The redundancy is $R=\bar{\ell }-H(X)$, measured in bits per source symbol. Write $s_v={\sigma }_0(v)+{\sigma }_1(v)$ for the total mass passing through interior node $v$. Counting each codeword's length as one for each interior node its route visits givesThe entropy can be split into the conditional binary decisions along that route. Algebraically each node contributesSumming telescopes: each non-root interior mass appears once positively as a parent and once negatively as a child. The root term is $1{\log }_{2}1=0$ and the remaining leaf terms sum to $-{\sum }_x{p}_x{\log }_2{p}_x=H(X)$. HenceSubtracting this identity from the expected-length identity provesThus redundancy is the sum of the probability-weighted imbalances of the binary branch decisions. Nodes with no mass contribute zero by continuity. The formula holds for every full prefix tree, even without optimality.

Let $\bar{\ell }=\mathbb{E}|c(X)|$. The Huffman encoder emits an average of $w\bar{\ell }$ binary digits per minute. Shannon's noisy-channel coding theorem states that for a discrete memoryless channel of capacity $K$, every information rate strictly below $K$ bits per use is achievable with error probability tending to zero as block length grows; rates above $K$ are not achievable for unrestricted messages.

Thus a channel code carrying the Huffman bitstream reliably with buffering and arbitrarily long blocks can be chosen wheneverThe strict inequality supplies positive coding and buffering slack. Huffman's expected-length bound is $H(X)\le \bar{\ell }<H(X)+1$ for a nontrivial finite source. Therefore the simpler bound $r>w(H(X)+1)/K$ is sufficient without knowing its exact Huffman lengths.

Distinguish this bitstream construction from the absolute source-channel limit. The noiseless source coding theorem permits long source blocks to be compressed to any rate above $H(X)$ bits per source symbol, with arbitrarily small overhead. Together with channel coding, this gives the sharper limiting requirement $rK>wH(X)$; $rK<wH(X)$ is impossible by the source-channel converse. Further compression can exploit redundancy remaining in a single-symbol Huffman stream, so $w\bar{\ell }/K$ need not be the universal minimum channel symbol rate. At $K=0$ a positive-entropy source cannot be sent reliably at a finite rate.

The entropy isTo construct a Huffman code, work with integer weights $2,2,3,4,5$. Merge the two twos to make four. Of the remaining weights $3,4,4,5$, merge three with the original four to make seven, and merge the new four with five to make nine. Finally merge seven with nine. Resolving the tie in this way gives, for example,This is a prefix code and Huffman's optimality theorem makes it optimal. Its expected length isDifferent choices at a tied merge can give different optimal codewords with the same expected length.

Unique decipherability makes the map from $(X_1,\dots ,X_m)$ to the concatenated binary string $M$ injective. Entropy is unchanged by an injective relabelling, and memorylessness gives independence, henceGiven $L=\ell$, the string has at most $2^{\ell }$ possibilities. The entropy bound for finite alphabets therefore gives $H(M\mid L=\ell )\le \ell$. Averaging,The random length cannot simply be discarded: it can also carry information. Let ${\ell }_{\max }$ be the maximum codeword length, finite since the alphabet is finite. Since $L$ is a function of $M$ and takes at most $m{\ell }_{\max }+1$ values,Divide by $m$ and let $m\to \infty$. This proves $\boxed{\mathbb{E}|c(X)|\ge H(X)}$. The conclusion also covers a deterministic source; the source's lowercase $x$ in its final expectation denotes the random source letter.

Define $H(X\mid Y)={\sum }_y\mathbb{P}(Y=y)H(X\mid Y=y)$. The joint entropy identity $H(X,Y)=H(Y)+H(X\mid Y)$ follows by writing $p_{xy}=q_y{p}_{x\mid y}$ and expanding its logarithm. If $Y=g(X)$, then $H(Y\mid X)=0$ and $H(X,Y)=H(X)$, giving the desired decomposition.

Group the first two symbols together and keep the third separate. If $c<1$, the identity becomeswhere $h(t)=-t{\log }_{2}t-(1-t){\log }_2(1-t)$. Binary entropy is at most one, with equality precisely at $1/2$. HenceEquality holds exactly when $a=b=(1-c)/2$. When $c=1$, $a=b=0$, so the same conclusion holds at the boundary.

For a finite probability distribution,with $0{\log }_{2}0=0$. If the positive support has size $s\le n$, apply the given strict Jensen inequality to the random variable $Z=1/p_X$:Equality requires $s=n$ and $Z$ constant almost surely, so all probabilities are $1/n$. Conversely the uniform distribution attains this bound. Thus $\boxed{H(X)\le {\log }_{2}n}$, with exactly that equality case.

The conditional output distributions are permutations of the same triple, so both conditional entropies equalWith input probability $p$, the three output probabilities are $q_1=(2-p)/6$, $q_2=(1+2p)/6$, and $q_3=(3-p)/6$. ThereforeDifferentiating, the constant terms cancel because $\sum q_j^{\prime }=0$:Thus there is a unique maximum. Its stationary equation reduces to $3p^2+9p-5=0$, whose root in $[0,1]$ isThe other root is negative; strict concavity proves this is the global maximum.