Generalized Grubbs test on moments of order 1 (G1)

Let's consier a series of observed cummulative frequencies f_i obtained testing an known distribution function against a series of n data points (like in the following example).

i	xi	fi
1	-1.310	0.095
2	-1.029	0.152
3	-0.793	0.214
4	+2.052	0.980

(n = 4; (x_i)_i=1,2,3,4 = Sort{Random[Normal(0,1)]}; f_i = CDF(x_i; Normal(0,1))
Let's consider only \[ g0 \gets \mathop{\max}\limits_{1 \leq i \leq n} |f_i-0.5| \] It's associated probability (to be observed, CDF) is (see (Jäntschi, 2019)):\[ p(g0) \gets \left( 2 \cdot g0 \right) ^{n} \] The risk of being in error assuming that (f_i)_{i=1, ..., n} is coming from uniform distribution(the risk of being in error assuming that (x_i)_{i=1, ..., n} is coming from assumed distribution) is 1 - p.
For the above given example, g0 = 0.4799, p(g0) = 0.85, 1 - p = 15%.
Let's consider now \[ g1 = \mathop{\max} \left( \frac{|f_1 - 0.5|}{\sum_{i=1}^{n}{|f_i-0.5|}},\frac{|f_n-0.5|}{\sum_{i=1}^{n}{|f_i-0.5|}} \right) \] It's associated probability (to be observed, CDF) is to be taken from (see (Jäntschi, 2020)):
\[ p(g1) = 1 - \sum_{k=0}^{\lfloor n - \frac{1}{g1} \rfloor} (-1)^{k} \frac{(n - \frac{1}{g1} - k)^{n-1}}{k!(n-1-k)!} \]For the above given example, g1 = 0.3159, 1/g1 = 3.1655, p(g1) = 1 - 0.097, 1 - p = 9.7%.

Refs:

• Lorentz JÄNTSCHI, 2019. A test detecting the outliers for continuous distributions based on the cumulative distribution function of the data being tested. Symmetry 11(6): 835. DOI 10.3390/sym11060835. (15p. + 7p. supp. mat.)
• Lorentz JÄNTSCHI, 2020. Detecting extreme values with order statistics in samples from continuous distributions. Mathematics 8(2): 216. DOI 10.3390/math8020216. (21p. + 21p. supp. mat.)

Last Update: August 10, 2023.