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The binary entropy function is the information entropy of a Bernoulli random variable:with endpoint values zero. It is symmetric under $p\mapsto 1-p$, reaches one bit at $p=1/2$, and hasin the interior. Its strict concavity expresses why an even mixture of two outcomes has more uncertainty than a strong bias toward either one. The curve in mutual information also shows how this same uncertainty becomes the loss term in binary symmetric channel capacity.

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.

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].

At crossover probability $p=1/2$, the two rows of the transition matrix of a binary symmetric channel coincide. Its output is then independent of its input and its capacity is zero.

Let $q=\mathrm{Pr}(X=1)$ for the input bit of a binary symmetric channel with crossover probability $p$. The law of total probability givesConsequently its mutual information isHere $p$ specifies the noise and $q$ specifies the input bias; they are different choices. At $p=0.1$ and $q=0.2$, the output has $\mathrm{Pr}(Y=1)=0.26$, giving $H(Y)\approx 0.82674637$ and $I(X;Y)\approx 0.35775078$ bits. With the same noise and the fair input $q=1/2$, the output becomes fair and $I(X;Y)\approx 0.53100441$ bits, attaining binary symmetric channel capacity.

For $p\ne 1/2$, binary entropy reaches one only at output mass $1/2$, and the equation $p+(1-2p)q=1/2$ forces $q=1/2$. Thus the maximizing input is unique in those cases. For $p=1/2$, every input produces the same fair output and zero mutual information, so every input bias is optimal and none transmits information. A fixed input $q=0$ or $q=1$ also carries zero mutual information for every $p$: noise can make its output unpredictable without creating an unknown message to recover.

For finite discrete random variables, mutual information is the reduction of information entropy provided by observing the other variable:The last expression makes the symmetry $I(X;Y)=I(Y;X)$ visible. It is nonnegative and is zero exactly when $X$ and $Y$ are independent random variables. One way to see both facts iswhere $D$ is the Kullback-Leibler divergence with base-two logarithms. The Gibbs inequality gives nonnegativity; equality means that the joint probability distribution factors. The statement extends to more general distributions via relative entropy, but differences of infinite information entropies must not be used as if they were finite numbers.

A binary symmetric channel gives a useful example. Let $X$ be a fair bit, let $N$ be an independent Bernoulli random variable with error probability $q$, and set $Y=X\oplus N$, where $\oplus$ is exclusive or. The output is also a fair bit, but knowing the input leaves just the error uncertainty:Here $h_2$ is binary entropy. At $q=0.1$, $h_2(q)\approx 0.469$, so the output carries about $0.531$ bits of mutual information about the input per use. This is an average uncertainty reduction, not a claim that each observed symbol can be divided into fractional physical bits.

At $q=1/2$, every output is equally likely for either input, so $I(X;Y)=0$. At $q=1$, the output is always the inverted input: knowing this rule lets the receiver recover the input exactly, so $I(X;Y)=1$ again. A high raw error rate need not mean low mutual information when the errors are predictable. For this particular channel the fair input is optimal, giving the binary symmetric channel capacity $1-h_2(q)$.

Mutual information detects dependence that ordinary correlation can miss. Let $X$ be uniform on $\{-1,0,1\}$ and $Y=X^2$. Symmetry gives $\mathbb{E}[X]=\mathbb{E}[XY]=0$, hence zero covariance and zero correlation, while $Y$ is completely determined by $X$. Thus $I(X;Y)=H(Y)=h_2(1/3)>0$. This distinction is useful whenever a relationship is nonlinear: knowing one variable can reveal another without a linear trend.

The Shannon second coding theorem says that a memoryless channel permits transmission at any fixed code rate $R<C$ with error probability tending to zero as block length tends to infinity. Conversely, a sequence of codes with error probability tending to zero cannot have a limiting rate above $C$. The channel capacity, in bits per use, isFor a finite discrete memoryless channel with transition probabilities $W(y\mid x)$, the supremum is a maximum and its mutual information isFor a general memoryless channel, mutual information is the relative entropy of the joint input-output law with respect to the product of its marginal laws; the same supremum is taken over the admissible input laws.

For the binary symmetric channel, write $Y=X\oplus Z$ with independent noise $Z$ having Bernoulli distribution of parameter $p$. Given either input bit, the output conditional entropy is $h_2(p)$. ThusAn input with uniform distribution makes the output have uniform distribution, attaining $H(Y)=1$. The binary symmetric channel capacity is therefore

A reliable transmission rate $R$ means that length-$n$ codes can carry $M_n$ messages with $n^{-1}{\log }_2{M}_n$ tending to at least $R$ while the probability measure of the decoding-error event tends to zero. By the Shannon second coding theorem, the supremum of reliable rates is the binary symmetric channel capacityFor small positive $p$, the binary entropy function has the asymptotic expansionso $C(p)$ starts at one bit per channel use and decreases rapidly from one as errors are introduced.

A binary symmetric channel independently flips each input bit with probability $p$, so, with rows indexed by the input and columns by the output, its channel matrix isIf $p>1/2$, complementing every received bit converts the channel into one with crossover probability $1-p<1/2$ without changing its information-carrying ability. The case $p=1/2$ has identical rows and capacity zero, so it suffices to study $p<1/2$.

The Shannon second coding theorem states that for every rate below the channel capacity there are arbitrarily long codes whose decoding error tends to zero, whereas no sequence of codes with rate above capacity can have vanishing error. It also identifiesFor the present channel,The output entropy is at most one bit, with equality for a uniform input. Hence the binary symmetric channel capacity isbits per channel use.

For a discrete memoryless channel with $m$ output symbols, suppose every row of the transition matrix is a permutation of every other row and all column sums agree. Such a communication channel hasThe row Shannon entropy is the same for every input, so $H(Y\mid X)$ has that fixed value for every input probability distribution. The remaining term $H(Y)$ is at most ${\log }_{2}m$ by maximum entropy on a finite alphabet.

A uniform input attains the bound. If there are $\ell$ input symbols, the total of all transition probabilities in the matrix is $\ell$; equal column sums must therefore each be $\ell /m$. Multiplication by the uniform input mass $1/\ell$ gives output mass $1/m$ in each column. Hence $H(Y)={\log }_{2}m$, which proves the displayed formula. The cyclic ternary channel capacity, binary symmetric channel capacity, and ternary symmetric channel capacity are concrete applications of this symmetry calculation.