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?

  • $\begingroup$ 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


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.