柯爾莫哥洛夫-斯米爾諾夫檢驗（Колмогоров-Смирнов檢驗）基于累計分布函數，用以檢驗兩個經驗分布是否不同或一個經驗分布與另一個理想分布是否不同。

在進行cumulative probability統計(如下圖)的時候，你怎么知道組之間是否有顯著性差異？有人首先想到單因素方差分析或雙尾檢驗（2 tailed TEST）。其實這些是不準確的，最好采用Kolmogorov-Smirnov test（柯爾莫諾夫-斯米爾諾夫檢驗）來分析變量是否符合某種分布或比較兩組之間有無顯著性差異。

Kolmogorov-Smirnov test原理：尋找最大距離（Distance）， 所以常簡稱為D法。 適用于大樣本。KS test checks if two independent distributions are similar or different, by generating cumulative probability plots for two distributions and finding the distance along the y-axis for a given x values between the two curves. From all the distances calculated for each x value, the maximum distance is searched.

如何分析結果呢？This maximum distance or maximum difference is then plugged into KS probability function to calculate the probability value. The lower the probability value is the less likely the two distributions are similar. Conversely, the higher or more close to 1 the value is the more similar the two distributions are.極端情況：如果P值為１的話，說明兩給數據基本相同，如果P值無限接近０，說明兩組數據差異性極大。

有一個網站可以進行在線的統計，你只需要輸入數據就可以了。地址如下：http://www.physics.csbsju.edu/stats/KS-test.n.plot_form.html

當然還有更多的軟件支持這個統計，如SPSS,SAS,MiniAnalysis,Clampfit10

根據軟件統計出來后給出的結果決定有沒有顯著性差異，如果D_max值>D_0.05。則認為有顯著性差異。D_0.05的經驗算法：1.36/SQRT(N) 其中SQRT為平方要，N為樣本數。D_0.0１經驗算法1.64/SQRT(N) 。當然最準確的辦法還是去查KS檢定表。不過大多數軟件如CLAMPFIT,MINIANALYSIS統計出來的結果都是直接有P值。根據這個值（alpha=0.05）就可以斷定有沒有差異了。

在統計學中，柯爾莫可洛夫-斯米洛夫檢驗基于累計分布函數，用以檢驗兩個經驗分布是否不同或一個經驗分布與另一個理想分布是否不同。

在進行累計概率（cumulative probability）統計的時候，你怎么知道組之間是否有顯著性差異？有人首先想到單因素方差分析或雙尾檢驗（2 tailedTEST）。其實這些是不準確的，最好采用Kolmogorov-Smirnovtest（柯爾莫諾夫-斯米爾諾夫檢驗）來分析變量是否符合某種分布或比較兩組之間有無顯著性差異。

分類：

1、Single sample Kolmogorov-Smirnov goodness-of-fit hypothesis test.

采用柯爾莫諾夫-斯米爾諾夫檢驗來分析變量是否符合某種分布，可以檢驗的分布有正態分布、均勻分布、Poission分布和指數分布。指令如下：

>> H = KSTEST(X,CDF,ALPHA,TAIL) % X為待檢測樣本，CDF可選：如果空缺，則默認為檢測標準正態分布；

如果填寫兩列的矩陣，第一列是x的可能的值，第二列是相應的假設累計概率分布函數的值G(x)。ALPHA是顯著性水平（默認0.05）。TAIL是表示檢驗的類型（默認unequal,不平衡）。還有larger，smaller可以選擇。

如果，H=1 則否定無效假設； H=0，不否定無效假設（在alpha水平上）

例如，

x = -2:1:4
x =
-2-101234

[h,p,k,c] = kstest(x,[],0.05,0)
h =
0
p =
0.13632
k =
0.41277
c =
0.48342

The test fails to reject the null hypothesis that the values come from a standard normal distribution.

2、Two-sample Kolmogorov-Smirnov test

檢驗兩個數據向量之間的分布的。

>>[h,p,ks2stat] = kstest2(x1,x2,alpha,tail)

% x1,x2都為向量，ALPHA是顯著性水平（默認0.05）。TAIL是表示檢驗的類型（默認unequal,不平衡）。

例如，x = -1:1:5
y = randn(20,1);
[h,p,k] = kstest2(x,y)
h =
0
p =
0.0774
k =
0.5214

wiki翻譯起來太麻煩，還有可能曲解本意，最好看原版解釋。

Instatistics, theKolmogorov–Smirnovtest (K–S test)is a form ofminimum distance estimationused as anonparametric testof equality of one-dimensionalprobability distributionsused to compare asamplewith a reference probability distribution (one-sample K–S test), or to compare two samples (two-sample K–S test). The Kolmogorov–Smirnov statistic quantifies adistancebetween theempirical distribution functionof the sample and thecumulative distribution functionof the reference distribution, or between the empirical distribution functions of two samples. Thenull distributionof this statistic is calculated under the null hypothesis that the samples are drawn from the same distribution (in the two-sample case) or that the sample is drawn from the reference distribution (in the one-sample case). In each case, the distributions considered under the null hypothesis are continuous distributions but are otherwise unrestricted.

The two-sample KS test is one of the most useful and general nonparametric methods for comparing two samples, as it is sensitive to differences in both location and shape of the empirical cumulative distribution functions of the two samples.

The Kolmogorov–Smirnov test can be modified to serve as agoodness of fittest. In the special case of testing fornormalityof the distribution, samples are standardized and compared with a standard normal distribution. This is equivalent to setting the mean and variance of the reference distribution equal to the sample estimates, and it is known that using the sample to modify the null hypothesis reduces thepowerof a test. Correcting for this bias leads to theLilliefors test. However, even Lilliefors' modification is less powerful than theShapiro–Wilk testorAnderson–Darling testfor testing normality.^[1]

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