Proving that the intercept OLS estimator is unbiased

Solution

a) We can begin by splitting the \(E\left( \bar{y}-b_2\bar{x}|x \right)\) into two parts:

\[E\left( \bar{y}-b_2\bar{x}|x \right)=E\left( \bar{y}|x \right)-E\left( b_2\bar{x}|x \right)\]

Do not take \(b_2\) outside, \(b_2\) is not a constant (it is random) and it is not known even if we condition on all the \(x\) -data (it also depends on the \(y\) -data). However, \(\bar{x}\) is known if we condition on \(x\) so that guy can go outside.

\[E\left( \bar{y}|x \right)-E\left( b_2\bar{x}|x \right)=E\left( \bar{y}|x \right)-\bar{x}E\left( b_2|x \right)\]

b) Using the formula for \(\bar{y}\)

\[E\left( \bar{y}|x \right)=E\left( β_1+β_2\bar{x}+\bar{ε}|x \right)\]

This can be split into three:

\[E\left( β_1+β_2\bar{x}+\bar{ε}|x \right)=E\left( β_1|x \right)+E\left( β_2\bar{x}|x \right)+E\left( \bar{ε}|x \right)\]

\(β_1\) is a constant so \(E\left( β_1|x \right)=β_1\) . \(β_2\) is a constant and conditionally on \(x\) , so is \(\bar{x}\) . Thus, \(E\left( β_2\bar{x}|x \right)=β_2\bar{x}\) . For the final term we use

\[\bar{ε}= \frac{1}{n}\sum_{i=1}^{n}{ ε_i }\]

We can the find

\[E\left( \bar{ε}|x \right)=E\left( \frac{1}{n}\sum_{i=1}^{n}{ ε_i }|x \right)\]

\(1/n\) is a constant we can take outside the expectation. We then have an expected value of a sum which we can split into \(n\) expected values:

\[ \frac{1}{n}\sum_{i=1}^{n}{ E\left( ε_i|x \right) }\]

By exogeneity, all terms in the sum are zero and \(E\left( \bar{ε}|x \right)=0\) . This shows that \(E\left( \bar{y}|x \right)=β_1+β_2\bar{x}\) .

c) From a) we have

\[E\left( b_1 \mid x \right)=E\left( \bar{y}|x \right)-\bar{x}E\left( b_2|x \right)\]

Part b) gave us \(E\left( \bar{y}|x \right)=β_1+β_2\bar{x}\) . Since \(b_2\) is unbiased, \(E\left( b_2|x \right)=β_2\) . Combining

\[E\left( b_1 \mid x \right)=β_1+β_2\bar{x}-\bar{x}β_2=β_1\]

c) Take one more expectation

\[E\left( b_1 \right)=E\left( E\left( b_1 \mid x \right) \right)=E\left( β_1 \right)=β_1\]

Unbiased!