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The empty family satisfies the bound, so assume $\mathcal{F}$ is nonempty and choose a graph according to the uniform distribution on a finite set $F\in \mathcal{F}$. Let $Y$ be its random vector of edge indicators, indexed by $E=\left(\genfrac{}{}{0ex}{}{[n]}{2}\right)$. Then its information entropy is $H(Y)={\log }_2|\mathcal{F}|$.

For each vertex $v$, put $S_v=\{e\in E:v\in e\}$, so $Y_{S_v}$ records the graph neighbourhood of $v$. The possible graph neighbourhoods form an intersecting family of subsets of $[n]\setminus \{v\}$: the graph intersection of any two members of $\mathcal{F}$ has no isolated vertex at $v$. A subset and its complement cannot both occur. Pairing the $2^{n-1}$ subsets into complementary pairs gives at most $2^{n-2}$ possible graph neighbourhoods. The maximum entropy on a finite alphabet therefore givesThis argument applies for $n\ge 2$. For $n=1$ the assumed nonempty family cannot exist, since its only possible graph has an isolated vertex.

Each edge belongs to exactly two of the sets $S_v$. To obtain the needed instance of Shearer inequality directly from the preceding compression argument, repeatedly apply a union-intersection compression to incomparable members of the multiset $(S_v{)}_{v=1}^n$. Each such step strictly increases ${\sum }_S|S{|}^2$, byThe potential is bounded and integer-valued, so the process terminates in a chain under inclusion. Coordinate multiplicities stay equal to two, so, for $E\ne \varnothing$, the terminal chain consists of two copies of $E$ and $n-2$ empty sets. The compression of an entropy sum now givesExponentiating proves the entropy bound for graphs with isolated-vertex-free intersections:

An elementary union-intersection compression replaces two occurrences $A_1,A_2$ in a multiset by $A_1\cup A_2,A_1\cap A_2$. By entropy submodularity, this changes the sum of information entropies byA compression of an entropy sum is obtained by iterating these elementary union-intersection compressions. Summing the inequalities over the sequence, with repetitions counted according to their multiset multiplicities, proves the required monotonicity:Neither the number of occurrences of an individual coordinate nor the total number of sets changes under union-intersection compression.

Take information entropy in bits and put $H(X_{\varnothing })=0$. Write $C=A\cap B$, $U=A\setminus B$, and $V=B\setminus A$. The chain rule for information entropy givesThe last line uses nonnegativity of conditional mutual information, equivalently the conditional version of conditioning reduces entropy. All random variables are finite-valued, so every conditional entropy here is finite. Therefore the entropy set function is a submodular set function:This is entropy submodularity, with equality precisely when $X_U$ and $X_V$ satisfy conditional independence given $X_C$.

On a probability measure-preserving system, for a finite measurable partition $\xi =\{A_1,\dots ,A_r\}$, the entropy of a finite measurable partition isUse natural logarithms, so information entropy is measured in nats; another fixed logarithm base rescales all answers. The join of measurable partitions is their common refinement, and write ${\xi }_a^b={\bigvee }_{j=a}^b{T}^{-j}\xi$, with an empty join the trivial partition. The entropy rate of a measurable partition and Kolmogorov-Sinai entropy are respectivelyThe block entropies form a subadditive sequence, so the first limit exists by the Fekete lemma. For finite partitions the conditional entropy of finite measurable partitions is $H(\eta \mid \zeta )=H(\eta \vee \zeta )-H(\zeta )$.

Put $c_0=H(\xi )$ and $c_k=H(\xi \mid {\xi }_1^k)$ for $k\ge 1$. The chain rule for information entropy, applied from the last coordinate backwards, and measure preservation giveThe second equality uses invariance of the joint partition atom probabilities under the common pullback $T^{-j}$; invertibility is unnecessary. Since conditioning reduces entropy, $c_k$ decreases to a nonnegative limit $c$. The Cesaro convergence of a sequence of this convergent sequence has the same limit. ThereforeEquivalently $h_{\mu }(T,\xi )=H_{\mu }(\xi \mid {\mathcal{F}}_1)$, where ${\mathcal{F}}_1=\sigma ({\bigvee }_{j\ge 1}{T}^{-j}\xi )$. Here conditional entropy of a countable measurable partition conditioned on a sigma-algebra is computed using conditional partition atom probabilities; the Martingale convergence theorem gives continuity under increasing conditioning sigma-algebras. This is the infinite-future formula for partition entropy rate.

