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    filename: simple_linear_regression.go
    path of file: ./repos/cockroach/pkg/sql/stats
    
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//
// Use of this software is governed by the Business Source License
// included in the file licenses/BSL.txt.
//
// As of the Change Date specified in that file, in accordance with
// the Business Source License, use of this software will be governed
// by the Apache License, Version 2.0, included in the file
// licenses/APL.txt.

package stats

import "github.com/cockroachdb/errors"

// float64SimpleLinearRegression fits a simple linear model y = α + βx to a set
// of observations of x and y, and uses this model to predict ŷₙ given an
// xₙ. The R² (goodness of fit) measurement of the model is also returned.
func float64SimpleLinearRegression(x, y []float64, xₙ float64) (yₙ, r2 float64) {
	// We use the ordinary least squares method to find the best fitting α̂ and β̂
	// for linear model y = α + βx. This allows us to solve for α̂ and β̂ directly:
	//
	//   α̂ = y̅ - β̂x̅
	//
	//       ∑ (xᵢ - x̅) (yᵢ - y̅)
	//   β̂ = ---------------------
	//           ∑ (xᵢ - x̅)²
	//
	//            ∑ (yᵢ - ŷᵢ)²
	//   R² = 1 - ------------
	//            ∑ (yᵢ - y̅)²
	//
	// where:
	// * all ∑ are from i=0 to i=n-1
	// * x̅ = (1/n) ∑ xᵢ
	// * y̅ = (1/n) ∑ yᵢ
	// * ŷᵢ = α̂ + β̂xᵢ
	//
	// See https://en.wikipedia.org/wiki/Simple_linear_regression and
	// https://en.wikipedia.org/wiki/Coefficient_of_determination for
	// background.
	//
	// Note that this uses a naive two-pass algorithm. We expect there to be less
	// than a dozen observations, so this is fine.
	// TODO(michae2): Switch to a one-pass algorithm based on Welford-Knuth, see:
	// https://www.johndcook.com/blog/running_regression/

	if len(x) != len(y) {
		panic(errors.AssertionFailedf("x and y had different numbers of observations"))
	}
	if len(x) == 0 {
		panic(e
    
func float64SimpleLinearRegression(x, y []float64, xₙ float64) (yₙ, r2 float64) {
	// We use the ordinary least squares method to find the best fitting α̂ and β̂
	// for linear model y = α + βx. This allows us to solve for α̂ and β̂ directly:
	//
	//   α̂ = y̅ - β̂x̅
	//
	//       ∑ (xᵢ - x̅) (yᵢ - y̅)
	//   β̂ = ---------------------
	//           ∑ (xᵢ - x̅)²
	//
	//            ∑ (yᵢ - ŷᵢ)²
	//   R² = 1 - ------------
	//            ∑ (yᵢ - y̅)²
	//
	// where:
	// * all ∑ are from i=0 to i=n-1
	// * x̅ = (1/n) ∑ xᵢ
	// * y̅ = (1/n) ∑ yᵢ
	// * ŷᵢ = α̂ + β̂xᵢ
	//
	// See https://en.wikipedia.org/wiki/Simple_linear_regression and
	// https://en.wikipedia.org/wiki/Coefficient_of_determination for
	// background.
	//
	// Note that this uses a naive two-pass algorithm. We expect there to be less
	// than a dozen observations, so this is fine.
	// TODO(michae2): Switch to a one-pass algorithm based on Welford-Knuth, see:
	// https://www.johndcook.com/blog/running_regression/

	if len(x) != len(y) {
		panic(errors.AssertionFailedf("x and y had different numbers of observations"))
	}
	if len(x) == 0 {
		panic(errors.AssertionFailedf("zero observations"))
	}

	n := len(x)

	// Calculate x̅ and y̅ (means of x and y).
	var Σx, Σy float64
	for i := 0; i < n; i++ {
		Σx += x[i]
		Σy += y[i]
	}
	xM, yM := Σx/float64(n), Σy/float64(n)

	// Solve for α̂ and β̂.
	var Σxx, Σxy, Σyy float64
	for i := 0; i < n; i++ {
		xD, yD := x[i]-xM, y[i]-yM
		Σxx += xD * xD
		Σxy += xD * yD
		Σyy += yD * yD
	}
	var α, β float64
	// Σxx could be zero if our x's are all the same. In this case Σxy is also
	// zero, and more generally y is not dependent on x, so β̂ is zero.
	if Σxx != 0 {
		β = Σxy / Σxx
	}
	α = yM - β*xM

	// Now use the model to calculate R² and predict ŷₙ given xₙ.
	var Σεε float64
	for i := 0; i < n; i++ {
		yHat := α + β*x[i]
		ε := y[i] - yHat
		Σεε += ε * ε
	}
	yₙ = α + β*xₙ
	// Σyy could be zero if our y's are all the same. In this case we will have a
	// perfect fit and Σεε will also be zero.
	if Σyy == 0 {
		r2 = 1
	} else {
		r2 = 1 - Σεε/Σyy
	}
	return yₙ, r2
}