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1 Sectio 13 KolmogorovSmirov test. Suppose that we have a i.i.d. sample X 1,..., X with some ukow distributio P ad we would like to test the hypothesis that P is equal to a particular distributio P 0, i.e. decide betwee the followig hypotheses: H 0 : P = P 0, H 1 : P = P 0. We already kow how to test this hypothesis usig chisquared goodessoffit test. If distributio P 0 is cotiuous we had to group the data ad cosider a weaker discretized ull hypothesis. We will ow cosider a differet test for H 0 based o a very differet idea that avoids this discretizatio all data ormal fit 0.8 me data me fit wome data wome fit Cumulative probability Cumulative probability Data Data Figure 13.1: (a) Normal fit to the etire sample. (b) Normal fit to me ad wome separately. Example.(KS test) Let us agai look at the ormal body temperature dataset. Let all be a vector of all 130 observatios ad me ad wome be vectors of legth 65 each correspodig to me ad wome. First, we fit ormal distributio to the etire set all. MLE ˆµ ad ˆ are 83
2 mea(all) = , std(all,1) = We see i figure 13.1 (a) that this distributio fits the data very well. Let us perfom KS test that the data comes from this distributio N(ˆµ, ˆ2). To ru the test, first, we have to create a vector of N(ˆµ, ˆ2) c.d.f. values o the sample all (it is a required iput i Matlab KS test fuctio): CDFall=ormcdf(all,mea(all),std(all,1)); The we ru Matlab kstest fuctio [H,P,KSSTAT,CV] = kstest(all,[all,cdfall],0.05) which outputs H = 0, P = , KSSTAT = , CV = We accept H 0 sice the pvalue is CV is a critical value such that H 0 is rejected if statistic KSSTAT > CV. Remark. KS test is desiged to test a simple hypothesis P = P 0 for a give specified distributio P 0. I the example above we estimated this distributio, N(ˆµ, ˆ2) from the data so, formally, KS is iaccurate i this case. There is a versio of KS test, called Lilliefors test, that tests ormality of the distributio by comparig the data with a fitted ormal distributio as we did above, but with a correctio to give a more accurate approximatio of the distributio of the test statistic. Example. (Lilliefors test.) We use Matlab fuctio [H,P,LSTAT,CV] = lillietest(all) that outputs H = 0, P = , LSTAT = , CV = We accept the ormality of all with pvalue Example. (KS test for two samples.) Next, we fit ormal distributios to me ad wome separately, see figure 13.1 (b). We see that they are slightly differet so it is a atural questio to ask whether this differece is statistically sigificat. We already looked at this problem i the lecture o ttests. Uder a reasoable assumptio that body temperatures of me ad wome are ormally distributed, all ttests  paired, with equal variaces ad with uequal variaces  rejected the hypothesis that the mea body temperatures are equal µ me = µ wome. I this sectio we will describe a KS test for two samples that tests the hypothesis H 0 : P 1 = P 2 that two samples come from the same distributio. Matlab fuctio kstest2 [H,P,KSSTAT] = kstest2(me, wome) 84
3 outputs H = 0, P = , KSSTAT = It accepts the ull hypothesis sice pvalue > 0.05 =  a default value of the level of sigificace. Accordig to this test, the differece betwee two samples is ot sigificat eough to say that they have differet distributio. Let us ow explai some ideas behid these tests. Let us deote by F (x) = P(X 1 x) a c.d.f. of a true uderlyig distributio of the data. We defie a empirical c.d.f. by 1 F (x) = P (X x) = I(X i x) i=1 that couts the proportio of the sample poits below level x. For ay fixed poit x R the law of large umbers implies that F (x) = 1 I(X i x) EI(X 1 x) = P(X 1 x) = F (x), i=1 i.e. the proportio of the sample i the set (, x] approximates the probability of this set. It is easy to show from here that this approximatio holds uiformly over all x R: sup F (x) F (x) 0 xr i.e. the largest differece betwee F ad F goes to 0 i probability. The key observatio i the KolmogorovSmirov test is that the distributio of this supremum does ot deped o the ukow distributio P of the sample, if P is cotiuous distributio. Theorem 1. If F (x) is cotiuous the the distributio of does ot deped o F. Proof. Let us defie the iverse of F by sup F (x) F (x) xr F 1 (y) = mi{x : F (x) y}. The makig the chage of variables y = F (x) or x = F 1 (y) we ca write P(sup F (x) F (x) t) = P( sup F (F 1 (y)) y t). xr 0y1 Usig the defiitio of the empirical c.d.f. F we ca write F (F 1 (y)) = 1 I(X i F 1 (y)) = 1 I(F (X i ) y) i=1 i=1 85
