Linear regression implementation in pure C# with example of Bulgarian population prediction

LinearRegression.cs

namespace LinearRegression
{
    using System;
    using System.Diagnostics;

    public static class Program
    {
        public static void Main()
        {
            var xValues = new double[]
                              {
                                  1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004,
                                  2005, 2006, 2007, 2008, 2009
                              };
            var yValues = new double[]
                              {
                                  8669269, 8595500, 8484900, 8459800, 8427400, 8384700, 8340900, 8283200, 8230400, 8190900,
                                  8149468, 7932984, 7845841, 7801273, 7761049, 7720000, 7679290, 7640238, 7606551,
                                  7563710
                              };

            double rSquared, intercept, slope;
            LinearRegression(xValues, yValues, out rSquared, out intercept, out slope);

            Console.WriteLine($"R-squared = {rSquared}");
            Console.WriteLine($"Intercept = {intercept}");
            Console.WriteLine($"Slope = {slope}");

            var predictedValue = (slope * 2017) + intercept;
            Console.WriteLine($"Prediction for 2017: {predictedValue}");
        }

        /// <summary>
        /// Fits a line to a collection of (x,y) points.
        /// </summary>
        /// <param name="xVals">The x-axis values.</param>
        /// <param name="yVals">The y-axis values.</param>
        /// <param name="rSquared">The r^2 value of the line.</param>
        /// <param name="yIntercept">The y-intercept value of the line (i.e. y = ax + b, yIntercept is b).</param>
        /// <param name="slope">The slop of the line (i.e. y = ax + b, slope is a).</param>
        public static void LinearRegression(
            double[] xVals,
            double[] yVals,
            out double rSquared,
            out double yIntercept,
            out double slope)
        {
            if (xVals.Length != yVals.Length)
            {
                throw new Exception("Input values should be with the same length.");    
            }

            double sumOfX = 0;
            double sumOfY = 0;
            double sumOfXSq = 0;
            double sumOfYSq = 0;
            double sumCodeviates = 0;

            for (var i = 0; i < xVals.Length; i++)
            {
                var x = xVals[i];
                var y = yVals[i];
                sumCodeviates += x * y;
                sumOfX += x;
                sumOfY += y;
                sumOfXSq += x * x;
                sumOfYSq += y * y;
            }

            var count = xVals.Length;
            var ssX = sumOfXSq - ((sumOfX * sumOfX) / count);
            var ssY = sumOfYSq - ((sumOfY * sumOfY) / count);

            var rNumerator = (count * sumCodeviates) - (sumOfX * sumOfY);
            var rDenom = (count * sumOfXSq - (sumOfX * sumOfX)) * (count * sumOfYSq - (sumOfY * sumOfY));
            var sCo = sumCodeviates - ((sumOfX * sumOfY) / count);

            var meanX = sumOfX / count;
            var meanY = sumOfY / count;
            var dblR = rNumerator / Math.Sqrt(rDenom);

            rSquared = dblR * dblR;
            yIntercept = meanY - ((sCo / ssX) * meanX);
            slope = sCo / ssX;
        }
    }
}

Hi,
If we removed the last value of each of the above arrays (2009 & 7563710).
What formulae would I use to calculate the x-result (approx. 2009) when the y-input is 7563710?

This Version will work but it has some unnecessary calculations as well as too much Division. For example we are computing sumOfX * sumOfY in order to get therDenom but then we are computing sumOfX * sumOfY and storing it in the intermediate variable sCo.

A more readable way to do this with increased performance is to compute the sum of the codeviates while we are in the loop so that we can get the the sum of squares first then store them in intermediary variables. this allows us to get the answer without recalculating the exponents we can also avoid the overhead of a function call to the Math Library.

The time Complexity is the same in theory but in practice the code below is significantly faster.

       ``
        public static void LinearRegression(float[] xVals, float[] yVals, out double rSquared, out double yIntercept, out double slope)
        {
            double sumOfX = 0;
            double sumOfY = 0, 
            double sumOfXSq = 0, 
            double sumOfYSq = 0,
            double sumCodeviates = 0;
            int count = xVals.Length;

            // Compute all required sums in a single loop
            for (var i = 0; i < count; i++)
            {
                var x = xVals[i];
                var y = yVals[i];
                sumOfX += x;
                sumOfY += y;
                sumOfXSq += x * x;
                sumOfYSq += y * y;
                sumCodeviates += x * y;
            }

            // Precompute the reciprocal of count
            double reciprocalCount = 1.0 / count;

            // Precompute means using the reciprocal
            double meanX = sumOfX * reciprocalCount;
            double meanY = sumOfY * reciprocalCount;

            // Precompute terms for covariance and variance
            double ssX = sumOfXSq - (sumOfX * meanX);       // Variance of X
            double ssY = sumOfYSq - (sumOfY * meanY);       // Variance of Y
            double covariance = sumCodeviates - meanY * sumOfX; // Covariance of X and Y

            // Precompute reciprocals
            double reciprocalSSX = 1.0 / ssX;

            // Compute slope and y-intercept
            slope = covariance * reciprocalSSX;
            yIntercept = meanY - slope * meanX;

            // Compute r-squared using precomputed reciprocals
            rSquared = covariance * covariance * reciprocalSSX / ssY;
        }
       ``
        30% speed increase on 12 Core AMD Ryzen.