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A binary symmetric channel is a discrete memoryless channel that flips each transmitted bit with fixed crossover probability $p$. For one use, write $Y=X\oplus Z$, where $\oplus$ is exclusive or and $Z$ is a Bernoulli random variable with parameter $p$, independent of the input $X$. Thus $Z=0$ preserves the bit and $Z=1$ flips it. Different uses have independent noise under this model.

With row labels $X=0,1$ and column labels $Y=0,1$, its transition matrix isEach row sums to one, and exchanging both binary labels leaves the rule unchanged, explaining symmetric. For $p=0.1$, a prescribed three-bit input $010$ produces $010$ with probability $0.9^3=0.729$ and $011$ with probability $0.9^2(0.1)=0.081$. The received $011$ does not itself announce the changed position: it could also arise from input $011$ with no flips.

The binary symmetric channel capacity quantifies how well a block code can overcome this uncertainty, and input distribution optimization for a binary symmetric channel shows why a fair input attains the best information rate. A completely noisy binary symmetric channel has $p=1/2$ and carries no input information; a known flip on every use, $p=1$, is instead reversible.

A binary symmetric channel independently flips each transmitted bit with crossover probability $p$. Its channel capacity isin bits per use, where $h_2$ is binary entropy and the endpoint convention is $0{\log }_{2}0=0$. Knowing the input leaves just the uncertainty of the flip, so $H(Y\mid X)=h_2(p)$ for every input probability distribution. Since the output is a bit, $H(Y)\le 1$, and thus $I(X;Y)\le 1-h_2(p)$. A fair input produces a fair output and reaches the bound. This is the two-symbol case of weakly symmetric channel capacity; Eytan Modiano, MIT 16.36, Channel Capacity and Coding, slides 4–5 gives this calculation.

For crossover probabilities $p=0.01,0.1,0.25$, the resulting channel capacities are approximately $0.91920686$, $0.53100441$, and $0.18872188$ bits per use. At $p=0$, the communication channel is noiseless; at $p=1/2$, the output is independent of the input and the capacity is zero. At $p=1$, every bit is inverted, a known reversible rule, and capacity is one again. More generally $C(p)=C(1-p)$, because inverting every received bit exchanges the two crossover probabilities. The existing mutual information figure shows this full symmetric curve.

An uncoded transmission illustrates why the raw bit-error probability is not itself channel capacity. At $p=0.1$, transmitting 100 uncoded bits succeeds on all of them with probability $0.9^{100}\approx 0.00002656$. The probability of at least one error is therefore approximately $0.99997344$. A suitable block code changes which received words count as recoverable evidence of a message; simply repeating uncoded transmission at rate one does not obtain the reliability promised at rates below $C$.

A block code assigns source messages to words of a fixed length over an alphabet. Its codewords form a subset of that fixed-length word space; a binary block code uses the two-symbol finite field ${\mathbb{F}}_2$.

Channel capacity is the best reliable information rate available through a specified communication channel. For a discrete memoryless channel with finite input and output alphabets, fixed transition probabilities $W(y\mid x)$, and no input-cost constraint, it isin bits per channel use, when the mutual information uses base-two logarithms. The maximization varies the input probability distribution, while keeping the communication channel fixed. A large input alphabet or a large output information entropy alone does not determine how much information gets through: the receiver needs information about the input, rather than merely unpredictable output.

The operational meaning comes from the Shannon second coding theorem. A block code represents one of $M$ possible messages by a word of length $n$, giving a rate ${\log }_{2}M/n$ bits per use. Take the message label from the uniform distribution on a finite set of those $M$ choices when measuring average decoding-error probability. For every fixed rate $R<C$, there are sequences of such block codes whose decoding-error probability tends to zero as $n$ grows; a rate above $C$ cannot have vanishing decoding-error probability. This is an asymptotic limit, rather than a promise that an arbitrary finite block code at a rate below $C$ works well. See MIT 18.310, Shannon's Noisy Coding Theorem, sections 1–2.

For a binary symmetric channel with crossover probability $p=0.1$, the binary symmetric channel capacity is approximately $0.53100441$ bits per use. Thus a binary code rate $R=1/2$ lies below the limit, while $R=3/4$ lies above it. At $R=1/2$, two transmitted symbols represent one message bit on average; the redundancy helps distinguish messages after noise. This does not split individual physical symbols into fractional pieces.

