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;
        }
    }
}

Thanks a lot! :)

If anyone wants to calculate the sample standard deviation (sigma) with this, just put this at the bottom and also add it as another out parameter: stdDev = Math.Sqrt(ssY / (count - 1));

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?