How to tell if an estimator is unbiased? How to find expected value of an estimator?

Question

You come up with a great idea of an estimator for $\beta_1$ in the SLR model which satisfies SLR.1 to SLR.4:$$y_i=\beta_0+\beta_1x_i+u_i$$ Given a sample $\left\{(x_i,y_i),i=1,2,3,\dots,n\right\}$, you connect the first sample observation to the second sample observation and compute the slope of the line: $\frac{y_2-y_1}{x_2-x_1}$. Then you connect the first sample observation to the third sample observation and compute the slope of the line: $\frac{y_3-y_2}{x_3-x_1}$. Repeat the same procedure to the rest of the observations in the sample and you estimate $\beta_1$ as the average of all $n-1$ slopes. In terms of a formula, your estimator is: $\tilde\beta_1=\frac1{n-1}\sum_{i=2}^n\left(\frac{y_i-y_1}{x_i-x_1}\right)$.

1. Is $\tilde\beta_1$ a linear estimator? i.e. can you express $\tilde\beta_1$ in the format of $\sum_{i=1}^nw_iy_i$ where all $w_i$ are functions of $x$ only. Clearly write down the $w_i$ in your answer.

2. Is $\tilde\beta_1$ an unbiased estimator conditional on a set of sample values of the independent variable $X=\left\{x_1,x_2,\dots,x_n\right\}$? i.e. Is $E(\tilde\beta_1|X)=\beta_1$?

I am stuck with part 2. specifically. I am not sure how to find the expected value of $\tilde\beta_1$. I thought it might be just using the given $\tilde\beta_1$ formula and the sample of observations. I need some pointers on what I should think about while answering this question.

Answer 1

First note that by definition, $y_i - y_1 = \beta_1(x_i-x_1) + u_i - u_1$, hence $$\frac{y_i-y_1}{x_i-x_1} = \beta_1 + \frac{u_i-u_1}{x_i-x_1} $$ By applying the "pull-out property" of conditional expectation, we have $$\mathbb E\left[\frac{u_i-u_1}{x_i-x_1}\Big|\ X\right] =\frac{1}{x_i-x_1}\mathbb E\big[u_i-u_1\mid X\big] = 0 $$ Where the last inequality follows from assuming that the $u_i$ have zero mean given $X$.

We can finally put it all together by applying the linearity of expectation : $$\begin{align*}\mathbb E\left[\tilde\beta_1 \big|\ X\right] &:= \mathbb E\left[\frac1{n-1}\sum_{i=2}^n\left(\frac{y_i-y_1}{x_i-x_1}\right) \Big|\ X\right]\\ &= \frac1{n-1}\sum_{i=2}^n\mathbb E\left[\left(\beta_1 + \frac{u_i-u_1}{x_i-x_1} \right) \Big|\ X\right]\\ &= \frac1{n-1}\sum_{i=2}^n \beta_1 = \beta_1 \end{align*} $$

And we conclude that $\tilde\beta_1$ is indeed a consistent estimator of $\beta_1$.