Apakah ada versi uji ekivalensi sederhana dari tes Kolmogorov-Smirnov?

Ok, ini usaha pertamaku. Tutup pengawasan dan komentar dihargai!

Hipotesa Dua Sampel
Jika kita dapat membingkai tes hipotesis Kolmogorov-Smirnov satu sisi dua sampel , dengan hipotesis nol dan alternatif di sepanjang baris berikut:

H , dan $_{0}\text{: }F_{Y}\left(t\right) \geq F_{X}\left(t\right)$

H , untuk setidaknya satu , di mana: $_{\text{A}}\text{: }F_{Y}\left(t\right) < F_{X}\left(t\right)$ $t$

statistik uji $D^{-}=\left|\min_{t}\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right|$ sesuai dengan H ; $_0\text{: }F_{Y}\left(t\right) \geq F_{X}\left(t\right)$
statistik uji $D^{+}=\left|\max_{t}\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right|$ sesuai dengan H ; dan $_0\text{: }F_{Y}\left(t\right) \leq F_{X}\left(t\right)$
$F_{Y}\left(t\right)$ & $F_{X}\left(t\right)$ adalah empirisdari sampel $Y$ dan $X$ ,

maka harus masuk akal untuk membuat hipotesis interval umum untuk uji kesetaraan di sepanjang garis ini (dengan asumsi bahwa interval kesetaraan simetris untuk saat ini):

H , dan $^{-}_0\text{: }\left|F_{Y}\left(t\right) - F_{X}\left(t\right)\right| \geq \Delta$

H , setidaknya untuk satu . $^{-}_{\text{A}}\text{: }\left|F_{Y}\left(t\right) - F_{X}\left(t\right)\right| < \Delta$ $t$

Ini akan menerjemahkan ke spesifik dua satu sisi "negativist" nol hipotesis untuk menguji untuk kesetaraan (dua hipotesis ini mengambil bentuk yang sama, karena kedua dan secara ketat non-negatif): $D^{+}$ $D^{-}$

H , atau $^{-}_{01}\text{: }D^{+} \geq \Delta$

H . $^{-}_{02}\text{: }D^{-} \geq \Delta$

Menolak kedua H dan H akan membawa kita untuk menyimpulkan bahwa . Tentu saja, interval ekivalensi tidak perlu simetris, dan dan dapat diganti dengan (lebih rendah) dan (atas) untuk masing-masing hipotesis nol satu sisi. $^{-}_{01}$ $^{-}_{02}$ $-\Delta < F_{Y}\left(t\right) - F_{X}\left(t\right) < \Delta$ $-\Delta$ $\Delta$ $\Delta_{2}$ $\Delta_{1}$

Statistik Uji (Diperbarui: Delta berada di luar tanda nilai absolut)
Statistik uji dan (meninggalkan dan tersirat) masing-masing sesuai dengan H dan H , dan adalah: $D^{+}_{1}$ $D^{-}_{2}$ $n_{Y}$ $n_{X}$ $^{-}_{01}$ $^{-}_{02}$

$D^{+}_{1} = \Delta - D^{+} = \Delta - \left|\max_{t}\left[\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right]\right|$ , and

$D^{-}_{2} = \Delta - D^{-} = \Delta - \left|\min_{t}\left[\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right]\right|$

The Equivalence / Relevance Threshold
Interval —atau , jika menggunakan interval ekivalensi asimetris — dinyatakan dalam satuan dan , atau besarnya probabilitas yang dibedakan. Ketika dan mendekati tak terhingga, CDF atau untuk mendekati untuk $[-\Delta, \Delta]$ $[\Delta_{2}, \Delta_{1}]$ $D^{+}$ $D^{-}$ $n_{Y}$ $n_{X}$ $D^{+}$ $D^{-}$ $n_{Y},n_{X}$ $0$ $t<0$ , and for $t \ge 0$ :

lim n Y, n X \to \infty p + = P (n Y n X n Y + n X - - - - - - - - \sqrt D + \leq t) = 1 - e - 2 t 2

$\lim_{n_{Y},n_{X}\to \infty}p^{+} = \text{P}\left(\sqrt{\frac{n_{Y}n_{X}}{n_{Y}+n_{X}}}D^{+} \le t\right) = 1 - e^{-2t^{2}}$

$CDF of $D^{+}$ (or $D^{-}$)$

So it seems to me that the PDF for sample size-scaled $D^{+}$ (or sample size-scaled $D^{-}$ ) must be $0$ for $t<0$ , and for $t \ge 0$ :

f (t) = 1 - e - 2 t 2 d d t = 4 t e - 2 t 2

$f(t) = {1 - e^{-2t^{2}}}\frac{d}{dt} = 4te^{-2t^{2}}$

$PDF of $D^{+}$ (or $D^{-}$)$

Glen_b points out that this is a Rayleigh distribution with $\sigma=\frac{1}{2}$ . So the large sample quantile function for sample size-scaled $D^{+}$ and $D^{-}$ is:

CDF - 1 = Q (p) = - ln ( 1 - p ) 2 - - - - - - - - - - \sqrt

$\text{CDF}^{-1} = Q\left(p\right) = \sqrt{\frac{-\ln{\left(1 - p\right)}}{2}}$

and a liberal choice of $\Delta$ might be the critical value $Q_{\alpha}+\sigma/2 = Q_{\alpha}+\frac{1}{4}$ , and a more strict choice the critical value $Q_{\alpha}+\sigma/4=Q_{\alpha}+\frac{1}{8}$ .

