Markov chain: if $X\rightarrow Y\rightarrow Z$, then why is $Z\rightarrow Y\rightarrow X$ true?

Question

in a Markov chain, given three random variables $X,Y,Z$, we have $X\rightarrow Y\rightarrow Z$, which means $p(x,y,z) = p(x)p(y|x)p(z|y)$.

The right arrow symbol $\rightarrow$ is used to denote a relationship of (in)dependence between the two variables, so that if $X\rightarrow Y$, it means that whatever value $Y$ has, this has no influence on the distribution of $X$, while the value of $X$ has an influence on the distribution of $Y$.

my textbook Elements of Information Theory, in section 2.8 (page 34) gives this statement without a demonstration: $Z\rightarrow Y\rightarrow X$

is that always true? How can this be demonstrated? Or, do you have any intuitive explanation?

Answer 1

With $X \rightarrow Y \rightarrow Z$, we know that $X,Z$ are conditionally independent given $Y$. The reverse chain, $Z \rightarrow Y \rightarrow X$, follows from the symmetry of that independence:

\begin{eqnarray*} p(x,y,z) &=& p(x \mid y,z)\;p(y \mid z)\;p(z) \qquad\text{by the Chain Rule} \\ &=&p(x \mid y)\;p(y \mid z)\;p(z) \qquad\text{by conditional independence of $X,Z$ given $Y$}. \end{eqnarray*}

As with Equation 2.117, where $\;p(x,y,z) = p(x)\;p(y \mid x)\;p(z \mid y)\;$ implies $X \rightarrow Y \rightarrow Z$ , the result above implies $Z \rightarrow Y \rightarrow X$.