Existe uma versão simples do teste de equivalência do teste Kolmogorov

Ok, aqui está minha primeira tentativa. Apreciação minuciosa e comentários apreciados!

As hipóteses de duas amostras
Se podemos enquadrar testes de hipótese de Kolmogorov-Smirnov de duas amostras com uma face , com hipóteses nulas e alternativas ao longo destas linhas:

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

H , por pelo menos um , em que: $_{\text{A}}\text{: }F_{Y}\left(t\right) < F_{X}\left(t\right)$ $t$

a estatística de teste $D^{-}=\left|\min_{t}\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right|$ corresponde a H ; $_0\text{: }F_{Y}\left(t\right) \geq F_{X}\left(t\right)$
a estatística de teste $D^{+}=\left|\max_{t}\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right|$ corresponde a H ; e $_0\text{: }F_{Y}\left(t\right) \leq F_{X}\left(t\right)$
$F_{Y}\left(t\right)$ e $F_{X}\left(t\right)$ são osCDFs empíricosdas amostras $Y$ e $X$ ,

deve ser razoável criar uma hipótese geral de intervalo para um teste de equivalência ao longo destas linhas (assumindo que o intervalo de equivalência seja simétrico no momento):

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

H , por pelo menos um . $^{-}_{\text{A}}\text{: }\left|F_{Y}\left(t\right) - F_{X}\left(t\right)\right| < \Delta$ $t$

Isto se traduziria para o específico dois unilateral "negativista" hipóteses nulas para teste de equivalência (estas duas hipóteses assumir a mesma forma, uma vez que ambos e são estritamente não-negativo): $D^{+}$ $D^{-}$

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

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

Rejeitando ambos H e H iria levar à conclusão de que . Obviamente, o intervalo de equivalência não precisa ser simétrico, e e podem ser substituídos por (inferior) e (superior) para as respectivas hipóteses nulas unilaterais. $^{-}_{01}$ $^{-}_{02}$ $-\Delta < F_{Y}\left(t\right) - F_{X}\left(t\right) < \Delta$ $-\Delta$ $\Delta$ $\Delta_{2}$ $\Delta_{1}$

As estatísticas de teste (atualizadas: Delta estão fora do sinal de valor absoluto)
As estatísticas de teste e (deixando implícitos e ) correspondem a H e H , respectivamente, e são: $D^{+}_{1}$ $D^{-}_{2}$ $n_{Y}$ $n_{X}$ $^{-}_{01}$ $^{-}_{02}$

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

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

O limiar de equivalência / relevância
O intervalo - ou , se estiver usando um intervalo de equivalência assimétrico - é expresso em unidades de e ou a magnitude das probabilidades diferenciadas. À medida que e aproximam do infinito, o CDF de ou para aproxima de para $[-\Delta, \Delta]$ $[\Delta_{2}, \Delta_{1}]$ $D^{+}$ $D^{-}$ $n_{Y}$ $n_{X}$ $D^{+}$ $D^{-}$ $n_{Y},n_{X}$ $0$ $t<0$ , and for $t \ge 0$ :

lim_{n_{Y}, n_{X} \to \infty} p^{+} = P (\sqrt{\frac{n_{Y} n_{X}}{n_{Y} + n_{X}}} 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}} \frac{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) = \sqrt{\frac{- \ln (1 - p)}{2}}

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

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
fonte

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

K_{n_{Y}, n_{X}} = \sqrt{\frac{n_{Y} n_{X}}{n_{Y} + n_{X}}} 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_{\infty, \infty} \leq t) = 1 - e^{- 2 t^{2}}

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

f_{K} (t) = \frac{d}{d t} 1 - e^{- 2 t^{2}} = 4 t e^{- 2 t^{2}}

$f_K(t) = \frac{d}{dt} 1 - e^{-2t^2} = 4t\, e^{-2t^2}\quad$ (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

D_{1}

$D_{1}$ and

D_{2}

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

Q_{α}

$Q_{\alpha}$ 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.

Q_{α} = \sqrt{\frac{- \ln (1 - (1 - α))}{2}}

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

Δ

$\Delta$ depends on how far past

Q_{α}

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

H_{0}

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

\frac{1}{4}

$\frac{1}{4}$

σ

$\sigma$ beyond

Q_{α}

$Q_{\alpha}$ ).

Alexis

@stochazesthai (Continuing) So if both

D_{1} \geq Δ

$D_{1} \ge \Delta$ and

D_{2} \geq 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

D_{1} \geq Δ

$D_{1} \ge \Delta$ , then reject

H_{01}^{-}

$H^{-}_{01}$ , if

D_{1} < Δ

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

H_{01}^{-}

$H^{-}_{01}$ . If

D_{2} \geq Δ

$D_{2} \ge \Delta$ , then reject

H_{02}^{-}

$H^{-}_{02}$ , if

D_{2} < 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
fonte

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

D_{n_{1}, n_{2}}

$D_{n_{1},n_{2}}$ (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

Existe uma versão simples do teste de equivalência do teste Kolmogorov – Smirnov?

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