Rémi Coulom wrote:
to Magnus: If you consider the example of section 2.2: 1,2,3 wins
against 4,2 and 1,5,6,7. The probability is
P=c1c2c3/(c1c2c3+c4c2+c1c5c6c7). For this example:
N1j=c2c3,B1j=0,M1j=c2c3+c5c6c7,A1j=c4c2
N2j=c1c3,B2j=0
N3j=c1c2,B3j=0
N4j=0,B4j=c1c2c3
I will add this example to the paper if it makes things clearer.
I'd recommend not using N_ij as an arbitrary constant when N is used for
a completely different purpose in the paper.
My other stumbling block was the jump to the f(x) equation. Calling it
f(y_i) and putting the definition of W_i near there would help some.
Putting log(N_ij*y_i +B_ij) = delta(N_ij)*[log(N_ij)+log(y_i)] +
delta(B_ij)*log(N_ij) would make the jump very easy. by delta(x) = 1
when x=0 and 0 for anything else.
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