Today we're gonna talk about an Econometrics problem raised by one of my supervisors, Dr. Bekker. It was both interesting and unexpectedly tricky, took me quite some days, and is now summarized as below.

## Question

Assume model $\boldsymbol{y}=\boldsymbol{\iota}\beta_1+\boldsymbol{x}\beta_2+\boldsymbol{u}$ and $\{u_i\}$ i.i.d. Let $\boldsymbol{x'x}/n\to c>0$ and $\boldsymbol{\iota'x}\to0$ as $n\to\infty$. Suppose an estimator $\hat\gamma$ is available for the ratio $\gamma=\beta_1/\beta_2$, which is distributed independently of $\boldsymbol{u}$ and $n^{1/2}(\hat\gamma-\gamma)\overset{A}{\sim}\mathcal{N}(0,\lambda^2).$ Define $\hat\beta_2 = \frac{(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)'\boldsymbol{y}}{(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)'(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)},\tag{1}$ so $\boldsymbol{y}$ is regressed on $\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma$. Derive the asymptotic distribution of $\hat\beta_2$.

## Solution

By subtracting $\beta_2$ on both sides of (1) we have $\hat\beta_2-\beta_2 = \frac{(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)'\boldsymbol{y}}{(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)'(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)}-\beta_2 = \frac{(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)'[\beta_2(\gamma-\hat\gamma)\boldsymbol{\iota}+\boldsymbol{u}]}{(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)'(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)}.\tag{2}$ For the denominator of the RHS in (2), since $n^{1/2}(\hat\gamma-\gamma)\overset{A}{\sim}\mathcal{N}(0,\lambda^2),$ we have $\hat\gamma\overset{d}{\to}\gamma$ and thus $\hat\gamma^2\overset{d}{\to}\gamma^2$. So the following convergence holds $(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)'(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)/n= \boldsymbol{x}'\boldsymbol{x}/n+2\hat\gamma\boldsymbol{x}'\boldsymbol{\iota}/n+\hat\gamma^2\boldsymbol{\iota}'\boldsymbol{\iota}/n\overset{d}{\to} c+\gamma^2>0.$ For the numerator of the RHS in (2), we have \begin{align*} &(\boldsymbol{x}+\boldsymbol{\iota}\hat\gamma)'[\beta_2(\gamma-\hat\gamma)\boldsymbol{\iota}+\boldsymbol{u}]/n^{1/2}\\ =&\beta_2(\gamma-\hat\gamma)(\boldsymbol{x}+\boldsymbol{\iota}\gamma)'\boldsymbol{\iota}/n^{1/2} + (\boldsymbol{x}+\boldsymbol{\iota}\gamma)'\boldsymbol{u}/n^{1/2} - n^{1/2}(\hat\gamma-\gamma)^2\beta_2 + (\hat\gamma-\gamma)\boldsymbol{\iota}'\boldsymbol{u}/n^{1/2}.\tag{3} \end{align*} Now we derive the asymptotic distributions of each term in the half equation (3).

First term in (3): $\beta_2(\gamma-\hat\gamma)(\boldsymbol{x}+\boldsymbol{\iota}\gamma)'\boldsymbol{\iota}/n^{1/2} = n^{1/2}(\hat\gamma-\gamma)\beta_2(\boldsymbol{x}'\boldsymbol{\iota}/n+\gamma)\overset{A}{\sim}\mathcal{N}(0,\beta_2^2\gamma^2\lambda^2).$ Second term in (3): Define $\boldsymbol{a}$ as $\boldsymbol{a} = \frac{\boldsymbol{x}+\boldsymbol{\iota}\gamma}{[(\boldsymbol{x}+\boldsymbol{\iota}\gamma)'(\boldsymbol{x}+\boldsymbol{\iota}\gamma)]^{1/2}},$ then we have $\boldsymbol{a}'\boldsymbol{a}=1,\quad \boldsymbol{a}'\boldsymbol{\iota}/n\to 0.$ Notice that $\boldsymbol{x}'\boldsymbol{x}/n\to c$ implies $x_n^2/n\to 0$ and thus $x_n/n\to 0$, so $\boldsymbol{e}_n'\boldsymbol{a}(\boldsymbol{a}'\boldsymbol{a})^{-1}\boldsymbol{a}'\boldsymbol{e}_n =\boldsymbol{e}_n'\boldsymbol{a}\boldsymbol{a}'\boldsymbol{e}_n=\frac{(x_n+\gamma)^2}{(\boldsymbol{x}+\boldsymbol{\iota}\gamma)'(\boldsymbol{x}+\boldsymbol{\iota}\gamma)}\to 0$ which implies $\max_{1\le i\le n}a_i^2=\max_{1\le i\le n}\boldsymbol{e}_i'\boldsymbol{a}\boldsymbol{a}'\boldsymbol{e}_i\to 0.$ Hence, by Hajek-Sidak it holds that $\boldsymbol{a}'\boldsymbol{u}\overset{A}{\sim}\mathcal{N}(0,\sigma^2)$ and thus the asymptotic distribution of the second term is given by $(\boldsymbol{x}+\boldsymbol{\iota}\gamma)'\boldsymbol{u}/n^{1/2}=[(\boldsymbol{x}+\boldsymbol{\iota}\gamma)'(\boldsymbol{x}+\boldsymbol{\iota}\gamma)/n]^{1/2}\boldsymbol{a}'\boldsymbol{u}\overset{A}{\sim}\mathcal{N}(0,(c+\gamma^2)\sigma^2).$ Third term in (3): $t_3 = -n^{1/2}(\hat\gamma-\gamma)^2\beta_2\overset{d}{\to}0.$ Last term in (3): $(\hat\gamma-\gamma)\boldsymbol{\iota}'\boldsymbol{u}/n^{1/2}\overset{d}{\to}0.$ Therefore, by independence of $\hat\gamma$ and $\boldsymbol{u}$ we have $n^{1/2}(\hat\beta_2-\beta_2)\overset{A}{\sim}\mathcal{N}\left(0,\frac{\beta_2^2\gamma^2\lambda^2 + (c+\gamma^2)\sigma^2}{(c+\gamma^2)^2}\right).$

## Remarks

Therefore we have succesfully derived the asymptotic normality of $\hat\beta_2$. Notice here compared with the normal OLS estimation of $\beta_1$ and $\beta_2$ together, here by introducing extra information $\hat\gamma$ we achieve better efficiency in estimating $\beta_2$ solely.

On the other hand, there is actually a problem in the specification of the question settings. Can you find that? I'll soon upload another post on this.