# 2.4 — OLS: Goodness of Fit and Bias — Class Notes

## Contents

Tuesday, September 15, 2019

## Overview

Today we continue looking at basic OLS regression. We will cover how to measure if a regression line is a good fit (using $$R^2$$ and $$\sigma_u$$ or SER), and whether OLS estimators are biased. These will depend on four critical assumptions about $$u$$.

In doing so, we begin an ongoing exploration into inferential statistics, which will finally become clear in another week. The most confusing part is recognizing that there is a sampling distribution of each OLS estimator. We want to measure the center of that sampling distribution, to see if the estimator is biased. Next class we will measure the spread of that distribution.

We continue the extended example about class sizes and test scores, which comes from a (Stata) dataset from an old textbook that I used to use, Stock and Watson, 2007. Download and follow along with the data from today’s example:Note this is a .dta Stata file. You will need to (install and) load the package haven to read_dta() Stata files into a dataframe.

I have also made a RStudio Cloud project documenting all of the things we have been doing with this data that may help you when you start working with regressions (next class):

## Live Class Session on Zoom

The live class Zoom meeting link can be found on Blackboard (see LIVE ZOOM MEETINGS on the left navigation menu), starting at 11:30 AM.

If you are unable to join today’s live session, or if you want to review, you can find the recording stored on Blackboard via Panopto (see Class Recordings on the left navigation menu).

## Problem Set

Problem Set 1 answers are posted on that page in various formats.

Problem set 2 answers are posted on that page in various formats.

## Appendix: OLS Estimators

### Deriving the OLS Estimators

The population linear regression model is:

$Y_i=\beta_0+\beta_1 X_i + u _i$

The errors $$(u_i)$$ are unobserved, but for candidate values of $$\hat{\beta_0}$$ and $$\hat{\beta_1}$$, we can obtain an estimate of the residual. Algebraically, the error is:

$\hat{u_i}= Y_i-\hat{\beta_0}-\hat{\beta_1}X_i$

Recall our goal is to find $$\hat{\beta_0}$$ and $$\hat{\beta_1}$$ that minimizes the sum of squared errors (SSE):

$SSE= \sum^n_{i=1} \hat{u_i}^2$

So our minimization problem is:

$\min_{\hat{\beta_0}, \hat{\beta_1}} \sum^n_{i=1} (Y_i-\hat{\beta_0}-\hat{\beta_1}X_i)^2$

Using calculus, we take the partial derivatives and set it equal to 0 to find a minimum. The first order conditions are:

\begin{align*} \frac{\partial SSE}{\partial \hat{\beta_0}}&=-2\displaystyle\sum^n_{i=1} (Y_i-\hat{\beta_0}-\hat{\beta_1} X_i)=0\\ \frac{\partial SSE}{\partial \hat{\beta_1}}&=-2\displaystyle\sum^n_{i=1} (Y_i-\hat{\beta_0}-\hat{\beta_1} X_i)X_i=0\\ \end{align*}

#### Finding $$\hat{\beta_0}$$

Working with the first FOC, divide both sides by $$-2$$:

$\displaystyle\sum^n_{i=1} (Y_i-\hat{\beta_0}-\hat{\beta_1} X_i)=0$

Then expand the summation across all terms and divide by $$n$$:

$\underbrace{\frac{1}{n}\sum^n_{i=1} Y_i}_{\bar{Y}}-\underbrace{\frac{1}{n}\sum^n_{i=1}\hat{\beta_0}}_{\hat{\beta_0}}-\underbrace{\frac{1}{n}\sum^n_{i=1} \hat{\beta_1} X_i}_{\hat{\beta_1}\bar{X}}=0$

Note the first term is $$\bar{Y}$$, the second is $$\hat{\beta_0}$$, the third is $$\hat{\beta_1}\bar{X}$$.From the rules about summation operators, we define the mean of a random variable $$X$$ as $$\bar{X}=\frac{1}{n}\displaystyle\sum_{i=1}^n X_i$$. The mean of a constant, like $$\beta_0$$ or $$\beta_1$$ is itself.

