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WfccmAlgorithmFisher
(09 Dec 2004,
WillGray
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---+ Fisher Algorithm Performed on a set of non-negative numbers and labels of size N. There must be two specific labels. Label 1 must be 1 and label 2 must be 2. There can also be any other labels that will be ignored. Consider all numbers greater than 0 as 1. Construct the following table: ||Column1|Column2|Total| |Row1| a | b | r1 | |Row2| c | d | r2 | |Total| s1 | s2 | N | Where: * a is the number of 1 values in group 1. * b is the number of 1 values in group 2. * c is the number of 0 values in group 1. * d is the number of 0 values in group 2. * r1 = a + b * r2 = c + d * s1 = a + c * s2 = b + d * N = r1 + r2 = s1 + s2 Rearrage the table so a is the smallest value. There are two possible rearranging moves: * Swap Row1 and Row2. * Swap Column1 and Column2. ---++++ Fisher's Exact value 1 Chi1: %$\frac{(a - E)^2}{E}$% * E: %$\frac{r1 s1}{N}$% 1 Chi2: %$\frac{(b - E)^2}{E}$% * E: %$\frac{r1 s2}{N}$% 1 Chi3: %$\frac{(c - E)^2}{E}$% * E: %$\frac{r2 s1}{N}$% 1 Chi4: %$\frac{(d - E)^2}{E}$% * E: %$\frac{r2 s2}{N}$% 1 Chi-Square: Chi1 + Chi2 + Chi3 + Chi4 ---++++ Fisher's Exact probability The probability for a table is calculated by %$ \frac{r1!r2!s1!s2!}{N!a!b!c!d!}$%. The following table is called "Set 0": ||Column1|Column2|Total| |Row1| 0 | B | r1 | |Row2| C | D | r2 | |Total| s1 | s2 | N | 1 Determine the number of table probabilities to calculate by finding the smallest marginal total (least value among r1, r2, s1, s2) and adding 1. 1 Generate the table for "Set 0" * A = 0 * B = b + a * C = c + a * D = d - a * Totals remain unchanged. 1 Calculate the probability of "Set 0" * %$P_0$%: %$\frac{s2!r2!}{N!D!}$% * This is because in "Set 0" A! = 1, %$\frac{r1!}{B!}$% = 1, and %$\frac{s1!}{C!}$% = 1. 1 Calculate probabilities for other tables where A = {1, 2, ..., smallest marginal total} * %$P_i$%: %$P_{i-1}\frac{(B - (i - 1))(C - (i - 1))}{(D + i)(i)}$% 1 P-value: %$\sum{P_i}$% where %$P_i$% <= %$P_a$% + %$\delta$% * %$\delta=1.0*10^{-15}$% ---+++ Remarks * When the P-value gets very close to 1 and should in fact be 1, it is necessary to make a correction.
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Topic revision: r5 - 09 Dec 2004,
WillGray
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