Re^2: recursive formula.

[BioGeek]

Your results look sound to me.

The above formula is used in extreme value statistics, where outliers are important. You wouldn't catch this when you just take the mean of the values:

M() = mean

M(0.11, 0.07, 0.19) = 0.12
M(0.43, 0.31, 0.37) = 0.37
M(0.93, 0.78, 0.82) = 0.84
M(0.91, 0.12, 0.15) = 0.39
M(0.52, 0.18, 0.32) = 0.34

Now, with our formula: all low scores, also give a low P-value:
P(0.11, 0.07, 0.19) = 0.001595and
P(0.52, 0.18, 0.32) = 0.010192
all high scores give a high P-value:
P(0.93, 0.78, 0.82) = 0.548235
both as expected. But now: low scores with a high outlier are also important for us,
and indeed it also gives a high P-value:
P(0.91, 0.12, 0.15) = 0.439894
Only for P(0.43, 0.31, 0.37) = 0.046225 I would intitutivly have tougth of a lower value,
but it's probably correct.

Thanks for the effort you put into it.