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]

[Back to Text]

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

Open access to the SEP is made possible by a world-wide funding initiative.
Please Read How You Can Help Keep the Encyclopedia Free


The SEP would like to congratulate the National Endowment for the Humanities on its 50th anniversary and express our indebtedness for the five generous grants it awarded our project from 1997 to 2007.