maths.free › Statistics & Probability › Inference › Simple linear regression
Simple linear regression
In statistics, simple linear regression (SLR) is a linear regression model with a single explanatory variable.
Simple linear regression
In statistics, simple linear regression (SLR) is a linear regression model with a single explanatory variable. That is, it concerns two-dimensional sample points with one independent variable and one dependent variable (conventionally, the x and y coordinates in a Cartesian coordinate system) and finds a linear function (a non-vertical straight line) that, as accurately as possible, predicts the dependent variable values as a function of the independent variable. The adjective simple refers to the fact that the outcome variable is related to a single predictor.
It is common to make the additional stipulation that the ordinary least squares (OLS) method should be used: the accuracy of each predicted value is measured by its squared residual (vertical distance between the point of the data set and the fitted line), and the goal is to make the sum of these squared deviations as small as possible. In this case, the slope of the fitted line is equal to the correlation between y and x corrected by the ratio of standard deviations of these variables. The intercept of the fitted line is such that the line passes through the center of mass (x, y) of the data points.
Formulation and computation
Consider the model function \[y = \alpha + \beta x,\] which describes a line with slope β and y-intercept α. In general, such a relationship may not hold exactly for the largely unobserved population of values of the independent and dependent variables; we call the unobserved deviations from the above equation the errors. Suppose we observe n data pairs and call them {(xi, yi), i = 1, ..., n}. We can describe the underlying relationship between yi and xi involving this error term εi by
\[y_i = \alpha + \beta x_i + \varepsilon_i.\]
This relationship between the true (but unobserved) underlying parameters α and β and the data points is called a linear regression model.
The goal is to find estimated values \(\widehat\alpha\) and \(\widehat\beta\) for the parameters α and β which would provide the "best" fit in some sense for the data points. As mentioned in the introduction, in this article the "best" fit will be understood as in the least-squares approach: a line that minimizes the sum of squared residuals (see also Errors and residuals) \(\widehat\varepsilon_i\) (differences between actual and predicted values of the dependent variable y), each of which is given by, for any candidate parameter values \(\alpha\) and \(\beta\),
\[\widehat\varepsilon_i =y_i-\alpha -\beta x_i.\]
In other words, \(\widehat\alpha\) and \(\widehat\beta\) solve the following minimization problem:
\[(\hat\alpha,\, \hat\beta) = \operatorname{argmin}\left( Q(\alpha, \beta) \right),\] where the objective function Q is: \[Q(\alpha, \beta) = \sum_{i=1}^n\widehat\varepsilon_i^{\,2} = \sum_{i=1}^n (y_i -\alpha - \beta x_i)^2\ .\]
- \(\bar x\) and \(\bar y\) as the average of the xi and yi, respectively
- \(\Delta x_i\) and \(\Delta y_i\) as the deviations in xi and yi with respect to their respective means.
Condensed: the full section is in Wikipedia.
Expanded formulas
The above equations are efficient to use if the mean of the x and y variables (\(\bar{x} \text{ and } \bar{y}\)) are known. If the means are not known at the time of calculation, it may be more efficient to use the expanded version of the \(\widehat\alpha\text{ and }\widehat\beta\) equations. These expanded equations may be derived from the more general polynomial regression equations by defining the regression polynomial to be of order 1, as follows.
\[\begin{bmatrix} n & \sum_{i=1}^n x_i \\[1ex] \sum_{i=1}^n x_i & \sum_{i=1}^n x_i^2 \end{bmatrix} \begin{bmatrix} \widehat\alpha \\[1ex] \widehat\beta \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^n y_i \\[1ex] \sum_{i=1}^n y_i x_i \end{bmatrix}\]
The above system of linear equations may be solved directly, or stand-alone equations for \(\widehat\alpha\text{ and }\widehat\beta\) may be derived by expanding the matrix equations above. The resultant equations are algebraically equivalent to the ones shown in the prior paragraph, and are shown below without proof.