4 Cumulative probability sup F (x) F (x) x frag replacemets Figure 13.2: KolmogorovSmirov test statistic. ad, therefore, 1 P( sup F (F 1 (y)) y t) = P sup I(F (X i ) y) y t. 0y1 0y1 The distributio of F (X i ) is uiform o the iterval [0, 1] because the c.d.f. of F (X 1 ) is P(F (X 1 ) t) = P(X 1 F 1 (t)) = F (F 1 (t)) = t. Therefore, the radom variables U i = F (X i ) for i are idepedet ad have uiform distributio o [0, 1], so we proved that 1 P(sup F (x) F (x) t) = P sup I(U i y) y t xr which is clearly idepedet of F. 0y1 i=1 i=1 86
5 To motivate KS test, we will eed oe more result which we will formulate without proof. First of all, let us ote that for a fixed poit x the CLT implies that d (F (x) F (x)) N 0, F (x)(1 F (x)) because F (x)(1 F (x)) is the variace of I(X 1 x). If turs out that if we cosider sup F (x) F (x) it will also coverge i distributio. Theorem 2. We have, xr P sup F (x) F (x) t H(t) = 1 2 ( 1) i 1 xr where H(t) is the c.d.f. of KolmogorovSmirov distributio. e 2i2 t Let us reformulate the hypotheses i terms of cumulative distributio fuctios: H 0 : F = F 0 vs. H 1 : F = F 0, where F 0 is the c.d.f. of P 0. Let us cosider the followig statistic D = sup F (x) F 0 (x). xr If the ull hypothesis is true the, by Theorem 1, we distributio of D ca be tabulated (it will deped oly o ). Moreover, if is large eough the the distributio of D is approximated by KolmogorovSmirov distributio from Theorem 2. O the other had, suppose that the ull hypothesis fails, i.e. F = F 0. Sice F is the true c.d.f. of the data, by law of large umbers the empirical c.d.f. F will coverge to F ad as a result it will ot approximate F 0, i.e. for large we will have sup F (x) F 0 (x) > x for some small eough. Multiplyig this by implies that If H 0 fails the D > + as. Therefore, to test H 0 we will cosider a decisio rule H = 0 : D c H 1 : D > c i=1 D = sup F (x) F 0 (x) >. xr The threshold c depeds o the level of sigificace ad ca be foud from the coditio = P( = H 0 H 0 ) = P(D c H 0 ). 87
6 Sice uder H 0 the distributio of D ca be tabulated for each, we ca fid the threshold c = c from the tables. I fact, most statistical table books have these distributios for up to 100. Seems like Matlab has these tables built i the kstest but the distributio of D is ot available as a separate fuctio. Whe is large the we ca use KS distributio to fid c sice = P(D c H 0 ) 1 H(c). ad we ca use the table for H to fid c. KS test for two samples. KolmogorovSmirov test for two samples is very similar. Suppose that a first sample X 1,..., X m of size m has distributio with c.d.f. F (x) ad the secod ssmple Y 1,..., Y of size has distributio with c.d.f. G(x) ad we wat to test H 0 : F = G vs. H 1 : F = G. If F m (x) ad G (x) are correspodig empirical c.d.f.s the the statistic m 1/2 D m = sup F m (x) G (x) m + satisfies Theorems 1 ad 2 ad the rest is the same Example. Let us cosider a sample of size 10: 0.58, 0.42, 0.52, 0.33, 0.43, 0.23, 0.58, 0.76, 0.53, 0.64 ad let us test the hypothesis that the distributio of the sample is uiform o [0, 1] i.e. H 0 : F (x) = F 0 (x) = x. The figure 13.3 shows the c.d.f. F 0 ad empirical c.d.f. F (x). To compute D we otice that the largest differece betwee F 0 (x) ad F (x) is achieved either before or after oe of the jumps, i.e. F (X i ) F (X i )  before the ith jump sup F (x) F (x) = max 1i F (X i ) F (X i )  after the ith jump. 0x1 Writig these differeces for our data we get before the jump after the jump The largest value will be achieved at = 0.26 ad, therefore, D = sup F (x) x = = x1 x 88
7 1 Empirical CDF sup F (x) x 0x1 = 0.26 F(x) frag replacemets x Figure 13.3: F ad F 0 i the example. If we take the level of sigificace = 0.05 ad use KS approximatio of Theorem 2 to fid threshold c: 1 H(c) = 0.05 c = 1.35, the accordig to KS test = H 1 : D 1.35 H 2 : D > 1.35 we accept the ull hypothesis H 0 sice D = 0.82 < c = However, we have oly = 10 observatios so the approximatio of Theorem 2 might be iaccurate. We could use the advaced statistical tables to fid the distibutio of D for = 10 or let Matlab do it. Ruig [H,P,KSSTAT,CV] = kstest(x,[x,x],0.05) (remark 1 ) outputs H = 0, P = , KSSTAT = , CV = Here the secod iput of kstest should be a 2 matrix where the first colum is the data X ad the secod colum is the correspodig values of c.d.f. F 0 (x). But sice we test with F 0 (x) = x, the secod colum is equal to X ad, thus, we iput [X,X] 89
8 Sice Matlab fuctio kstest does ot scale the statistic by sice it is usig the exact distributio of sup x F (x) F (x) istead of approximatio of Theorem 2, the critical value CV mupliplied by, i.e = will be exactly our threshold such that P(D > c H 0 ) = = It is slightly differet from c = 1.35 give by the approximatio of Theorem 2. So for small sample sizes it is better to use the exact distributio of D. 90
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