So we can rewrite as: $\bar{Y}-\hat{\beta_0}-\beta_1=0$

Rearranging:

$\hat{\beta_0}=\bar{Y}-\bar{X}\beta_1$

#### Finding $$\hat{\beta_1}$$

To find $$\hat{\beta_1}$$, take the second FOC and divide by $$-2$$:

$\displaystyle\sum^n_{i=1} (Y_i-\hat{\beta_0}-\hat{\beta_1} X_i)X_i=0$

From the formula for $$\hat{\beta_0}$$, substitute in for $$\hat{\beta_0}$$:

$\displaystyle\sum^n_{i=1} \bigg(Y_i-[\bar{Y}-\hat{\beta_1}\bar{X}]-\hat{\beta_1} X_i\bigg)X_i=0$

Combining similar terms:

$\displaystyle\sum^n_{i=1} \bigg([Y_i-\bar{Y}]-[X_i-\bar{X}]\hat{\beta_1}\bigg)X_i=0$

Distribute $$X_i$$ and expand terms into the subtraction of two sums (and pull out $$\hat{\beta_1}$$ as a constant in the second sum:

$\displaystyle\sum^n_{i=1} [Y_i-\bar{Y}]X_i-\hat{\beta_1}\displaystyle\sum^n_{i=1}[X_i-\bar{X}]X_i=0$

Move the second term to the righthand side:

$\displaystyle\sum^n_{i=1} [Y_i-\bar{Y}]X_i=\hat{\beta_1}\displaystyle\sum^n_{i=1}[X_i-\bar{X}]X_i$

Divide to keep just $$\hat{\beta_1}$$ on the right:

$\frac{\displaystyle\sum^n_{i=1} [Y_i-\bar{Y}]X_i}{\displaystyle\sum^n_{i=1}[X_i-\bar{X}]X_i}=\hat{\beta_1}$

Note that from the rules about summation operators:

$\displaystyle\sum^n_{i=1} [Y_i-\bar{Y}]X_i=\displaystyle\sum^n_{i=1} (Y_i-\bar{Y})(X_i-\bar{X})$

and:

$\displaystyle\sum^n_{i=1} [X_i-\bar{X}]X_i=\displaystyle\sum^n_{i=1} (X_i-\bar{X})(X_i-\bar{X})=\displaystyle\sum^n_{i=1}(X_i-\bar{X})^2$

Plug in these two facts:

$\frac{\displaystyle\sum^n_{i=1} (Y_i-\bar{Y})(X_i-\bar{X})}{\displaystyle\sum^n_{i=1}(X_i-\bar{X})^2}=\hat{\beta_1}$

### Algebraic Properties of OLS Estimators

The OLS residuals $$\hat{u}$$ and predicted values $$\hat{Y}$$ are chosen by the minimization problem to satisfy:

1. The expected value (average) error is 0: $E(u_i)=\frac{1}{n}\displaystyle \sum_{i=1}^n \hat{u_i}=0$

2. The covariance between $$X$$ and the errors is 0: $\hat{\sigma}_{X,u}=0$

Note the first two properties imply strict exogeneity. That is, this is only a valid model if $$X$$ and $$u$$ are not correlated.

1. The expected predicted value of $$Y$$ is equal to the expected value of $$Y$$: $\bar{\hat{Y}}=\frac{1}{n} \displaystyle\sum_{i=1}^n \hat{Y_i} = \bar{Y}$

2. Total sum of squares is equal to the explained sum of squares plus sum of squared errors: \begin{align*}TSS&=ESS+SSE\\ \sum_{i=1}^n (Y_i-\bar{Y})^2&=\sum_{i=1}^n (\hat{Y_i}-\bar{Y})^2 + \sum_{i=1}^n {u}^2\\ \end{align*}

Recall $$R^2$$ is $$\frac{ESS}{TSS}$$ or $$1-SSE$$

1. The regression line passes through the point $$(\bar{X},\bar{Y})$$, i.e. the mean of $$X$$ and the mean of $$Y$$.

### Bias in $$\hat{\beta_1}$$

Begin with the formula we derived for $$\hat{\beta_1}$$:

$\hat{\beta_1}=\frac{\displaystyle \sum^n_{i=1} (Y_i-\bar{Y})(X_i-\bar{X})}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2}$

Recall from Rule 6 of summations, we can rewrite the numerator as

\begin{align*} =&\displaystyle \sum^n_{i=1} (Y_i-\bar{Y})(X_i-\bar{X})\\ =& \displaystyle \sum^n_{i=1} Y_i(X_i-\bar{X})\\ \end{align*}