The result is useful because it separates two tasks. Computing channel capacity establishes the target imposed by the transition probabilities; constructing an effective block code attempts to approach that target with acceptable delay and decoding effort. The surprise is that reliability can improve indefinitely without forcing every positive rate to disappear. In his 1948 paper, Shannon defines noisy channel capacity in section 12 and develops its operational significance in section 13, Theorem 11. His original formulation uses information per unit time; the fixed-duration channel-use convention here expresses the same rate question per symbol. ^[ref].

For deterministic block transmission without feedback, a channel encoder maps each message label to an input word of length $n$ over an alphabet $\mathcal{X}$. These words are the codewords of a block code. Distinct messages should have distinct codewords: if two labels give the same word, even a noiseless communication channel cannot tell them apart. The channel encoder may add redundancy, so the transmitted word can be longer than the message representation.

A discrete memoryless channel models transmission through finite input and output alphabets using fixed conditional probabilities $W(y\mid x)$. For each input $x$, these nonnegative numbers sum to one over the output symbols. Discrete refers to the symbols, and memoryless means that the noise of one use does not remember earlier uses.

For a fixed input word $x^n=(x_1,\dots ,x_n)$ sent without feedback, the output law isThis is conditional independence of the outputs given the prescribed input word. It does not require an encoder to choose independent input symbols: the codewords of a block code deliberately impose relationships among them. Those relationships let a decoder distinguish possible messages after noise.

The binary symmetric channel and binary erasure channel make two different memoryless noise models with a binary input. One hides flipped bits; the other marks missing bits. Their different channel capacities show why the whole transition matrix matters, as the flips versus erasures in a binary channel demonstrate.

In the terminology used here, a block code assigns a distinct word of the same length to every source symbol. For a binary code with $m$ symbols this common length must satisfy $2^{\ell }\ge m$. The equal-length property is what matters in this question; elsewhere the term can also refer more generally to coding blocks of source symbols.

Every internal node of an optimal binary prefix tree has two children. If one had only one child, remove that one-edge detour from all descendant words. Prefix-freeness is preserved and at least one positive-probability word becomes shorter, strictly lowering the expected length. Thus the tree is full.

Take a leaf at maximum depth $L$. Its sibling exists because the tree is full. That sibling cannot be an internal node, since a descendant would be deeper than $L$. Hence it is another leaf at depth $L$, and the corresponding codewords differ only in their final bit. This is the deepest sibling property of an optimal prefix code.

A finite full binary tree satisfies Kraft equality, ${\sum }_x{2}^{-{\ell }_x}=1$, by dividing mass equally down each branch from the root. If all $2^n$ lengths were at least $n$, with at least one strictly greater, that sum would be less than $2^n{2}^{-n}=1$, a contradiction. Since the code is not a block code, the alternative that all lengths equal $n$ is excluded. Therefore some word has length below $n$. Also a longest word must have length above $n$: if all lengths were at most $n$ and some shorter, the Kraft sum would exceed one.

Suppose every sum of two distinct probabilities exceeds every third probability. If an optimal code were not a block code, part (b) gives two deepest sibling leaves $x,y$ at depth $L>n$ and a leaf $z$ at depth $\ell <n$. Replace the sibling pair by a leaf labelled $z$, and replace the former $z$ leaf by two children labelled $x,y$. The resulting tree is still a prefix tree, and other word lengths are unchanged. The change in expected length isHere $L-\ell -1\ge 1$. This contradicts optimality. Thus every optimal code is a block code.

Conversely, suppose distinct $x,y,z$ satisfy $p_x+p_y\le p_z$. Starting with the complete length-$n$ code, arrange $x,y$ as a sibling pair, which is possible by permuting the source labels. Its parent at depth $n-1$ does not contain $z$. Label that parent $z$, and split the old $z$ leaf into children $x,y$. This gives lengths $n-1$ for $z$, $n+1$ for $x,y$, and $n$ otherwise, with expected lengthIf the optimal mean length is below $n$, no optimal code can be a block code: any injective binary block code for $2^n$ symbols has length at least $n$. If the optimal mean is $n$, the constructed non-block code itself is optimal. In either case there is an optimal non-block code. This proves the reverse implication, including the equality case, and hence the strict inequality in the criterion is necessary.

This is the optimality of a fixed-length code for a power-of-two alphabet.

