Supplement to Inductive Logic

Proof that the EQI for cn is the sum of EQI for the individual ck

Theorem: The EQI Decomposition Theorem:

 When the Independent Evidence Conditions are satisfied,

EQI[cn | hi/hj | b] = n

k = 1
EQI[ck | hi/hj | b].

[Back to Text]

Proof:

EQI[cn | hi/hj | b]
 =  {en} QI[en | hi/hj | b·cn] × P[en | hi·b·cn]
 =  {en} log[P[en | hi·b·cn]/P[en | hj·b·cn]] × P[en | hi·b·cn]
 =  {en−1}{en}(log[P[en | hi·b·cn·(cn−1·en−1)]/P[en | hj·b·cn·(cn−1·en−1)]]
         + log[P[en−1 | hi·b·cn·cn−1]/P[en−1 | hj·b·cn·cn−1]]) ×
          P[en | hi·b·cn·(cn−1·en−1)]  × P[en−1 | hi·b·cn·cn−1]
 =  {en−1}{en} (log[P[en | hi·b·cn]/P[en | hj·b·cn]]
         + log[P[en−1 | hi·b·cn−1]/P[en−1 | hj·b·cn−1]])  ×
          P[en | hi·b·cn]  × P[en−1 | hi·b·cn−1]
 =  (∑{en} log[P[en | hi·b·cn]/P[en | hj·b·cn]] × P[en | hi·b·cn]  ×
                 {en−1} P[en−1 | hi·b·cn−1])
+ (∑{en−1} log[P[en−1 | hi·b·cn−1]/P[en−1 | hj·b·cn−1]] × P[en−1 | hi·b·cn−1] ×
                 {en} P[en | hi·b·cn])
 =  EQI[cn | hi/hj | b] + EQI[cn−1 | hi/hj | b]
 =  …   (iterating this decomposition process)

 = 
n

k = 1
EQI[ck | hi/hj | b].

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. Readers who have benefited from the SEP are encouraged to examine the NEH’s anniversary page and, if inspired to do so, send a testimonial to neh50@neh.gov.