Question

我正在考虑将Johnson SU分布拟合到一套经验性的S＆amp; P 500指数回报中。我的理解（免责声明：不是数学家）是这个分布包含第三和第四个时刻（倾斜和峰度）。除loc（平均值）和scale（标准偏差）外，johnsonsu还需要另外两个参数a和b。但这些参数的顺序和规格令人困惑。

这就是我的困惑所在：如果我回到SPDR S＆amp; P 500 ETF Trust（SPY），我会得到以下经验统计数据：

from pandas_datareader.data import DataReader as dr
r = dr('SPY', 'google', start='2000')['Close'].pct_change().dropna()
mean, var, std, skew, kurt = r.mean(), r.var(0), r.std(0), r.skew(), r.kurt() # ddof = 0
# mean: 0.00027732907268771364
# var: 0.00014416720067485022
# std: 0.012006964673673785

现在，如果我符合此经验数据的正态分布，.fit应该返回loc和scale参数。（正常分发所需的一切。）检查出来：

import scipy.stats as scs

normmean, normstd = scs.norm.fit(r)
print(np.allclose(normmean, mean))
print(np.allclose(normstd, std))

True
True

但scs.johnsonsu.fit：

返回的内容不太清楚

print(scs.johnsonsu.fit(r))
(0.098009661042083682, 1.022060362199493, 0.0013471690867203458, 0.0072653444313926403)

这些应该是分布的four parameters： xi，gamma，delta，lam 。

但我不能让他们回到经验的意思，应该是：

I.e。：

def johnsonmean(gamma, xi, delta, lam):
    mean = xi - lam * np.exp(delta ** -2 / 2) * np.sinh(gamma / delta)
    return mean
gamma, xi, delta, lam = scs.johnsonsu.fit(r) # correct order?
print(johnsonmean(gamma, xi, delta, lam))
-inf

和

mean, var, skew, kurt = scs.johnsonsu.stats(loc=xi, scale=lam, 
                                            a=gamma, b=delta, moments='msvk')

获得了一堆NaN s。

Answer 1

它们是Johnson SU的参数。记住，你得到的样本的平均值与分布的平均值不同。这是平均值的表达式

这里是方差的表达式：

在您的代码中，ξ为Set wApp = CreateObject("Word.Application") wApp.Visible = True Set wDoc = wApp.Documents.Open(filename:=ThisWorkbook.path & "\TestAccount.docx") With wDoc .Shapes("InvoiceXLS").OLEFormat.Edit ' ??? ' How do I get a Worksheet object that I can work with?? ' Or, just paste in a whole table over top? End With，λ为loc，γ为scale，δ为a。 sinh ^-1（x）等于log（x + sqrt（1 + x ²））。

因此，检查拟合的返回值，为所有四个参数赋值，然后计算分布均值并与样本均值进行比较。如果有效，请重复练习以获得差异

更新

我尝试了您的代码，建议检查均值和方差，但效果很好，请查看下面的

并产生了输出：

import sys
import math

from pandas_datareader.data import DataReader as dr
import scipy.stats as scs

def read_data():
    return dr('SPY', 'google', start='2000')['Close'].pct_change().dropna()

def johnsonsu_mean(a, b, loc, scale):
    """
    Johnson SU mean according to https://en.wikipedia.org/wiki/Johnson%27s_SU-distribution
    """
    v = loc - scale * math.exp(0.5 / b**2) * math.sinh(a/b)
    return v

def johnsonsu_var(a, b, loc, scale):
    """
    Johnson SU variance according to https://en.wikipedia.org/wiki/Johnson%27s_SU-distribution
    """
    t = math.exp(1.0 / b**2)
    v = 0.5*scale**2 * (t - 1.0) * (t * math.cosh(2.0*a/b) + 1.0)
    return v

def johnsonsu_median(a, b, loc, scale):
    """
    Johnson SU median according to https://en.wikipedia.org/wiki/Johnson%27s_SU-distribution
    """
    v = loc + scale * math.sinh(-a/b)
    return v

def main(r):
    sample_mean, sample_med, sample_var, sample_std, sample_skew, sample_kurt = r.mean(), r.median(), r.var(0), r.std(0), r.skew(), r.kurt()

    a, b, loc, scale = scs.johnsonsu.fit(r) # fit the data and get distribution parameters back

    # distribution mean and variance according to SciPy
    dist_mean = scs.johnsonsu.mean(a, b, loc, scale)
    dist_med  = scs.johnsonsu.median(a, b, loc, scale)
    dist_var  = scs.johnsonsu.var(a, b, loc, scale)

    # distribution mean, var vs sample ones
    print("{0} {1}".format(sample_mean, dist_mean))
    print("{0} {1}".format(sample_med, dist_med))
    print("{0} {1}".format(sample_var, dist_var))
    print("")

    # distribution mean and variance according to Wiki vs SciPy
    print("{0} {1}".format(dist_mean, johnsonsu_mean(a, b, loc, scale)))
    print("{0} {1}".format(dist_var, johnsonsu_var(a, b, loc, scale)))
    print("{0} {1}".format(dist_med, johnsonsu_median(a, b, loc, scale)))

if __name__ == "__main__":
    r = read_data()
    main(r)

    sys.exit(0)

scipy.stats.johnsonsu中的a和b参数是什么？

1 个答案: