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Let $y_i = \beta_1 x_i + u_i$ for $i=1,2,..,n$. If I define $$\hat \beta_1 = \frac{y_1 + y_n}{x_1 + x_n}$$ then whether my $\hat \beta_1$ will be consistent or not in this setup?

For my estimator to be consistent, I just need to show $\text{plim}(\hat \beta_1) = \beta_1$

To solve this, I have substituted $y_1$ and $y_n$ as $\beta_1 x_1 + u_1$ and $\beta_1 x_n + u_n$ respectively such that

\begin{align} \text{plim}(\hat \beta_1) &= \text{plim}\left(\frac{\beta_1 x_1 + u_1 + \beta_1 x_n + u_n}{x_1+x_n}\right) \\& =\beta_1 + \text{plim}\left(\frac{u_1 + u_n}{x_1+x_n}\right) \end{align}

Now, I was thinking to substitute $u_1 + u_n = \sum_{i=1}^n u_i - \sum_{i=2}^{n-1} u_i$ and $x_1 + x_n = \sum_{i=1}^n x_i - \sum_{i=2}^{n-1} x_i$ to proceed further but I don't think it will take me anywhere to show that

$$\text{plim}\left(\frac{u_1 + u_n}{x_1+x_n}\right) = 0.$$

such that the estimator turns out to be consistent. Any help on how can I proceed further?

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  • $\begingroup$ Hint: compute the variance of $\hat\beta_1$ and compare that to the variance of the purported limit $\beta_1.$ $\endgroup$
    – whuber
    Commented Nov 23, 2022 at 15:01

1 Answer 1

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Hint: Consider the possibility that your conjecture may be false. As a simple illustration, consider the case where $x_1+x_n \neq 0$ and find the resulting variance for your estimator. Show that this variance does not (necessarily) vanish as $n \rightarrow \infty$ and use this to draw an appropriate conclusion.

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  • $\begingroup$ Thanks Ben! Voted :) $\endgroup$
    – Ujjwal
    Commented Nov 23, 2022 at 18:36

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