The Kolmogorov-Sinai generator theorem states that if a finite or countable measurable partition $\eta$ has finite entropy of a countable measurable partition and its iterates generate the whole completed sigma-algebra modulo null sets, then $h_{\mu }(T)=h_{\mu }(T,\eta )$. For an invertible system, generating means $\sigma ({\bigvee }_{j\in \mathbb{Z}}{T}^{-j}\eta )=\mathcal{B}$ modulo null sets. For a noninvertible system a one-sided generator, using $j\ge 0$, suffices. The two-sided and one-sided versions must not be confused.

For a Bernoulli shift with discrete symbol probabilities $(p_i)$, the coordinate-zero measurable partition has independent coordinate iterates. ThusFor a finite alphabet the coordinate partition is a generator of finite entropy of a finite measurable partition; on the two-sided sequence space use all integer coordinate iterates, and on the one-sided space use the nonnegative ones. The Kolmogorov-Sinai generator theorem proves the displayed answer in both cases. The same calculation applies to countably many symbols when their Shannon entropy is finite, using the countable finite-entropy version of the theorem. If the Shannon entropy is infinite, merge all but the first $r$ symbols into one cell. These finite coordinate partitions have entropy rate $-{\sum }_{i\le r}{p}_i\log p_i-p_{>r}\log p_{>r}\to \infty$, so the system entropy is infinite. Zero-probability symbols contribute zero. In particular a fair $r$-symbol shift has entropy $\log r$.

For the final assertion, let ${\mathcal{F}}_0=\sigma ({\bigvee }_{j\ge 0}{T}^{-j}\xi )$ and complete it modulo null sets. The approximation property forces ${\mathcal{F}}_0=\mathcal{B}$ modulo null sets. Indeed for each $A\in \mathcal{B}$, choose approximating sets from finite blocks with error tending to zero. Their indicators approach $1_A$ in $L^2$, and $L^2({\mathcal{F}}_0)$ is a closed vector subspace, so $A$ is ${\mathcal{F}}_0$-measurable modulo a null set.

Invertibility now gives ${\mathcal{F}}_1=T^{-1}{\mathcal{F}}_0=T^{-1}\mathcal{B}=\mathcal{B}$ modulo null sets. In particular the present partition $\xi$ is measurable with respect to its entire future, so $H(\xi \mid {\mathcal{F}}_1)=0$. The infinite-future formula gives $h_{\mu }(T,\xi )=0$. Since the given one-sided generator is also a two-sided generator, the Kolmogorov-Sinai generator theorem finishes the proof:This is finite one-sided generator of an invertible system forces zero entropy. Invertibility is essential: a fair binary one-sided Bernoulli shift has a finite one-sided generator and Kolmogorov-Sinai entropy $\log 2$.

Take every logarithm in base two. The information entropy of $X$ iswith $0{\log }_{2}0=0$. If the codeword $c({\mu }_i)$ has length $s_i$, then its expected codeword length is

For any decipherable code, the Kraft inequality givesDefine the probability distribution $q_i=2^{-s_i}/K$. By Gibbs inequality,Since ${\log }_{2}K\le 0$,

Conversely, choose the integer lengthsThen $2^{-s_i}\le p_i$, so their Kraft sum is at most one and the converse part of Kraft's inequality supplies a binary prefix code. Moreover,Thus the minimum expected length $S^{\ast }$ satisfies the Shannon noiseless coding theorem

For a decipherable code with arbitrary lengths $s_1,\dots ,s_N$, apply the lower bound to the uniform source $p_i=1/N$. Its entropy is ${\log }_{2}N$ and its expected length is $(s_1+\cdots +s_N)/N$. Hencethe total length lower bound for a decipherable binary code.

Now consider the cumulative construction. Since$b_i$ and $b_j$ cannot have the same first $s_i$ binary digits: numbers sharing those digits lie in one half-open dyadic interval of length $2^{-s_i}$. Also, $p_i\ge p_j$ implies $s_i\le s_j$. Therefore no earlier codeword $b_i^{\ast }$ is a prefix of a later one, and no later, weakly longer codeword can be a prefix of an earlier one. The cumulative Shannon code is thus a prefix code, hence decipherable.

