# 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?

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.