## 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].

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].