Exercise 2.8 - Conditional independence iff joint factorizes

Answers

We prove that (2.129) is tantamount to (2.130). One direction is trivial by denoting:

g (x, z) = p (x | z),

h (y, z) = p (y | z) .

Conversely, we have:

\begin{aligned} p (x | z) = & \sum_{y} p (x, y | z) \\ = & \sum_{y} g (x, z) h (y, z) \\ = & g (x, z) \sum_{y} h (y, z) . \end{aligned}

And vice versa,

p (y | z) = h (y, z) \sum_{x} g (x, z) .

Moreover, for any $z$ :

\begin{align} 1 = & \sum_{x, y} p (x, y | z) \\ = & (\sum_{x} g (x, z)) (\sum_{y} h (y, z)) . \end{align}

Thus:

\begin{align} p (x | z) p (y | z) = & g (x, z) h (y, z) (\sum_{x} g (x, z)) (\sum_{y} h (y, z)) \\ = & g (x, z) h (y, z) \\ = & p (x, y | z) . \end{align}

solour_lfq

2021-03-24 13:42