public void GetInverseCdfValue_ResultOfCdfValue_InputValue([Range(-5.0, 5.0, 0.1)] double x)
{
double expected = x;
double probability = StandardNormalDistribution.GetCdfValue(x);
double actual = StandardNormalDistribution.GetInverseCdfValue(probability);
Assert.That(actual, Is.EqualTo(expected).Within(1E-8));
}
public void GetCdfValue_ResultOfInverseCdfValue_InputValue([Range(0.0, 1.0, 0.01)] double probability)
{
Assume.That(probability, Is.LessThanOrEqualTo(1.0));
double expected = probability;
double x = StandardNormalDistribution.GetInverseCdfValue(probability);
double actual = StandardNormalDistribution.GetCdfValue(x);
Assert.That(actual, Is.EqualTo(expected).Within(1E-9));
}
// [TestCase(0.99999999999999999, 8.4937932241095980744447188132289548161213991737094E+0)]
public void GetInverseCdfValue_TestCase_BenchmarkResult(double probability, double expected)
{
double actual = StandardNormalDistribution.GetInverseCdfValue(probability);
Assert.That(actual, Is.EqualTo(expected).Within(1E-7)); // low tolerance only!
}
/// <summary>Gets a specific value of the inverse of the cumulative distribution function.
/// </summary>
/// <param name="probability">The probability where to evaluate.</param>
/// <returns>The specified value of the inverse of the cumulative distribution function.</returns>
public double GetInverseCdfValue(double probability)
{
return(Math.Exp(StandardNormalDistribution.GetInverseCdfValue(probability) * Sigma + Mu) + Shift);
}