Simple linear regression

By multiplying all members of the summation in the numerator by : ${\displaystyle {\begin{aligned}{\frac {(x_{i}-{\bar {x}})}{(x_{i}-{\bar {x}})}}=1\end{aligned}}}$ (thereby not changing it):

${\displaystyle {\begin{aligned}{\widehat {\beta }}&={\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}{\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}}}={\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}{\frac {(y_{i}-{\bar {y}})}{(x_{i}-{\bar {x}})}}}{\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}}}=\sum _{i=1}^{n}{\frac {(x_{i}-{\bar {x}})^{2}}{\sum _{j=1}^{n}(x_{j}-{\bar {x}})^{2}}}{\frac {(y_{i}-{\bar {y}})}{(x_{i}-{\bar {x}})}}\\[6pt]\end{aligned}}}$

We can see that the slope (tangent of angle) of the regression line is the weighted average of ${\displaystyle {\frac {(y_{i}-{\bar {y}})}{(x_{i}-{\bar {x}})}}}$ that is the slope (tangent of angle) of the line that connects the i-th point to the average of all points, weighted by ${\displaystyle (x_{i}-{\bar {x}})^{2}}$ because the further the point is the more "important" it is, since small errors in its position will affect the slope connecting it to the center point more.

${\displaystyle {\begin{aligned}{\widehat {\alpha }}&={\bar {y}}-{\widehat {\beta }}\,{\bar {x}},\\[5pt]\end{aligned}}}$

Given ${\displaystyle {\widehat {\beta }}=\tan(\theta )=dy/dx\rightarrow dy=dx*{\widehat {\beta }}}$ with ${\displaystyle \theta }$ the angle the line makes with the positive x axis, we have ${\displaystyle y_{\rm {intersection}}={\bar {y}}-dx*{\widehat {\beta }}={\bar {y}}-dy}$

In the above formulation, notice that each ${\displaystyle x_{i}}$ is a constant ("known upfront") value, while the ${\displaystyle y_{i}}$ are random variables that depend on the linear function of ${\displaystyle x_{i}}$ and the random term ${\displaystyle \varepsilon _{i}}$. This assumption is used when deriving the standard error of the slope and showing that it is unbiased.

In this framing, when ${\displaystyle x_{i}}$ is not actually a random variable, what type of parameter does the empirical correlation ${\displaystyle r_{xy}}$ estimate? The issue is that for each value i we'll have: ${\displaystyle E(x_{i})=x_{i}}$ and ${\displaystyle Var(x_{i})=0}$. A possible interpretation of ${\displaystyle r_{xy}}$ is to imagine that ${\displaystyle x_{i}}$ defines a random variable drawn from the empirical distribution of the x values in our sample. For example, if x had 10 values from the natural numbers: [1,2,3...,10], then we can imagine x to be a Discrete uniform distribution. Under this interpretation all ${\displaystyle x_{i}}$ have the same expectation and some positive variance. With this interpretation we can think of ${\displaystyle r_{xy}}$ as the estimator of the Pearson's correlation between the random variable y and the random variable x (as we just defined it).

Sometimes it is appropriate to force the regression line to pass through the origin, because x and y are assumed to be proportional. For the model without the intercept term, y = βx, the OLS estimator for β simplifies to

${\displaystyle {\widehat {\beta }}={\frac {\sum _{i=1}^{n}x_{i}y_{i}}{\sum _{i=1}^{n}x_{i}^{2}}}={\frac {\overline {xy}}{\overline {x^{2}}}}}$

Substituting (x − h, y − k) in place of (x, y) gives the regression through (h, k):

${\displaystyle {\begin{aligned}{\widehat {\beta }}&={\frac {\sum _{i=1}^{n}(x_{i}-h)(y_{i}-k)}{\sum _{i=1}^{n}(x_{i}-h)^{2}}}={\frac {\overline {(x-h)(y-k)}}{\overline {(x-h)^{2}}}}\\[6pt]&={\frac {{\overline {xy}}-k{\bar {x}}-h{\bar {y}}+hk}{{\overline {x^{2}}}-2h{\bar {x}}+h^{2}}}\\[6pt]&={\frac {{\overline {xy}}-{\bar {x}}{\bar {y}}+({\bar {x}}-h)({\bar {y}}-k)}{{\overline {x^{2}}}-{\bar {x}}^{2}+({\bar {x}}-h)^{2}}}\\[6pt]&={\frac {\operatorname {Cov} (x,y)+({\bar {x}}-h)({\bar {y}}-k)}{\operatorname {Var} (x)+({\bar {x}}-h)^{2}}},\end{aligned}}}$

where Cov and Var refer to the covariance and variance of the sample data (uncorrected for bias).

The last form above demonstrates how moving the line away from the center of mass of the data points affects the slope.

The estimators ${\displaystyle {\widehat {\alpha }}}$ and ${\displaystyle {\widehat {\beta }}}$ are unbiased.

To formalize this assertion we must define a framework in which these estimators are random variables. We consider the residuals ε_i as random variables drawn independently from some distribution with mean zero. In other words, for each value of x, the corresponding value of y is generated as a mean response α + βx plus an additional random variable ε called the error term, equal to zero on average. Under such interpretation, the least-squares estimators ${\displaystyle {\widehat {\alpha }}}$ and ${\displaystyle {\widehat {\beta }}}$ will themselves be random variables whose means will equal the "true values" α and β. This is the definition of an unbiased estimator.