\[\begin{aligned} \widehat\alpha &= \frac{\sum\limits_{i=1}^n y_i \sum\limits_{i=1}^n x_i^2 - \sum\limits_{i=1}^n x_i \sum\limits_{i=1}^n x_i y_i }{n \sum\limits_{i=1}^n x_i^2 - \left(\sum\limits_{i=1}^n x_i\right)^2 } \\[2ex] \widehat\beta &= \frac {n \sum\limits_{i=1}^n x_i y_i - \sum\limits_{i=1}^n x_i \sum\limits_{i=1}^n y_i }{ n \sum\limits_{i=1}^n x_i^2 - \left(\sum\limits_{i=1}^n x_i\right)^2 } \end{aligned}\]
Relationship with the sample covariance matrix
The solution can be reformulated using elements of the covariance matrix: \[\widehat\beta = \frac{ s_{x, y} }{ s^2_{x} } = r_{xy} \frac{s_y}{s_x}\]
where
- rxy is the sample correlation coefficient between x and y
- sx and sy are the uncorrected sample standard deviations of x and y
- \(s^2_x\) and \(s_{x, y}\) are the sample variance and sample covariance, respectively
Substituting the above expressions for \(\widehat{\alpha}\) and \(\widehat{\beta}\) into the original solution yields
\[\frac{ y - \bar{y}}{s_y} = r_{xy} \frac{ x - \bar{x}}{s_x} .\]
This shows that rxy is the slope of the regression line of the standardized data points (and that this line passes through the origin). Since \(-1 \leq r_{xy} \leq 1\) then we get that if x is some measurement and y is a followup measurement from the same item, then we expect that y (on average) will be closer to the mean measurement than it was to the original value of x. This phenomenon is known as regressions toward the mean.
Generalizing the \(\bar x\) notation, we can write a horizontal bar over an expression to indicate the average value of that expression over the set of samples. For example:
\[\overline{xy} = \frac{1}{n} \sum_{i=1}^n x_i y_i.\]
Condensed: the full section is in Wikipedia.
Interpretation about the slope
By multiplying all members of the summation in the numerator by : \(\frac{x_i - \bar{x}}{x_i - \bar{x}} = 1\) (thereby not changing it):
\[\begin{aligned} \widehat\beta &= \frac{ \sum_{i=1}^n \left(x_i - \bar{x}\right) \left(y_i - \bar{y}\right) }{ \sum_{i=1}^n \left(x_i - \bar{x}\right)^2 } \\[1ex] &= \frac{ \sum_{i=1}^n \left(x_i - \bar{x}\right)^2 \frac{y_i - \bar{y}}{x_i - \bar{x}} }{ \sum_{i=1}^n \left(x_i - \bar{x}\right)^2 } \\[1ex] &= \sum_{i=1}^n \frac{ \left(x_i - \bar{x}\right)^2}{ \sum_{j=1}^n \left(x_j - \bar{x}\right)^2 } \frac{y_i - \bar{y}}{x_i - \bar{x}} \\[6pt] \end{aligned}\]
We can see that the slope (tangent of angle) of the regression line is the weighted average of \(\frac{y_i - \bar{y}}{x_i - \bar{x}}\) that is the slope (tangent of angle) of the line that connects the i-th point to the average of all points, weighted by \((x_i - \bar{x})^2\) 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.
Interpretation about the intercept
The parameter \(\widehat\alpha\) is the intercept of the linear function \(\begin{aligned} {y} & = \widehat\alpha\ + \widehat\beta\,{x}, \\[5pt] \end{aligned}\). Therefore, the \({y}\)-intercept of the function found with simple linear regression is
\[y_{\rm intercept} = \widehat\alpha = \bar{y} - \widehat\beta\,\bar{x}\].
Because \(\widehat\beta\) is the slope of the linear function, \(\widehat\beta = \tan(\theta)\). Therefore, the angle \(\theta\) the graph of the function makes with the \({x}\) axis is equal to
\[\theta = \arctan( \widehat\beta)\].
Interpretation about the correlation
In the above formulation, notice that each \(x_i\) is a constant ("known upfront") value, while the \(y_i\) are random variables that depend on the linear function of \(x_i\) and the random term \(\varepsilon_i\). This assumption is used when deriving the standard error of the slope and showing that it is unbiased.
In this framing, when \(x_i\) is not actually a random variable, what type of parameter does the empirical correlation \(r_{xy}\) estimate? The issue is that for each value i we'll have: \(E(x_i)=x_i\) and \(Var(x_i)=0\). A possible interpretation of \(r_{xy}\) is to imagine that \(x_i\) 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 \(x_i\) have the same expectation and some positive variance. With this interpretation we can think of \(r_{xy}\) as the estimator of the Pearson's correlation between the random variable y and the random variable x (as we just defined it).