An optimal code minimizes expected codeword length over all decipherable codes for the specified source probabilities. The cumulative construction need not be optimal. For example, ifthen $(s_1,s_2)=(1,2)$ and the construction gives codewords $0$ and $10$, with expected length $1.4$. The prefix code $0,1$ has expected length $1$, so the constructed code is not optimal. In general, Huffman coding produces an optimal prefix code.

Because $X$ and $Y$ are independent random variables, conditioning on $Y$ merely translates $X$. The conditional entropy under a deterministic change of variables and the fact that conditioning reduces entropy therefore giveInterchanging $X$ and $Y$ similarly gives $H(X+Y)\ge H(Y)$, and henceThe proof applies to countable alphabets whenever the displayed information entropies are finite.

On $B$, part b gives ${\log }_2{P}_n={\log }_2{Q}^n-{\log }_2{Q}^n(B)$. Hence the information entropy of the conditional law satisfiesParts d and e implyExponentiation proves $Q^n(B)\le 2^{-nD(\overline{P}\Vert Q)}$.

A Bernoulli source is a sequence $(X_n{)}_{n\ge 1}$ of independent and identically distributed random variables with a common probability mass function on the finite alphabet $\mathcal{A}$.

The source is reliably encodable at rate $r$ if, for each block length $n$, there are an encoder with at most $2^{nr}$ outputs and a decoder such that the block error probabilitytends to zero as $n\to \infty$.

Its information rate is the limiting information entropy per source symbol,when this limit exists.

The unicity distance is the least ciphertext length for which the key is determined, or in the usual approximation the length at which the expected number of spurious keys falls to about zero. Assume equiprobable keys, a stationary plaintext source of information entropy rate $H_E$, and ciphertext symbols that are approximately uniform on $A$. The language redundancy per symbol isThe key equivocation is then approximated byHence the classical closed-form estimate is

The axioms for information entropy give the formula $H(X)=-{\sum }_{x}p(x)\log p(x)$ and hence the chain rule for information entropyBecause conditioning reduces entropy, $H(Y\mid X)\le H(Y)$, and thereforeThis is subadditivity of information entropy.

For the entropy submodularity rule, apply the chain rule for information entropy twice:The last quantity is conditional mutual information, whose nonnegativity again expresses that conditioning reduces entropy.

Let $\mathcal{Q}$ be the set of valid quintuples and choose $(A_1,\dots ,A_5)$ uniformly from $\mathcal{Q}$. ThenApply Shearer's inequality to the ten pairs $\{i,j\}\subseteq [5]$. Every index occurs in four pairs, so

Put $U_{ij}=A_i\cup A_j$. Since $U_{ij}\in \mathcal{A}$, it has cardinality $k$. For each element of $U_{ij}$, its membership bits in $(A_i,A_j)$ are one of $(1,0),(0,1),(1,1)$; outside $U_{ij}$ they are forced to be $(0,0)$. Thus at most $3^k$ ordered pairs have any prescribed union. The chain rule for information entropy, conditional entropy, and the support bound for information entropy givebecause every $U_{ij}$ belongs to $\mathcal{A}$. There are ten pairs, henceExponentiating proveswhich is the five-variable case of the entropy bound for pairwise-union tuples.

For feasible mass functions $Q_1,Q_2$ at distortion levels $d_1,d_2$, the mixture $Q=\lambda Q_1+(1-\lambda )Q_2$ has nonzero-symbol mass at most $\lambda d_1+(1-\lambda )d_2$. Concavity of information entropy givesTaking maximizing sequences proves

Let $X_1,X_2,Y$ be independent, with $X_1,X_2$ distributed as $X$. Apply part (b) to $X_1,-Y,X_2$:Adding an independent random variable cannot decrease information entropy, soIn terms of Entropic Ruzsa distance, this is $d_R(X,-X)\le 2d_R(X,Y)$. The Entropic Ruzsa triangle inequality and invariance under simultaneous negation now givewhich is the Entropic Ruzsa sum-difference inequality.

For independent $X,Y$, expand the Entropic Ruzsa sum-difference inequality from part (d):Collecting the information entropy terms givesReplacing $Y$ by $-Y$ interchanges sum and difference and preserves $H(Y)$, yielding the requested orientation

For $m=2^b$, assign every outcome one of the $b$-bit words. This complete fixed-length prefix code has expected length $b$ and meets the information entropy lower bound.