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 = 1EQI[ck | hi/hj | b].
Proof:
| = | ∑{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) | ||
= |
|