$\hat{\beta_1}=\frac{\displaystyle \sum^n_{i=1} Y_i(X_i-\bar{X})}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2}$

We know the true population relationship is expressed as:

$Y_i=\beta_0+\beta_1 X_i+u_i$

Substituting this in for $$Y_i$$ in equation 2:

$\hat{\beta_1}=\frac{\displaystyle \sum^n_{i=1} (\beta_0+\beta_1X_i+u_i)(X_i-\bar{X})}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2}$ Breaking apart the sums in the numerator:

$\hat{\beta_1}=\frac{\displaystyle \sum^n_{i=1} \beta_0(X_i-\bar{X})+\displaystyle \sum^n_{i=1} \beta_1X_i(X_i-\bar{X})+\displaystyle \sum^n_{i=1} u_i(X_i-\bar{X})}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2}$

We can simplify equation 4 using Rules 4 and 5 of summations

1. The first term in the numerator $$\left[\displaystyle \sum^n_{i=1} \beta_0(X_i-\bar{X})\right]$$ has the constant $$\beta_0$$, which can be pulled out of the summation. This gives us the summation of deviations, which add up to 0 as per Rule 4:

\begin{align*} \displaystyle \sum^n_{i=1} \beta_0(X_i-\bar{X})&= \beta_0 \displaystyle \sum^n_{i=1} (X_i-\bar{X})\\ &=\beta_0 (0)\\ &=0\\ \end{align*}

1. The second term in the numerator $$\left[\displaystyle \sum^n_{i=1} \beta_1X_i(X_i-\bar{X})\right]$$ has the constant $$\beta_1$$, which can be pulled out of the summation. Additionally, Rule 5 tells us $$\displaystyle \sum^n_{i=1} X_i(X_i-\bar{X})=\displaystyle \sum^n_{i=1}(X_i-\bar{X})^2$$:

\begin{align*} \displaystyle \sum^n_{i=1} \beta_1X_1(X_i-\bar{X})&= \beta_1 \displaystyle \sum^n_{i=1} X_i(X_i-\bar{X})\\ &=\beta_1\displaystyle \sum^n_{i=1}(X_i-\bar{X})^2\\ \end{align*}

When placed back in the context of being the numerator of a fraction, we can see this term simplifies to just $$\beta_1$$:

\begin{align*} \frac{\beta_1\displaystyle \sum^n_{i=1}(X_i-\bar{X})^2}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2} &=\frac{\beta_1}{1} \times \frac{\displaystyle \sum^n_{i=1}(X_i-\bar{X})^2}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2}\\ &=\beta_1 \\ \end{align*}

Thus, we are left with:

$\hat{\beta_1}=\beta_1+\frac{\displaystyle \sum^n_{i=1} u_i(X_i-\bar{X})}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2}$

Now, take the expectation of both sides:

$E[\hat{\beta_1}]=E\left[\beta_1+\frac{\displaystyle \sum^n_{i=1} u_i(X_i-\bar{X})}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2} \right]$

We can break this up, using properties of expectations. First, recall $$E[a+b]=E[a]+E[b]$$, so we can break apart the two terms.

$E[\hat{\beta_1}]=E[\beta_1]+E\left[\frac{\displaystyle \sum^n_{i=1} u_i(X_i-\bar{X})}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2} \right]$

Second, the true population value of $$\beta_1$$ is a constant, so $$E[\beta_1]=\beta_1$$.

Third, since we assume $$X$$ is also “fixed” and not random, the variance of $$X$$, $$\displaystyle\sum_{i=1}^n (X_i-\bar{X})$$, in the denominator, is just a constant, and can be brought outside the expectation.

$E[\hat{\beta_1}]=\beta_1+\frac{E\left[\displaystyle \sum^n_{i=1} u_i(X_i-\bar{X})\right] }{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2}$

Thus, the properties of the equation are primarily driven by the expectation $$E\bigg[\displaystyle \sum^n_{i=1} u_i(X_i-\bar{X})\bigg]$$. We now turn to this term.