Numerical properties
- The regression line goes through the center of mass point, \((\bar x,\, \bar y)\), if the model includes an intercept term (i.e., not forced through the origin).
- The sum of the residuals is zero if the model includes an intercept term: \[\sum_{i=1}^n \widehat{\varepsilon}_i = 0.\]
- The residuals and x values are uncorrelated (whether or not there is an intercept term in the model), meaning: \[\sum_{i=1}^n x_i \widehat{\varepsilon}_i \;=\; 0\]
- The relationship between \(\rho_{xy}\) (the correlation coefficient for the population) and the population variances of \(y\) (\(\sigma_y^2\)) and the error term of \(\varepsilon\) (\(\sigma_\varepsilon^2\)) is: \[\sigma_\varepsilon^2 = (1-\rho_{xy}^2)\sigma_y^2\] For extreme values of \(\rho_{xy}\) this is self evident. Since when \(\rho_{xy} = 0\) then \(\sigma_\varepsilon^2 = \sigma_y^2\). And when \(\rho_{xy} = 1\) then \(\sigma_\varepsilon^2 = 0\).
Statistical properties
Description of the statistical properties of estimators from the simple linear regression estimates requires the use of a statistical model. The following is based on assuming the validity of a model under which the estimates are optimal. It is also possible to evaluate the properties under other assumptions, such as inhomogeneity, but this is discussed elsewhere.
Unbiasedness
The estimators \(\widehat{\alpha}\) and \(\widehat{\beta}\) are unbiased.
To formalize this assertion we must define a framework in which these estimators are random variables. We consider the residuals εi 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 \(\widehat\alpha\) and \(\widehat\beta\) will themselves be random variables whose means will equal the "true values" α and β. This is the definition of an unbiased estimator.
Variance of the mean response
Since the data in this context is defined to be (x, y) pairs for every observation, the mean response at a given value of x, say xd, is an estimate of the mean of the y values in the population at the x value of xd, that is \(\hat{E}(y \mid x_d) \equiv\hat{y}_d\!\). The variance of the mean response is given by:
\[\operatorname{Var}\left(\hat{\alpha} + \hat{\beta}x_d\right) = \operatorname{Var}\left(\hat{\alpha}\right) + \left(\operatorname{Var} \hat{\beta}\right)x_d^2 + 2 x_d \operatorname{Cov} \left(\hat{\alpha}, \hat{\beta} \right) .\]
This expression can be simplified to
\[\operatorname{Var}\left(\hat{\alpha} + \hat{\beta}x_d\right) =\sigma^2\left(\frac{1}{m} + \frac{\left(x_d - \bar{x}\right)^2}{\sum (x_i - \bar{x})^2}\right),\]
where m is the number of data points.
To demonstrate this simplification, one can make use of the identity
\[\sum_i (x_i - \bar{x})^2 = \sum_i x_i^2 - \frac 1 m \left(\sum_i x_i\right)^2 .\]
Variance of the predicted response
The predicted response distribution is the predicted distribution of the residuals at the given point xd. So the variance is given by
\[\begin{aligned} \operatorname{Var}\left(y_d - \left[\hat{\alpha} + \hat{\beta} x_d \right] \right) &= \operatorname{Var} (y_d) + \operatorname{Var} \left(\hat{\alpha} + \hat{\beta}x_d\right) - 2\operatorname{Cov}\left(y_d,\left[\hat{\alpha} + \hat{\beta} x_d \right]\right)\\ &= \operatorname{Var} (y_d) + \operatorname{Var} \left(\hat{\alpha} + \hat{\beta}x_d\right). \end{aligned}\]
The second line follows from the fact that \(\operatorname{Cov}\left(y_d,\left[\hat{\alpha} + \hat{\beta} x_d \right]\right)\) is zero because the new prediction point is independent of the data used to fit the model. Additionally, the term \(\operatorname{Var} \left(\hat{\alpha} + \hat{\beta}x_d\right)\) was calculated earlier for the mean response.
