2
$\begingroup$

Let $P$ be a transition matrix of a random walk in an undirected (may not regular) graph $G$. Let $\pi$ be a distribution on $V(G)$. The Shannon entropy of $\pi$ is defined by

$$H(\pi)=-\sum_{v \in V(G)}\pi_v\cdot\log(\pi_v).$$

How do we prove that $H(P\pi)\ge H(\pi)$ ?

$\endgroup$
8
$\begingroup$

Is this even true? Consider an undirected graph which is a star. That is, a central vertex $V_0$ is connected to all other vertices $V_1, V_2, \dots, V_{n-1}$, and there are no other edges in the graph. Then, if you start with an equal distribution on $V_1, V_2, \ldots, V_{n-1}$, after one step all the weight is on the central vertex $V_0$. So in one step the entropy has gone from $\log (n-1)$ to $0$.

$\endgroup$
  • $\begingroup$ Thank you. It turns out that if a graph is regular, then we can prove it. $\endgroup$ – eig Nov 24 '12 at 10:03
  • 1
    $\begingroup$ Indeed. Section 4.4 of Elements of Information Theory, by Cover and Thomas, has this theorem as well as several generalizations that also cover the case of non-regular graphs. $\endgroup$ – Peter Shor Nov 24 '12 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.