Simple linear regression is a technique to analyze a linear relationship
between two variables.
A common generalization is to study relationships between two variables that
can be transformed into a linear relationship, which we will call linearized.
Simple linear regression is implemented by the
SimpleRegressionModel class,
and supports both linear and linearized regression.
Constructing Simple Regression Models
The SimpleRegressionModel
class has six constructors that come in pairs.
The first constructor takes two vectors and creates a linear regression model.
The first vector contains the data for the dependent variable.
The second vector contains the data for the independent variable.
The dependent variable is named y, while
the independent variable is named x.
double[] xData = {0.2, 337.4, 118.2, 884.6, 10.1,
226.5, 666.3, 996.3, 448.6, 777.0, 558.2, 0.4, 0.6,
775.5, 666.9, 338.0, 447.5, 11.6, 556.0, 228.1, 995.8,
887.6, 120.2, 0.3, 0.3, 556.8, 339.1, 887.2, 999.0,
779.0, 11.1, 118.3, 229.2, 669.1, 448.9, 0.5};
double[] yData = {0.1, 338.8, 118.1, 888.0, 9.2,
228.1, 668.5, 998.5, 449.1, 778.9, 559.2, 0.3, 0.1,
778.1, 668.8, 339.3, 448.9, 10.8, 557.7, 228.3, 998.0,
888.8, 119.6, 0.3, 0.6, 557.6, 339.3, 888.0, 998.5,
778.9, 10.2 , 117.6, 228.9, 668.4, 449.2, 0.2};
var y = Vector.Create(yData);
var x = Vector.Create(xData);
var model1 = new SimpleRegressionModel(y, x);
var model2 = new SimpleRegressionModel(yData, xData);
Dim xData = {0.2, 337.4, 118.2, 884.6, 10.1,
226.5, 666.3, 996.3, 448.6, 777.0, 558.2, 0.4, 0.6,
775.5, 666.9, 338.0, 447.5, 11.6, 556.0, 228.1, 995.8,
887.6, 120.2, 0.3, 0.3, 556.8, 339.1, 887.2, 999.0,
779.0, 11.1, 118.3, 229.2, 669.1, 448.9, 0.5}
Dim yData = {0.1, 338.8, 118.1, 888.0, 9.2,
228.1, 668.5, 998.5, 449.1, 778.9, 559.2, 0.3, 0.1,
778.1, 668.8, 339.3, 448.9, 10.8, 557.7, 228.3, 998.0,
888.8, 119.6, 0.3, 0.6, 557.6, 339.3, 888.0, 998.5,
778.9, 10.2, 117.6, 228.9, 668.4, 449.2, 0.2}
Dim y = Vector.Create(yData)
Dim x = Vector.Create(xData)
Dim model1 = New SimpleRegressionModel(y, x)
Dim model2 = New SimpleRegressionModel(yData, xData)
No code example is currently available or this language may not be supported.
let xData = [| 0.2; 337.4; 118.2; 884.6; 10.1;
226.5; 666.3; 996.3; 448.6; 777.0; 558.2; 0.4; 0.6;
775.5; 666.9; 338.0; 447.5; 11.6; 556.0; 228.1; 995.8;
887.6; 120.2; 0.3; 0.3; 556.8; 339.1; 887.2; 999.0;
779.0; 11.1; 118.3; 229.2; 669.1; 448.9; 0.5 |]
let yData = [| 0.1; 338.8; 118.1; 888.0; 9.2;
228.1; 668.5; 998.5; 449.1; 778.9; 559.2; 0.3; 0.1;
778.1; 668.8; 339.3; 448.9; 10.8; 557.7; 228.3; 998.0;
888.8; 119.6; 0.3; 0.6; 557.6; 339.3; 888.0; 998.5;
778.9; 10.2 ; 117.6; 228.9; 668.4; 449.2; 0.2 |]
let y = Vector.Create(yData)
let x = Vector.Create(xData)
let model1 = SimpleRegressionModel(y, x)
let model2 = SimpleRegressionModel(
Vector.Create(yData), Vector.Create(xData))
Note that, because arrays can be implicitly converted to vectors,
it is also possible to pass arrays instead of vectors.