Since \(\operatorname{Var}(y_d)=\sigma^2\) (a fixed but unknown parameter that can be estimated), the variance of the predicted response is given by
\[\begin{aligned} \operatorname{Var}\left(y_d - \left[\hat{\alpha} + \hat{\beta} x_d \right] \right) & = \sigma^2 + \sigma^2\left(\frac 1 m + \frac{\left(x_d - \bar{x}\right)^2}{\sum (x_i - \bar{x})^2}\right)\\[4pt] & = \sigma^2\left(1 + \frac 1 m + \frac{(x_d - \bar{x})^2}{\sum (x_i - \bar{x})^2}\right). \end{aligned}\]
Confidence intervals
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 do not tell us how precise the estimates are, i.e., how much the estimators \(\widehat{\alpha}\) and \(\widehat{\beta}\) 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:
- the errors in the regression are normally distributed (the so-called classic regression assumption), or
- 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.
Numerical example
This data set gives average masses for women as a function of their height in a sample of American women of age 30-39. Although the OLS article argues that it would be more appropriate to run a quadratic regression for this data, the simple linear regression model is applied here instead.
There are n = 15 points in this data set. Hand calculations would be started by finding the following five sums:
\[\begin{aligned} S_{x} &= \sum_i x_i \, = 24.76, &\qquad S_{y} &= \sum_i y_i \, = 931.17, \\[5pt] S_{xx} &= \sum_i x_i^2 = 41.0532, &\;\;\, S_{yy} &= \sum_i y_i^2 = 58498.5439, \\[5pt] S_{xy} &= \sum_i x_iy_i = 1548.2453 & \end{aligned}\]
These quantities would be used to calculate the estimates of the regression coefficients, and their standard errors.
\[\begin{aligned} \widehat\beta &= \frac{nS_{xy} - S_xS_y}{nS_{xx} - S_x^2} = 61.272 \\[8pt] \widehat\alpha &= \frac{1}{n}S_y - \widehat{\beta} \frac{1}{n}S_x = -39.062 \\[8pt] s_\varepsilon^2 &= \frac{1}{n(n - 2)} \left[ nS_{yy} - S_y^2 - \widehat\beta^2(nS_{xx} - S_x^2) \right] = 0.5762 \\[8pt] s_\widehat{\beta}^2 &= \frac{n s_\varepsilon^2}{nS_{xx} - S_x^2} = 3.1539 \\[8pt] s_\widehat{\alpha}^2 &= s_\widehat{\beta}^2 \frac{1}{n} S_{xx} = 8.63185 \end{aligned}\]
The 0.975 quantile of Student's t-distribution with 13 degrees of freedom is t13 = 2.1604, and thus the 95% confidence intervals for α and β are
\[\begin{aligned} & \alpha \in [\,\widehat{\alpha} \mp t^*_{13} s_\widehat{\alpha} \,] = [\,{-45.4},\ {-32.7}\,] \\[5pt] & \beta \in [\,\widehat{\beta} \mp t^*_{13} s_\widehat{\beta} \,] = [\, 57.4,\ 65.1 \,] \end{aligned}\]
Condensed: the full section is in Wikipedia.
Guhitamo iyi Rimwe, i Bya. Rimwe munsi, Cyangwa Ubwoko:.
A Kigenga Konti: & OYA & Ongera Ibisobanuro: ku, A Ibijyana Bya, in Rimwe, na A Kubaza Bigyanye iyi Ipaji:. Imibare ni Gufungura Kuri, in Cyangwa OYA.
Kwiyandikisha IfashayinjiraByakoreshejwe
Icyo ari cyo cyose Ibimenyetso ya: i Byuzuye Insobanuro:, A Ishusho, na buri Ibaruwa: in.
Abaturage
Mean or median, which should I use?
Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree. The gap is itself information.
What does a p-value actually say?
The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.
Why divide by n − 1 for the sample variance?
The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.
Bya iyi Ipaji: Bivuye Wikipedia (CC BY-SA 4.0). na -; Ikosa.
in Statistics & Probability
Sampling and dataDescribing data with graphsMean, median and modeProbabilityCounting: permutations and combinationsDiscrete random variablesContinuous random variablesThe normal distributionThe central limit theoremConfidence intervalsHypothesis testingComparing two samplesChi-square testsLinear regression and correlation