Supplement to Inductive Logic

Immediate Consequences of the Independent Evidence Conditions

When neither independence condition holds, we at least have:

P[en | hj·b·cn] = P[en | hj·b·cn·en−1] × P[en−1 | hj·b·cn]
=
=
n
Π
k = 1
P[ek | hj·b·cn·ek−1]

When condition-independence holds we have:

P[en | hj·b·cn] = P[en | hj·b·cn·(cn−1·en−1)] × P[en−1 | hj·b·cn·cn−1]
= P[en | hj·b·cn·(cn−1·en−1)] × P[en−1 | hj·b·cn−1]
=
=
n
Π
k = 1
P[ek | hj·b·ck·(ck−1·ek−1)]

If we add result-independence to condition-independence, the occurrences of ‘(ck−1·ek−1)’ may be removed from the previous formula, giving:

P[en | hj·b·cn] = n
Π
k = 1
P[ek | hj·b·ck]

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Copyright © 2012 by
James Hawthorne <hawthorne@ou.edu>

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