The second constructor takes 3 arguments. The first argument is an
IDataFrame
(a DataFrameR, C
or MatrixT) that
contains the variables to be used in the regression.
The second argument is a string containing the name of the dependent variable.
The third argument is a string containing the name of the independent variable.
The two names must exist in the data frame and their element type must be
convertible to Double.
var dataFrame = DataFrame.FromColumns(new Dictionary<string, object>()
{ { "y", y }, {"x", x }});
var model3 = new SimpleRegressionModel(dataFrame, "y", "x");
Dim frame = DataFrame.FromColumns(New Dictionary(Of String, Object)() From
{{"y", y}, {"x", x}})
Dim model3 = New SimpleRegressionModel(frame, "y", "x")
No code example is currently available or this language may not be supported.
let columns = Dictionary<string,obj>()
[ "x", x ; "y", y ] |> Seq.iter columns.Add
let dataFrame = DataFrame.FromColumns<string>(columns)
let model3 = SimpleRegressionModel(dataFrame, "y", "x")
The third and fourth constructors are similar to the first two, but add a third argument
that specifies the kind of relationship between the two variables. This argument is of type
SimpleRegressionKind,
which can take on the following values:
Value | Description |
---|
Exponential |
The logarithm of the dependent variable depends linearly on the independent variable:
Y = a * bX.
|
Inverse |
The dependent variable depends linearly on the inverse (reciprocal)
of the independent variable: Y = a / x + b.
|
Linear |
The dependent variable depends linearly on the independent variable:
Y = a X + b. This is the default.
|
Logarithmic |
The dependent variable depends linearly on the logarithm of the independent variable:
Y = a log X + b |
Power |
The logarithm of the dependent variable depends linearly on
the logarithm of the independent variable: Y = a * Xb.
|
var model4 = new SimpleRegressionModel(y, x,
SimpleRegressionKind.Exponential);
var model5 = new SimpleRegressionModel(dataFrame, "y", "x",
SimpleRegressionKind.Exponential);
Dim model4 = New SimpleRegressionModel(y, x,
SimpleRegressionKind.Exponential)
Dim model5 = New SimpleRegressionModel(frame, "y", "x",
SimpleRegressionKind.Exponential)
No code example is currently available or this language may not be supported.
let model4 = SimpleRegressionModel(y, x, SimpleRegressionKind.Exponential)
let model5 = SimpleRegressionModel(dataFrame, "y", "x",
SimpleRegressionKind.Exponential)
The fifth constructor is like the first but has 5 arguments in total:
The third argument is a vector containing weights for the observations.
The fourth argument is the kind of relationship between the variables.
The fifth argument is a boolean that indicates whether the intercept
(constant term) should be excluded from the model.
The sixth constructor is like the second and takes the same 3 additional arguments:
var weights = Vector.Reciprocal(x);
var model6 = new SimpleRegressionModel(y, x, weights,
kind: SimpleRegressionKind.Exponential, noIntercept: false);
Dim weights = Vector.Reciprocal(x)
Dim model6 = New SimpleRegressionModel(y, x, weights,
SimpleRegressionKind.Exponential, False)
No code example is currently available or this language may not be supported.
let weights = Vector.Reciprocal(x)
let model6 = SimpleRegressionModel(y, x, weights,
kind= SimpleRegressionKind.Exponential, noIntercept= false)
Computing the Regression Curve
The Compute
method performs the actual analysis. Most properties and methods throw an exception
when they are accessed before the
Compute
method is called. You can verify that the model has been calculated by inspecting the
Computed property.
