Immediate Consequences of the Independent Evidence Conditions
When neither independence condition holds, we at least
have:
P[e^{n}  h_{j}·b·c^{n}]

= 
P[e_{n}  h_{j}·b·c^{n}·e^{n−1}] ×
P[e^{n−1}  h_{j}·b·c^{n}]


= 
… 

= 
n
Π
k = 1

P[e_{k}  h_{j}·b·c^{n}·e^{k−1}]


When conditionindependence holds we have:
P[e^{n}  h_{j}·b·c^{n}]

= 
P[e_{n}  h_{j}·b·c_{n}·(c^{n−1}·e^{n−1})] × P[e^{n−1}  h_{j}·b·c_{n}·c^{n−1}]


= 
P[e_{n}  h_{j}·b·c_{n}·(c^{n−1}·e^{n−1})] × P[e^{n−1}  h_{j}·b·c^{n−1}]


= 
… 

= 
n
Π
k = 1

P[e_{k}  h_{j}·b·c_{k}·(c^{k−1}·e^{k−1})]


If we add resultindependence to
conditionindependence, the occurrences of
‘(c^{k−1}·e^{k−1})’
may be removed from the previous formula, giving:
P[e^{n}  h_{j}·b·c^{n}]

= 
n
Π
k = 1

P[e_{k}  h_{j}·b·c_{k}]

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