Consider a no intercept simple linear regression model $Y_i = \beta{X_i} + u_i$ This is the "true" model, which is unknown to the researcher.

Now consider an estimator $\beta^* = \frac{\bar Y}{\bar X}$. The researcher has decided for some unknown reason to use this estimator to regress Y on X using a sample from the true model above (perhaps calling it a model is wrong?). But, she first wants to re-assure herself that this is indeed a linear estimator, so that she can begin the work of showing that it is B.L.U.E - The Best Linear Unbiased Estimator.

What strategy would you recommend to prove that this is a linear estimator without using matrix notation, but only summation and expectation operators? That means, make no references to scalars and vectors.

The reason for this restriction is for the answer to comply with the baseline knowledge offered in many introductory linear regression textbooks in econometrics, in which all proofs are completed using only summation notation (Including the Gauss-Markov Theorem for the Classic Linear Regression Model).

To set a baseline definition of a linearity, consider a simple finite case of an estimator \hat\beta:

$\hat\beta = \sum\limits_{i=1}^{i=3}{a_ix_i} = a_1x_1 + a_2x_2 + a_3x_3 $

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    $\begingroup$ In simple model, $X$ is not considered random. Accordingly, your estimator has a random variable only in numerator which is a linear sum of the random variable $Y_i$. $\endgroup$
    – Dayne
    Commented Mar 10, 2021 at 8:04
  • $\begingroup$ Ah, I see. That's a good explanation. How would you write that linear relationship in a mathematical statement in the context of the specified model and estimator? $\endgroup$
    – Hexatonic
    Commented Mar 10, 2021 at 15:38
  • $\begingroup$ I don't understand that comment: you already have written an explicitly linear function of the observations $Y,$ understanding that $\bar Y$ is a linear combination. $\endgroup$
    – whuber
    Commented Mar 10, 2021 at 16:24
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    $\begingroup$ Linear in pure math language: $T(au+bv)=aT(u)+bT(v)$ for scalars $a,b$ and vectors $u,v$ $\endgroup$
    – Dave
    Commented Mar 11, 2021 at 2:40
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    $\begingroup$ The proof really is quite trivial, so there's clearly some kind of misunderstanding here. To help get at it, would you please share the most promising proof approach you've tried? This will help us better understand where you're coming from. $\endgroup$
    – user257566
    Commented Mar 11, 2021 at 18:15

1 Answer 1


I belive this solution only works if we assume sample values of X are non stochastic, and constant across samples.

$\beta^* = \frac{\bar{Y}}{\bar{X}}$

$= \frac{\frac{1}{n}\sum{Y_i}}{\frac{1}{n}\sum{X_i}} = \frac{\sum{Y_i}}{\sum{X_i}} $

$= \left( \frac{1}{\sum{X_i}} \right)\sum{y_i}$

$Let \ g = \left( \frac{1}{\sum{X_i}} \right)$

$\implies g\sum{y_i} = \sum{g\cdot{y_i}}$

  • $\begingroup$ That looks exactly right. Another way to think about it is that your same proof strategy goes through to prove that the sample mean $\bar{Y}$ is linear. With that, the linearity of your estimator follows from a simple property of linearity: a constant times a linear function is also linear. $\endgroup$
    – user257566
    Commented Mar 13, 2021 at 4:17
  • $\begingroup$ That's even better! $\endgroup$
    – Hexatonic
    Commented Mar 16, 2021 at 3:00

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