Statistics, regression model, unbiased estimator help

[lape99]

Homework Statement

Professor E.Z.Stuff has decided that the least squares estimator is too much trouble. Noting that two points determine a line, Dr. Stuff chooses two points from a sample of size N and draws a line between them, calling the slope of this line the E.Z. estimator of beta1 in the simple regression model. Algebraically, if the two points are (x1, y1) and (x2, y2 ) , the E.Z. estimation rule is beta1=(y2-y1)/(x2-x1)

I need to show:
(1) Show that beta1
EZ is an unbiased estimator.
To show this the lecturer said we need to calculate the mean: E(( beta0+beta1*xN+ epsilonN-(beta0 +beta1*x1+epsilon1))/(xN-x1), and this should be equal to beta1. How to do it?

(2) Find the probability distribution of beta1
EZ .
To show this we need to show that a*epsilonN + b*epsilon1~N(..,..) Normal distribution and calculate the mean value and the dispersion. But i don't know how.
Can someone help me?

 

[lippyka]

Homework Equations beta1=(y2-y1)/(x2-x1)The Attempt at a Solution For (1), we need to calculate the expectation value of the E.Z. estimator, which isE(( y2-y1 )/( x2-x1 ) ). Since y2 and y1 are random variables, we can rewrite this as E(( E( y2 | x2 ) - E( y1 | x1 ) )/( x2-x1 ) ). By the law of iterated expectations, we can rewrite this as E( E( E( y2 | x2 ) - E( y1 | x1 ) | x2, x1 ) )/( x2-x1 ). Now, by the definition of the conditional expectation, we have that E( E( y2 | x2 ) | x2, x1 ) = E( y2 | x2 ), and similarly E( E( y1 | x1 ) | x2, x1 ) = E( y1 | x1 ). Therefore, we have E(( y2-y1 )/( x2-x1 ) ) = E( ( E( y2 | x2 ) - E( y1 | x1 ) )/( x2-x1 ) ) = E( E( y2 | x2 ) - E( y1 | x1 ) )/( x2-x1 ) = E( E( y2 | x2 ) | x2, x1 ) - E( E( y1 | x1 ) | x2, x1 ) )/( x2-x1 ) = E( y2 | x2 ) - E( y1 | x1 ) )/( x2-x1 ) Now, since we are assuming that the regression model is given by y = beta0 + beta1*x + epsilon, we have that E( y2 | x2 ) = beta0 + beta1*x2 and