Alexis
sumber

In the line where you pass from the cdf to the pdf, I think you got that wrong. Let

KnY,nX=nYnXnY+nX−−−−−√D+ $K_{n_{Y},n_{X}}=\sqrt{\frac{n_{Y}n_{X}}{n_{Y}+n_{X}}}D^{+}$ , so (abusing notation), in the limit

P(K∞,∞≤t)=1−e−2t2 $P(K_{\infty,\infty}\leq t) = 1 - e^{-2t^{2}}$ . Then

fK(t)=ddt1−e−2t2=4te−2t2 $f_K(t) = \frac{d}{dt} 1 - e^{-2t^2} = 4t\, e^{-2t^2}\quad$ (note the

t $t$ after the

4 $4$ ). (note also a missing sign in the exponent in the line above the taking of the derivative. Also I'm not sure why you have an integral symbol there, but maybe I misunderstood something.)

Glen_b -Reinstate Monica

@stochazesthai

D1 $D_{1}$ and

D2 $D_{2}$ are two one-sided test statistics. Per TOST you need to reject both the null hypotheses to which these test statistics apply.

Qα $Q_{\alpha}$ is a critical value from CDF

−1 $^{-1}$ on the above line, and where you want to sub in

1−α $1-\alpha$ for

p $p$ (e.g.

Qα=−ln(1−(1−α))2−−−−−−−−−−√ $Q_{\alpha} = \sqrt{\frac{-\ln{(1-(1-\alpha))}}{2}}$ ). The choice of

Δ $\Delta$ depends on how far past

Qα $Q_{\alpha}$ (the critical rejection value for a plain old positivist

H0 $H_{0}$ ) you need to go, before you conclude relevant difference (e.g. liberal 'equivalence' is

14 $\frac{1}{4}$

σ $\sigma$ beyond

Qα $Q_{\alpha}$ ).

Alexis

@stochazesthai (Continuing) So if both

D1≥Δ $D_{1} \ge \Delta$ and

D2≥Δ $D_{2} \ge \Delta$ , then you reject

H−0 $H_{0}^{-}$ .

Alexis

@stochazesthai Whoops! I should have put the quotes around the word liberal rather than equivalence two comments back. :)

Alexis

@stochazesthai If

D1≥Δ $D_{1} \ge \Delta$ , then reject

H−01 $H^{-}_{01}$ , if

D1<Δ $D_{1} < \Delta$ , then fail to reject

H−01 $H^{-}_{01}$ . If

D2≥Δ $D_{2} \ge \Delta$ , then reject

H−02 $H^{-}_{02}$ , if

D2<Δ $D_{2} < \Delta$ , then fail to reject

H−02 $H^{-}_{02}$ . If reject both

H−01 $H^{-}_{01}$ and

H−02 $H^{-}_{02}$ , then reject

H−0 $H^{-}_{0}$ , otherwise fail to reject

H−0 $H^{-}_{0}$ .

Alexis

An alternative to TOST in equivalence testing is based on the confidence interval approach:

Let $\Delta$ denote the prespecified equivalence margin and

θ : = sup t | F X (t) - F Y (t) |

$\theta := \sup_t |F_X(t) - F_Y(t)|$ the Kolmogorov-Smirnov distance between the unknown underlying distribution functions.

Now, if a 90% confidence interval for $\theta$ is completely within $[-\Delta, \Delta]$ , then we may be 95% certain that $\theta$ is enough close to 0 to speak of "equivalence".

Without knowing the underlying distributions, it seems to be hopeless to derive an approximate analytic confidence interval, so we might need to rely on (bias corrected) bootstrap confidence intervals based on resampling from pairs $X$ and $Y$ . (I don't want to find conditions for their validity in this particular application though...)

Michael M
sumber

Excellent. Do you have a citation for anyone undertaking the CI of

$D_{n_{1},n_{2}}$ (bootstrap or otherwise)?

Alexis

Good point... The short paper tomswebpage.net/images/K-S_test.doc mentions the "Handbook of Parametric and Nonparametric Statistical Procedures, Fifth Edition by David J.Sheskin (Apr 27, 2011)." to offer a two-sample case construcion for D. But at the moment, I don't have access to this book.

Michael M

Apakah ada versi uji ekivalensi sederhana dari tes Kolmogorov-Smirnov?

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