X_{i2}]'$.
In first differences, this model can be written as
$$
\Delta Y_i = \beta_1 \Delta X_i + \Delta U_i
$$
The estimators under consideration, written out in full are
$$
\begin{alignat}{2}
&\widehat{\beta}_1 &=& \beta_1 &+& \dfrac{\sum_{i=1}^n \Delta U_i \left(\Delta X_i - \overline{\Delta X_i}\right)}{\sum_{i=1}^n \left(\Delta X_i - \overline{\Delta X_i}\right)^2}\\
&\widetilde{\beta}_1 &=& \beta_1 &+&\dfrac{\sum_i \Delta U_i \Delta X_i}{\sum_{i=1}^n (\Delta X_i)^2}
\end{alignat}
$$
The next thing to note is that everything is done conditionally on the regressors $(\boldsymbol{X}_1, \ldots, \boldsymbol{X}_n)$, although for these estimators, the first differences of the regressors are sufficient statistics. 

So, we can write
$$
\begin{align}
\mathbb{V}\left(\widetilde{\beta}_1 \mid \boldsymbol{X}_1, \ldots, \boldsymbol{X}_n\right) &= \dfrac{\mathbb{V}(\Delta U_i)\sum_{i=1}(\Delta X_i)^2}{\left(\sum_{i=1}^n (\Delta X_i)^2\right)^2} \\
&= \dfrac{2\sigma^2}{\sum_{i=1}(\Delta X_i)^2}
\end{align}
$$
If $(\Delta X_i)^2$ is bounded so that an LLN applies, we have that
$$
\dfrac{\sum_{i=1}^n (\Delta X_i)^2}{n}\rightarrow^{p} \mathbb{E}((\Delta X_i)^2)
$$
so that
$$
\sqrt{n}\left(\widetilde{\beta}_1-\beta_1\right)\mid  \boldsymbol{X}_1, \ldots, \boldsymbol{X}_n\overset{a}{\sim}\text{N}\left(0, \dfrac{2\sigma^2}{\mathbb{E}((\Delta X_i)^2)}\right)
$$

Similarly, we can write
$$
\begin{align}
\mathbb{V}\left(\widehat{\beta}_1 \mid \boldsymbol{X}_1, \ldots, \boldsymbol{X}_n\right) &= \dfrac{\mathbb{V}(\Delta U_i)\sum_{i=1}^n\left(\Delta X_i - \overline{\Delta X_i}\right)^2}{\left(\sum_{i=1}^n \left(\Delta X_i - \overline{\Delta X_i}\right)^2\right)^2} \\
&= \dfrac{2\sigma^2}{\sum_{i=1}\left(\Delta X_i - \overline{\Delta X_i}\right)^2}
\end{align}
$$
As before, using an LLN and an application of the Slutsky theorem, we get that
$$
\dfrac{\sum_{i=1}\left(\Delta X_i - \overline{\Delta X_i}\right)^2}{n}\rightarrow^{p} \mathbb{V}(\Delta X_i)
$$ 
So, we can write
$$
\sqrt{n}\left(\widehat{\beta}_1-\beta_1\right)\mid  \boldsymbol{X}_1, \ldots, \boldsymbol{X}_n\overset{a}{\sim}\text{N}\left(0, \dfrac{2\sigma^2}{\mathbb{V}(\Delta X_i)}\right)
$$
Using the identity, $\mathbb{V}(\Delta X_i) = \mathbb{E}((\Delta X_i)^2) - (\mathbb{E}(\Delta X_i))^2$, we easily see that
$$
\text{asy.}\mathbb{V}(\widehat{\beta}_1\mid \boldsymbol{X}_1, \ldots, \boldsymbol{X}_n) \geq \text{asy.}\mathbb{V}(\widetilde{\beta}_1\mid \boldsymbol{X}_1, \ldots, \boldsymbol{X}_n) 
$$
It has been a while I did these kinds of computations, so consume with due care.