Use the property of summation operators to expand the numerator term:

\begin{align*} \hat{\beta_1}&=\beta_1+\frac{\displaystyle \sum^n_{i=1} u_i(X_i-\bar{X})}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2} \\ \hat{\beta_1}&=\beta_1+\frac{\displaystyle \sum^n_{i=1} (u_i-\bar{u})(X_i-\bar{X})}{\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2} \\ \end{align*}

Now divide the numerator and denominator of the second term by $$\frac{1}{n}$$. Realize this gives us the covariance between $$X$$ and $$u$$ in the numerator and variance of $$X$$ in the denominator, based on their respective definitions.

\begin{align*} \hat{\beta_1}&=\beta_1+\cfrac{\frac{1}{n}\displaystyle \sum^n_{i=1} (u_i-\bar{u})(X_i-\bar{X})}{\frac{1}{n}\displaystyle\sum^n_{i=1} (X_i-\bar{X})^2} \\ \hat{\beta_1}&=\beta_1+\cfrac{cov(X,u)}{var(X)} \\ \hat{\beta_1}&=\beta_1+\cfrac{s_{X,u}}{s_X^2} \\ \end{align*}

By the Zero Conditional Mean assumption of OLS, $$s_{X,u}=0$$.

Alternatively, we can express the bias in terms of correlation instead of covariance:

$E[\hat{\beta_1}]=\beta_1+\cfrac{cov(X,u)}{var(X)}$

From the definition of correlation:

\begin{align*} cor(X,u)&=\frac{cov(X,u)}{s_X s_u}\\ cor(X,u)s_Xs_u &=cov(X,u)\\ \end{align*}

Plugging this in:

\begin{align*} E[\hat{\beta_1}]&=\beta_1+\frac{cov(X,u)}{var(X)} \\ E[\hat{\beta_1}]&=\beta_1+\frac{\big[cor(X,u)s_xs_u\big]}{s^2_X} \\ E[\hat{\beta_1}]&=\beta_1+\frac{cor(X,u)s_u}{s_X} \\ E[\hat{\beta_1}]&=\beta_1+cor(X,u)\frac{s_u}{s_X} \\ \end{align*}

### Proof of the Unbiasedness of $$\hat{\beta_1}$$

Begin with equation:Admittedly, this is a simplified version where $$\hat{\beta_0}=0$$, but there is no loss of generality in the results.

$\hat{\beta_1}=\frac{\sum Y_iX_i}{\sum X_i^2}$

Substitute for $$Y_i$$:

$\hat{\beta_1}=\frac{\sum (\beta_1 X_i+u_i)X_i}{\sum X_i^2}$

Distribute $$X_i$$ in the numerator:

$\hat{\beta_1}=\frac{\sum \beta_1 X_i^2+u_iX_i}{\sum X_i^2}$

Separate the sum into additive pieces:

$\hat{\beta_1}=\frac{\sum \beta_1 X_i^2}{\sum X_i^2}+\frac{u_i X_i}{\sum X_i^2}$

$$\beta_1$$ is constant, so we can pull it out of the first sum:

$\hat{\beta_1}=\beta_1 \frac{\sum X_i^2}{\sum X_i^2}+\frac{u_i X_i}{\sum X_i^2}$

Simplifying the first term, we are left with:

$\hat{\beta_1}=\beta_1 +\frac{u_i X_i}{\sum X_i^2}$

Now if we take expectations of both sides:

$E[\hat{\beta_1}]=E[\beta_1] +E\left[\frac{u_i X_i}{\sum X_i^2}\right]$

$$\beta_1$$ is a constant, so the expectation of $$\beta_1$$ is itself.

$E[\hat{\beta_1}]=\beta_1 +E\bigg[\frac{u_i X_i}{\sum X_i^2}\bigg]$

Using the properties of expectations, we can pull out $$\frac{1}{\sum X_i^2}$$ as a constant:

$E[\hat{\beta_1}]=\beta_1 +\frac{1}{\sum X_i^2} E\bigg[\sum u_i X_i\bigg]$

Again using the properties of expectations, we can put the expectation inside the summation operator (the expectation of a sum is the sum of expectations):

$E[\hat{\beta_1}]=\beta_1 +\frac{1}{\sum X_i^2}\sum E[u_i X_i]$

Under the exogeneity condition, the correlation between $$X_i$$ and $$u_i$$ is 0.

$E[\hat{\beta_1}]=\beta_1$