The formulas given in the previous section allow one to calculate the point estimates of α and β — that is, the coefficients of the regression line for the given set of data. However, those formulas don't tell us how precise the estimates are, i.e., how much the estimators ${\displaystyle {\widehat {\alpha }}}$ and ${\displaystyle {\widehat {\beta }}}$ vary from sample to sample for the specified sample size. Confidence intervals were devised to give a plausible set of values to the estimates one might have if one repeated the experiment a very large number of times.

The standard method of constructing confidence intervals for linear regression coefficients relies on the normality assumption, which is justified if either:

1. the errors in the regression are normally distributed (the so-called classic regression assumption), or
2. the number of observations n is sufficiently large, in which case the estimator is approximately normally distributed.

The latter case is justified by the central limit theorem.

Under the first assumption above, that of the normality of the error terms, the estimator of the slope coefficient will itself be normally distributed with mean β and variance ${\displaystyle \sigma ^{2}\left/\sum (x_{i}-{\bar {x}})^{2}\right.,}$ where σ² is the variance of the error terms (see Proofs involving ordinary least squares). At the same time the sum of squared residuals Q is distributed proportionally to χ² with n − 2 degrees of freedom, and independently from ${\displaystyle {\widehat {\beta }}}$. This allows us to construct a t-value

${\displaystyle t={\frac {{\widehat {\beta }}-\beta }{s_{\widehat {\beta }}}}\ \sim \ t_{n-2},}$

where

${\displaystyle s_{\widehat {\beta }}={\sqrt {\frac {{\frac {1}{n-2}}\sum _{i=1}^{n}{\widehat {\varepsilon }}_{i}^{\,2}}{\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}}}}}$

is the standard error of the estimator ${\displaystyle {\widehat {\beta }}}$.

This t-value has a Student's t-distribution with n − 2 degrees of freedom. Using it we can construct a confidence interval for β:

${\displaystyle \beta \in \left[{\widehat {\beta }}-s_{\widehat {\beta }}t_{n-2}^{*},\ {\widehat {\beta }}+s_{\widehat {\beta }}t_{n-2}^{*}\right],}$

at confidence level (1 − γ), where ${\displaystyle t_{n-2}^{*}}$ is the ${\displaystyle \scriptstyle \left(1\;-\;{\frac {\gamma }{2}}\right){\text{-th}}}$ quantile of the t_n−2 distribution. For example, if γ = 0.05 then the confidence level is 95%.

Similarly, the confidence interval for the intercept coefficient α is given by

${\displaystyle \alpha \in \left[{\widehat {\alpha }}-s_{\widehat {\alpha }}t_{n-2}^{*},\ {\widehat {\alpha }}+s_{\widehat {\alpha }}t_{n-2}^{*}\right],}$

at confidence level (1 − γ), where

${\displaystyle s_{\widehat {\alpha }}=s_{\widehat {\beta }}{\sqrt {{\frac {1}{n}}\sum _{i=1}^{n}x_{i}^{2}}}={\sqrt {{\frac {1}{n(n-2)}}\left(\sum _{i=1}^{n}{\widehat {\varepsilon }}_{i}^{\,2}\right){\frac {\sum _{i=1}^{n}x_{i}^{2}}{\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}}}}}}$

The US "changes in unemployment – GDP growth" regression with the 95% confidence bands.

The confidence intervals for α and β give us the general idea where these regression coefficients are most likely to be. For example, in the Okun's law regression shown here the point estimates are

${\displaystyle {\widehat {\alpha }}=0.859,\qquad {\widehat {\beta }}=-1.817.}$

The 95% confidence intervals for these estimates are

${\displaystyle \alpha \in \left[\,0.76,0.96\right],\qquad \beta \in \left[-2.06,-1.58\,\right].}$

In order to represent this information graphically, in the form of the confidence bands around the regression line, one has to proceed carefully and account for the joint distribution of the estimators. It can be shown[8] that at confidence level (1 − γ) the confidence band has hyperbolic form given by the equation

${\displaystyle (\alpha +\beta \xi )\in \left[\,{\widehat {\alpha }}+{\widehat {\beta }}\xi \pm t_{n-2}^{*}{\sqrt {\left({\frac {1}{n-2}}\sum {\widehat {\varepsilon }}_{i}^{\,2}\right)\cdot \left({\frac {1}{n}}+{\frac {(\xi -{\bar {x}})^{2}}{\sum (x_{i}-{\bar {x}})^{2}}}\right)}}\,\right].}$

When the model assumed the intercept is fixed and equal to 0 (${\displaystyle \alpha =0}$), the standard error of the slope turns into:

${\displaystyle s_{\widehat {\beta }}={\sqrt {{\frac {1}{n-1}}{\frac {\sum _{i=1}^{n}{\widehat {\varepsilon }}_{i}^{\,2}}{\sum _{i=1}^{n}x_{i}^{2}}}}}}$

With: ${\displaystyle {\hat {\varepsilon }}_{i}=y_{i}-{\hat {y}}_{i}}$

The alternative second assumption states that when the number of points in the dataset is "large enough", the law of large numbers and the central limit theorem become applicable, and then the distribution of the estimators is approximately normal. Under this assumption all formulas derived in the previous section remain valid, with the only exception that the quantile t*_n−2 of Student's t distribution is replaced with the quantile q* of the standard normal distribution. Occasionally the fraction 1/n−2 is replaced with 1/n. When n is large such a change does not alter the results appreciably.