Proposition 2.5 For random variable \(X,Y\), and \(Z\), \(X \perp Z\,|\,Y\) if and only if:

\[ \begin{aligned} p(x,y,z) = a(x,y)b(y,z) \\ \end{aligned} \]

Why ? The only-if part is rather trivial. To prove the "if" part, the key is to consider \(p(x,y), p(y,z)\), and \(p(y)\) individually, expressing each in terms of the functions \(a\), and \(b\). Here we go. Suppose \(\color{teal}{p(x,y,z) = a(x,y)b(y,z)}\).

\[ \begin{aligned} \color{blue}{p(x,y)} &= \sum_z \color{teal}{p(x,y,z)} = \sum_z \color{teal}{a(x,y) b(y,z)} = \color{blue}{a(x,y) \sum_z b(y,z)} \\ \color{orange}{p(y,z)} &= \sum_x \color{teal}{p(x,y,z)} = \sum_x \color{teal}{a(x,y) b(y,z)} = \color{orange}{b(y,z) \sum_x a(x,y)} \\ p(y) &= \sum_z \color{orange}{p(y, z)} = \sum_z \left( \color{orange}{b(y,z) \sum_x a(x,y)} \right) \\ &= \left(\sum_x a(x,y)\right) \left(\sum_z b(y,z)\right) \\ &= \color{orange}{p(y,z) \over b(y,z)} \color{blue}{p(x,y) \over a(x,y)} \\ &= \frac{p(x,y) p(y,z)}{\color{teal}{p(x,y,z)}} \\ \therefore p(x,y,z) &= \frac{p(x,y) p(y,z)}{p(y)} & \text{ie }X \perp Z\,|\,Y \\ \end{aligned} \]

\(\Box\)

Source: Information Theory.

# posted by rot13(Unafba Pune) @ 7:15 AM