Immediate Consequences of the Independent Evidence Conditions

When neither independence condition holds, we at least have:

$$\begin{align} P[e^n \pmid h_j\cdot b\cdot c^n] & = P[e_n \pmid h_j\cdot b\cdot c^n\cdot e^{n-1}] \times P[e^{n-1} \pmid h_j\cdot b\cdot c^n] \\ & = … \\ & = \prod^{n}_{k = 1} P[e_k \pmid h_j\cdot b\cdot c^n\cdot e^{k-1}] \end{align}$$

When condition-independence holds we have:

$$\begin{align} P[e^n \pmid h_j\cdot b\cdot c^n] & = P[ e_n \pmid h_j\cdot b\cdot c_n\cdot(c^{n-1}\cdot e^{n-1})] \\ & \qquad \times P[e^{n-1} \pmid h_j\cdot b\cdot c_n\cdot c^{n-1}] \\ & = P[e_n \pmid h_j\cdot b\cdot c_n\cdot(c^{n-1}\cdot e^{n-1})] \\ & \qquad \times P[e^{n-1} \pmid h_j\cdot b\cdot c^{n-1}] \\ & = … \\ & = \prod^{n}_{k = 1} P[e_k \pmid h_j\cdot b\cdot c_k\cdot(c^{k-1}\cdot e^{k-1})] \end{align}$$

If we add result-independence to condition-independence, the occurrences of ‘$(c^{k-1}\cdot e^{k-1})$’ may be removed from the previous formula, giving:

$$P[e^n \pmid h_j\cdot b\cdot c^n] = \prod^{n}_{k = 1} P[e_k \pmid h_j\cdot b\cdot c_k]$$

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