# Unbiased estimator in no-intercept regression model

On an assignment I've been tasked with finding whether $$\hat{\beta}_1=\frac{\sum_{i=1}^n x_iy_i}{\sum_{i=1}^n x_i^2},$$ the estimator for the slope of a no-intercept regression model $$Y_i=\beta_1 X_i+\epsilon_i$$, is unbiased. I'm not sure how this is done. I would like to use the traditional expected value definition $$\mathbb{E}(\hat{A})=A$$, but I encounter trouble at the beginning with

$$\mathbb{E}\left(\frac{\sum_{i=1}^n x_iy_i}{\sum_{i=1}^n x_i^2}\right),$$

given that (at least from what I've seen),

$$\mathbb{E}\left(\frac{\sum_{i=1}^n x_iy_i}{\sum_{i=1}^n x_i^2}\right) \neq \frac{\mathbb{E}\left(\sum_{i=1}^n x_iy_i\right)}{\mathbb{E}(\sum_{i=1}^n x_i^2)}.$$

As such, I'm stuck. We haven't learned about the Gauss-Markov theorem (so this answer couldn't be used here) or Jensen's inequality, and we are not given the distribution of $$X_i$$ nor told to treat $$\sum_{i=1}^n x_i^2$$ as constant.

Any help would be greatly appreciated.

It does not matter whether the $$X_i$$ are random or fixed (so long as the probability they are all zero is $$0$$).

$$\mathbb E \left[ \frac{\sum X_iY_i}{\sum X_i^2}\right] = \mathbb E \left[ \frac{\sum bX_i^2 +X_i \epsilon_i}{\sum X_i^2}\right]= \mathbb E \left[b\frac {\sum X_i^2}{\sum X_i^2}\right] + \sum\mathbb E\left[\frac{ X_i }{\sum X_i^2}\epsilon_i\right] \\=b+0=b$$ using the independence of the $$\epsilon_i$$ from the $$X_i$$ and $$\mathbb E \left[ \epsilon_i\right]=0$$.

• Thanks very much for the help. Clearly I faced the classic beginner mixup between $X_i$ and $x_i$... Sep 12, 2023 at 17:18

You don't need these theorems. $$x_i$$ is deterministic, and we have $$y_i=\beta_1x_i+\epsilon_i$$. So just substitute $$y_i$$ in the numerator with the above formula. Now just take expectation. The key is to just note that $$x_i$$ and $$y_i$$ are deterministic values and only $$\epsilon_i$$'s are random variables.

A major point is that the $$x_i$$ are NOT RANDOM and the $$Y_i$$ are random. (Note that below, I do not confuse $$Y_i$$ with $$y_i$$. The latter is an observed value; the former is a random variable.) So \begin{align} \operatorname E\left( \frac{\sum_{i=1}^n x_iY_i}{\sum_{i=1}^n x_i^2} \right) & = \frac{\operatorname E\left( \sum_{i=1}^n x_i Y_i \right)}{\sum_{i=1}^n x_i^2} \text{ because the xs are not random} \\[10pt] & = \frac{\sum_{i=1}^n \operatorname E(x_iY_i) }{\sum_{i=1}^n x_i^2} \text{ by linearity of expectation} \\[10pt] & = \frac{\sum_{i=1}^n x_i\operatorname E(Y_i)}{\sum_{i=1}^n x_i^2} \text{ because the xs are not random} \\[10pt] & = \frac{\sum_{i=1}^n x_i \big( \beta_1 x_i\big)}{\sum_{i=1}^n x_i^2} \\[10pt] & = \frac{\beta_1 \sum_{i=1}^n x_i \big( x_i\big)}{\sum_{i=1}^n x_i^2} \Large{ \begin{smallmatrix} \text{because \beta_1 does not} \\ \text{change as i goes from 1 to n} \end{smallmatrix}} \\[10pt] & = \beta_1. \end{align}

Explanatory variables $$x_i$$s are assumed to be known, fixed values. The randomness emanates from $$\boldsymbol\varepsilon.$$ Any linear estimator would be of the form $$\mathbf c^\top\mathbf Y= \sum_i c_iY_i.$$

If $$\mathbb E[\sum_i c_iY_i]=\beta_1,$$ this means $$\sum c_ix_i=1.$$ Imposing additional constraints and requirements for the estimator (viz. BLUE) would lead to particular value of $$\mathbf c.$$