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I read on Bishop Chap8 P 374 that:

sum(P(b|c)P(c|a)) = P(b|a) where the sum is over c.

Can you prove that?

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This is only true if $p(b\mid a, c) = p(b \mid c)$ i.e. if $a,b$ are conditionally independent given $c$. Bishop states that this is assumed.

Compute \begin{align} p(a, b) &= \sum p(a, b, c) = \sum p(b\mid a, c)p(a, c) \\ &= \sum p(b \mid c)p(c \mid a)p(a) = p(a) \sum p(b\mid c) p(c \mid a). \end{align}

Since $p(a, b)/p(a) = p(b\mid a)$ the equality holds: $\sum p(b\mid c) p(c \mid a) = p(b\mid a)$.

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