Jtdylktuy

Consider a vector $p =(p_1,...p_n)$, $p_i>0$, $sum p_i = 1$
and a matrix $M_{ij}$, which is doubly stochastic: $sum_i M_{ij} = 1, sum_j M_{ij} = 1, M_{ij} > 0$.

Question 1 Just apply matrix M to a vector $p$ , i.e. $q = Mp$ (i.e. $q_i = sum_j M_{ij} p_j$) is it true that entropy of new vector $q$ is greater then of original vector $p$ ? I.e.

$$ H(q) = -sum_i q_i ln(q_i) > -sum_i p_i ln(p_i) = H(p) $$

Question 2 Is there a simple proof of it ? (It might follow from the Gibb's inequality, but it does not seem obvious for me).

Question 3 What are generalizations,
in view of meta-principle "entropy always grows" that might be an example of some more general phenomena ?

Motivation: There is a meta-principle that entropy grows is some natural systems, matrix applied to a vector is probably the most simple system one can consider (Markov chain), so the question above arises. After some thinking one can restrict from all matrices to doubly stochastic, because $M^n v$ tends to a uniform distribution (which has maximal entropy of all) only for doubly stochastic $M$ , so it cannot be true for general $M$.

This is a particular case of the following general principle. On the one hand, a bistochastic matrix $M$ has the property that for every non-negative vector (say, a probability vector), $Mpsucc p$. On another hand, the order $succ$ can be defined by $xsucc y$ iff $sum_if(x_i)lesum_if(y_i)$ for every convex function $f$.

Now, use the fact that $f=-H$ is convex.

For proofs, see my book Matrices (Springer-Verlag GTM #216, second edition) at sections 6.5 (Proposition 6.4) and 8.5 (Theorem 8.5).

Edit (at Alexander's request.) Let $alphane1$ be a positive parameter. The Renyi entropy is
$$frac1{1-alpha}logsum_jp_j^alpha.$$
The map $pmapsto Mp$ does increase Renyi entropy. Proof: if $alpha>1$, the map $tmapsto t^alpha$ is convex, and $smapstofrac1{1-alpha}log s$ is increasing. If instead $alpha<1$, the map $tmapsto t^alpha$ is concave, and $smapstofrac1{1-alpha}log s$ is decreasing.

This is true, except that we have equality when $p$ is the uniform distribution, which is the limit distribution of your matrix. (Proof in here, 15.5, for example.)

Whenever $p_igeq 1/n$, it is not so hard to show that $q_ileq p_i$, and when $p_ileq 1/n$, we have $q_igeq p_i$. From there, it follows that $-sum_i p_i ln (q_i) leq -sum_i q_i ln (q_i) $, and this gives the result combined with Gibbs'.

This is a particular case of a well-known monotonicity property of the Kullback-Leibler divergence $D(cdot|cdot)$. Namely, if $alpha$ and $beta$ are any two distributions on the same (say, finite) space $X$, and $P$ is a Markov operator on $X$, then
$$
D(alpha|beta) ge D(alpha P|beta P) ;.
$$
In your case double stochasticity of $P$ means that its stationary distribution is the uniform distribution $m_X$ on $X$, whereas
$$
D(alpha | m_X) = log text{card} X - H(alpha) ;.
$$

The answer to Question 1 is yes as long as $p$ is not the uniform distribution $(frac{1}{n},frac{1}{n},ldots,frac{1}{n})$.

The proof (Question 2) is quite simple:

Recall from the Birkhoff–von Neumann theorem that every doubly stochastic matrix $M$ is a convex combination of permutation matrices. We can interpret this as saying that there is a distribution $theta$ on the set of all permutations of $A:={1,2,ldots,n}$ such that whenever $mathbf{x}$ is a random variable from $A$ and $pmb{pi}$ is a random permutation of $A$ with distribution $theta$ independent of $mathbf{x}$, and we set $mathbf{y}:=pmb{pi}(mathbf{x})$, then we have
begin{align}
mathbb{P}(mathbf{y}=i,|,mathbf{x}=j) &= M_{i,j} ;.
end{align}
Note that if $mathbf{x}$ is distributed as $p$, then $mathbf{y}$ is distributed as $q:=Mp$, so $H(mathbf{x})=H(p)$ and $H(mathbf{y})=H(q)$.

Now,
begin{align}
H(mathbf{y},pmb{pi}) &= H(mathbf{y}) + H(pmb{pi},|,mathbf{y}) ;, \
H(mathbf{y},pmb{pi}) &= H(pmb{pi}) + underbrace{H(mathbf{y},|,pmb{pi})}_{H(mathbf{x})} ;,
end{align}
which implies
begin{align}
H(mathbf{y}) &= H(mathbf{x}) + H(pmb{pi}) - H(pmb{pi},|,mathbf{y}) \
&= H(mathbf{x}) + I(mathbf{y};pmb{pi})
end{align}
where $I(mathbf{y};pmb{pi})$ is the mutual information between $mathbf{y}$ and $pmb{pi}$. We know that $I(mathbf{y};pmb{pi})geq 0$ with equality if and only if $mathbf{y}$ and $pmb{pi}$ are independent, which is the case if and only if $mathbf{x}$ has the uniform distribution. Q.E.D.

For Question 3, suppose that $M$ is merely a positive stochastic matrix (not doubly stochastic). Let $r$ denote the unique stationary distribution of $M$. Then we have a similar ``entropy increase'' principle if we replace entropy with (minus) the Kullback-Leibler divergence relative to $r$. There is a nice discussion on this in the book of Cover and Thomas.