The GetRegressionLine
method returns a Polynomial
that represents the regression line. For linearized regression, the
GetRegressionCurve
method returns a Curve
object that represents the fitted curve:
model1.Fit();
var line = model1.GetRegressionLine();
model4.Fit();
var curve = model4.GetRegressionCurve();
model1.Fit()
Dim line = model1.GetRegressionLine()
model4.Fit()
Dim curve = model4.GetRegressionCurve()
No code example is currently available or this language may not be supported.
model1.Fit()
let line = model1.GetRegressionLine()
model4.Fit()
let curve = model4.GetRegressionCurve()
The Predictions
property returns a vector that contains the values of the dependent variable as
predicted by the model. The
Residuals
property returns a vector containing the difference between the actual and
the predicted values of the dependent variable.
Both vectors contain one element for each observation.
The SimpleRegressionModel class'
Parameters
property returns a ParameterVectorT
that contains the parameters of the regression model.
The members of this collection are of type
ParameterT.
Regression parameters are created by the model. You cannot create
them directly.
A simple linear regression model takes two arguments. The first, with index 0, is the intercept:
the Y-value where the regression line crosses the Y-axis. The second parameter, with index 1,
is the slope of the regression line. For linearized
The Parameter class has four useful properties. The Value property returns the numerical value of the parameter, while the
StandardError property returns the standard deviation
of the parameter's distribution.
The Statistic
property returns the value of the t-statistic corresponding to the hypothesis
that the parameter equals zero. The
PValue property
returns the corresponding p-value. A high p-value indicates that the variable associated
with the parameter does not make a significant contribution to explaining the
data. The p-value always corresponds to a two-tailed test.
The following example prints the properties of the slope
parameter of our earlier example:
var slope = model1.Parameters[1];
Console.WriteLine("Name: {0}", slope.Name);
Console.WriteLine("Value: {0}", slope.Value);
Console.WriteLine("St.Err.: {0}", slope.StandardError);
Console.WriteLine("t-statistic: {0}", slope.Statistic);
Console.WriteLine("p-value: {0}", slope.PValue);
Dim slope = model1.Parameters(1)
Console.WriteLine("Name: {0}", slope.Name)
Console.WriteLine("Value: {0}", slope.Value)
Console.WriteLine("St.Err.: {0}", slope.StandardError)
Console.WriteLine("t-statistic: {0}", slope.Statistic)
Console.WriteLine("p-value: {0}", slope.PValue)
No code example is currently available or this language may not be supported.
let slope = model1.Parameters.[1]
Console.WriteLine("Name: 0}", slope.Name)
Console.WriteLine("Value: 0}", slope.Value)
Console.WriteLine("St.Err.: 0}", slope.StandardError)
Console.WriteLine("t-statistic: 0}", slope.Statistic)
Console.WriteLine("p-value: 0}", slope.PValue)
The Parameter class has one method: GetConfidenceInterval. This method takes one argument:
a confidence level between 0 and 1. A value of 0.95 corresponds to a confidence level of 95%. The method returns the
confidence interval for the parameter at the specified confidence level as an Interval structure.
Verifying the Quality of the Regression Line
The ResidualSumOfSquares property
gives the sum of the squares of the residuals. The regression line was found by minimizing this value. The StandardError property gives the standard deviation of
the data.
The RSquared property returns the coefficient
of determination. It is the ratio of the variation in the data that is explained by the model compared to the total
variation in the data. Its value is always between 0 and 1, where 0 means the model explains nothing and 1 means the
model explains the data perfectly.
An entirely different assessment is available through an analysis of variance.
Here, the variation in the data is decomposed into a component explained by the model,
and the variation in the residuals. The
FStatistic
property returns the F-statistic for the ratio of these two variances. The
PValue
property returns the corresponding p-value. A low p-value means that it is unlikely
that the variation in the model is the same as the
variation in the residuals. This means that the model is significant.
The results of the analysis of variance are also summarized in the regression model's
ANOVA table, returned by the
AnovaTable
property.