# 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?

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## 1 Answer

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.

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