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