Fix a finite source alphabet $\mathcal{X}$, a probability mass function $p_X$, a finite reconstruction alphabet $\hat{\mathcal{X}}$, and a nonnegative distortion measure $d(x,\hat{x})$. The information rate-distortion function isThe source probability distribution stays fixed; the conditional distribution of the reconstruction is what we optimize. Each feasible test channel must keep expected distortion at most $D$. The objective is mutual information, measured in bits when the logarithms have base two. If no test channel satisfies the constraint, the infimum is $+\infty$.^[ref]

Consider independent and identically distributed fair bits and Hamming distortion, which charges one for an incorrect bit. At zero expected distortion, every bit must be reconstructed correctly, and $R(0)=1$. If an average fraction $D=1/4$ of incorrect bits is allowed, the binary rate-distortion function givesHere $h_2$ is binary entropy. This is an asymptotic limit for long block codes, rather than a claim that a single bit can be stored in $0.18872$ physical bits. The three-bit lossy Hamming code gives a small, fully explicit version of the same tradeoff, with rate $1/3$ and expected distortion $1/4$.

Rate-distortion theory separates the question of what is possible from the design of a practical encoder and decoder. It also forces the fidelity criterion to be explicit. Hamming distortion treats all symbol mismatches alike; mean squared error instead penalizes larger numerical errors more strongly. Those objectives can favor different reconstructions of the same data. An average error criterion does not by itself control the error on every individual input.

Claude Shannon's Coding Theorems for a Discrete Source With a Fidelity Criterion by Claude Shannon was published in 1959. Its formulation fixes a cost for replacing one source letter by another and studies the communication rate compatible with the allowed average cost. The important shift is to quantify which distinctions may be discarded, rather than requiring exact recovery of every letter.^[ref]

For a block code, separable distortion is the average of the individual symbol costs. With Hamming distortion it is the fraction of incorrectly reconstructed symbols, rather than the unnormalized number of mismatches. This normalization makes expected distortion comparable across different block lengths.^[ref]

For a finite alphabet source with independent and identically distributed symbols, a finite reconstruction alphabet, and a bounded nonnegative separable distortion, the rate-distortion function is the asymptotic limit for fixed-length lossy codes. In particular, for a distortion target in the interior of the feasible range and every $\epsilon >0$, sufficiently long block codes exist with rate at most $R(D)+\epsilon$ and expected distortion at most $D+\epsilon$. Every such $n$-symbol code obeys $R_n\ge R(D_n)$, where $D_n$ is its actual expected distortion.^[ref]

The outline is instructive. The data processing inequality bounds the mutual information in a reconstruction by the information entropy of the transmitted message, which is at most ${\log }_2{M}_n$. Independence and the convexity of the rate-distortion function then give the per-symbol lower bound. For achievability, choose many reconstruction blocks using a suitable probability distribution, so that the source's typical blocks are likely to have an acceptable nearby reconstruction. This is a covering argument; it does not require recovering the original block exactly.

A test channel is a proposed conditional distribution of a reconstruction given the source symbol. It describes a candidate joint probability distribution for evaluating mutual information and expected distortion. It need not be a physical transmission channel, or an implementable one-symbol encoder; the Shannon rate-distortion theorem relates this optimization to long block codes.

Let $X_1,X_2,X_3$ be independent fair bits. The encoder sends their majority bit: zero when at least two inputs are zero, and one otherwise. The decoder reconstructs $000$ or $111$ by repeating that received bit three times. This is lossy compression: different original strings deliberately share a reconstruction.

There are eight equally likely inputs. The strings $000$ and $111$ have zero Hamming distance from their reconstructions; each of the other six has Hamming distance one. One transmitted bit represents three source symbols, soEvery input has average Hamming distortion at most $1/3$, whereas the population average is $1/4$. The actual block code therefore obeys $R\ge R(D)\approx 0.18872$, but does not reach the optimal long-block rate.

This construction uses the same reconstruction strings as the threefold binary repetition code, but the tasks differ. An error-correcting encoder expands one source bit into three transmitted bits; this lossy compression example compresses three source bits into one transmitted bit. The source distribution and the direction of the mapping matter.

The typical sequence theorem turns a probability concentration into a coding construction. Assign one binary number index to each weakly typical sequence, plus a reserved index for every word outside the set. The decoder recovers every typical word exactly; the reserved index has one predetermined output. The block code usesbits and has reconstruction-error probability at most $1-P(T_{\epsilon }^{(n)})$.

For this example and $\epsilon =0.1$, $n=40$ needs 34 bits but permits error up to about $0.3611$. At $n=640$, 570 bits suffice and the error bound is about $0.0002215$. These are existence calculations for an explicitly enumerated collection, rather than claims about an efficient practical encoder. The reduction comes from protecting the whole high-probability collection, rather than just its single most likely member.