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

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.

• Please type your question as text, do not just post a photograph or screenshot (see here). When you retype the question, add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. Commented Oct 2, 2023 at 6:13

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 $$\beta_1$$.
• Generally speaking, the pull out property tells you that if $Y$ and $Z$ are two random variables, $\mathcal F$ a $\sigma$-algebra and $Y$ is $\mathcal F$ measurable then $$\mathbb E[YZ\mid\mathcal F] = Y\mathbb E[Z\mid \mathcal F]$$ The idea being that if $Y$ is $\mathcal F$-measurable then, given $\mathcal F$, $Y$ is "constant" and can thus be taken out of the expectation. Here I applied it with $Y :=\frac{1}{x_i-x_1}$, $Z:=u_i-u_1$ and $\mathcal F := \sigma (X) \equiv \sigma(x_1,\ldots,x_n)$ Commented Oct 2, 2023 at 14:41
• Nothing special about $\beta_0$, it's simply that we have $y_1 = \beta_0 + \beta_1 x_1 + u_1$ and $y_i :=\beta_0 +\beta_1 x_i + u_i$ hence when we take the difference $y_i - y_1$ the $\beta_0$ cancel each other Commented Oct 2, 2023 at 14:43