使用MathNET的置信区间

Question

我有一个IEnumerable<double>数据样本.我想计算信号/数据的90%置信区间.我有MathNET库,但我对如何正确使用库感到困惑.根据我的数据,我们的想法是返回两个包含原始信号置信区间的附加数据数组

using MathNet.Numerics.Statistics;
using MathNet.Numerics.Distributions;

public static List<double[]> ConfidenceIntervals(IEnumerable<double> sample, double interval)
{
    Contract.Requires(interval > 0 && interval < 1.0);
    int sampleSize = sample.Count();
    double alpha = 1.0 - interval;
    double mean = sample.Mean();
    double sd = sample.StandardDeviation();

    double t, mu;
    double[] upper = new double[sampleSize];
    double[] lower = new double[sampleSize];
    StudentT studentT = new StudentT(mean, alpha, sampleSize - 1);
    int index = 0;
    foreach (double d in sample)
    {
        t = studentT.CumulativeDistribution(d);
        double tmp = t * (sd / Math.Sqrt(sampleSize));
        mu = mean - tmp;
        upper[index] = d + mu;
        lower[index] = d - mu;
    }
    return new List<double[]>() { upper, lower };
}

这在数学方面确实不复杂,我对如何正确使用MathNET库中可用的函数/方法感到困惑.

Answer 1

我不完全确定我理解信号的置信区间应该如何应用于信号的每个样本,但是我们可以如下计算样本集的置信区间:

public static Tuple<double, double> A(double[] samples, double interval)
{
    double theta = (interval + 1.0)/2;
    double mean = samples.Mean();
    double sd = samples.StandardDeviation();
    double T = StudentT.InvCDF(0,1,samples.Length-1,theta);
    double t = T * (sd / Math.Sqrt(samples.Length));
    return Tuple.Create(mean-t, mean+t);
}

除了我们计算T的行不能编译,因为遗憾的是StudentT.InvCDF当前的Math.NET Numerics还没有.但在此期间,我们仍然可以将其作为一种解决方法进行数值评估:

var student = new StudentT(0,1,samples.Length-1);
double T = FindRoots.OfFunction(x => student.CumulativeDistribution(x)-theta,-800,800);

例如,对于16个样本和alpha 0.05,我们按预期得到2.131.如果有超过60-100个样本,这也可以用正态分布近似:

double T = Nomal.InvCDF(0,1,theta);

总而言之:

public static Tuple<double, double> B(double[] samples, double interval)
{
    double theta = (interval + 1.0)/2;
    double T = FindRoots.OfFunction(x => StudentT.CDF(0,1,samples.Length-1,x)-theta,-800,800);

    double mean = samples.Mean();
    double sd = samples.StandardDeviation();
    double t = T * (sd / Math.Sqrt(samples.Length));
    return Tuple.Create(mean-t, mean+t);
}

这还不是完整的答案,因为我理解你想以某种方式将置信区间应用于每个样本,但希望它有助于实现目标.

PS:使用Math.NET Numerics v3.0.0-alpha7