Take the 2-minute tour ×
Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It's 100% free, no registration required.

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 $H(P\pi)\ge H(\pi)$ ?

share|improve this question
1  
cross-posted:math.stackexchange.com/questions/243298/… –  A.Schulz Nov 23 '12 at 17:50

1 Answer 1

up vote 8 down vote accepted

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 $0$.

share|improve this answer
    
Thank you. It turns out that if a graph is regular, then we can prove it. –  eig Nov 24 '12 at 10:03
1  
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. –  Peter Shor Nov 24 '12 at 14:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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