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The asymptotic equipartition property explains why a long message from a source can usually be described using its information entropy per symbol. Individual messages can have very different probabilities, but the messages carrying almost all the total probability have similar logarithmic costs per symbol. The Bernoulli typical-set example makes the distinction between an individual message and a large collection explicit.

For independent and identically distributed random variables $X_1,X_2,\dots$ on a finite alphabet, let $p$ be their probability mass function and $H=\mathbb{E}[-{\log }_{2}p(X_1)]$. ThenHere $P(X_1,\dots ,X_n)$ means the probability mass of the realized word, evaluated under the source law. The product rule for independent random variables turns its information content into a sum:The strong law of large numbers gives the limit. Zero-probability symbols do not occur; on the positive-probability support of a finite alphabet, these information costs are finite. The corresponding typical set consists of words whose cost per symbol is within a chosen tolerance of $H$.

For a strictly stationary process with a finite alphabet, the constant-limit extension requires ergodicity. The Shannon-McMillan-Breiman theorem then replaces the one-symbol information entropy with the information rate. Stationarity without ergodicity in an information source shows why stationarity alone is insufficient.^[ref]

Take an IID random variable source of bits with $P(X_j=1)=1/4$. Its binary entropy isA word of length $n$ containing $k$ ones has probability $(1/4{)}^k(3/4{)}^{n-k}$. Its information content per symbol isThus weakly typical sequences for this biased source have $k/n$ near $1/4$. With tolerance $\epsilon =0.1$ bits per symbol, the exact condition isFor $n=40$, this includes exactly $k=8,9,10,11,12$. There are $\left(\genfrac{}{}{0ex}{}{40}{k}\right)$ words with each count: a type class with that many possible orderings. Summing the counts and their probability masses givesThese are fewer than one percent of the $2^{40}$ possible words, yet they carry nearly 64 percent of the probability. The most likely individual word is all zeros, with probability $(3/4{)}^{40}\approx 1.006\times 10^{-5}$; it is outside this typical set. No contradiction is involved: the type classes near ten ones each contain many more individual words.

For a strictly stationary process on a finite alphabet, put $a_1=H(X_1)$ and $a_n=H(X_n\mid X_1,\dots ,X_{n-1})$ for $n\ge 2$. Each $a_n$ is the average remaining uncertainty in the next symbol after its history is known. More history can improve prediction, so this sequence is nonincreasing:The inequality uses conditioning reduces entropy; the last equality uses strict stationarity. Nonnegative $a_n$ therefore have a limit. By the chain rule for information entropy,The Cesàro mean converges to the same limit, proving the equality with the entropy rate. This is a statement about average information entropy. A constant almost-sure limit of the realized information content requires the further hypotheses in the Shannon-McMillan-Breiman theorem; stationarity without ergodicity in an information source explains the distinction.

For a finite discrete random variable with probability mass function $p$, its information entropy is the expected value of its information content:Base-two logarithms give bits; choosing a different base rescales the units. The convention $0{\log }_{2}0=0$ is justified by a limit, so outcomes of zero probability contribute nothing. The definition also applies to countable distributions, but the sum may be infinite.

A fair bit has $H=1$. A certain outcome has $H=0$. A Bernoulli random variable that is one with probability $0.1$ has $H=h_2(0.1)\approx 0.469$ bits, because most of its outcomes are predictable. A rare outcome still has large information content: the outcome of probability $0.1$ contributes $-{\log }_2(0.1)\approx 3.322$ bits of surprise when it occurs. Entropy is the average surprise, rather than the surprise of each outcome.

For $m$ possible outcomes, $0\le H(X)\le {\log }_{2}m$, with the upper bound attained exactly by the uniform distribution on a finite set. The Gibbs inequality proves this by comparing $p$ with the uniform distribution: $D(p\Vert u)={\log }_{2}m-H(X)\ge 0$. Thus increasing predictability reduces the required average description length. The four-symbol prefix code in information theory attains $H=7/4$ bits exactly; for general probabilities, integer codeword lengths prevent such a perfect match for every individual symbol. Shannon's source coding theorem explains how coding sufficiently long independent blocks approaches the entropy per symbol.

The logarithm is especially natural because independent outcomes have additive surprise: $-{\log }_2(pq)=-{\log }_{2}p-{\log }_{2}q$. Taking expected values gives additive information entropy for independent random variables; for dependent ones, the chain rule for information entropy replaces the second term by conditional entropy. This is why the same quantity can govern both a single observation and a long message.

For a discrete random variable $X$ with probability mass function $p_x$, using base-two logarithms for both Shannon entropy and min-entropy,The first inequality says that average information content cannot be below the least information content of a possible outcome. For a finite alphabet, equality $H(X)=H_{\infty }(X)$ holds exactly when the probability mass function is uniform on its positive-probability outcomes. Impossible outcomes do not need to share that positive probability. In particular, $(1/2,1/2,0)$ has both quantities equal to $1$ bit, while $(1,0,0)$ has both zero. The same equality characterization holds for a countable alphabet: equality forces only finitely many positive-probability outcomes, because all must have the same positive mass. The uniform maximum of Shannon entropy is discussed in Shannon's original treatment.^[ref]

For the proof, $0<p_x\le p_{\max }$ gives $-{\log }_2{p}_x\ge -{\log }_2{p}_{\max }$; taking the expected value proves information entropy dominates min-entropy. More precisely, the nonnegative gap isIt vanishes precisely when every positive term vanishes, namely $p_x=p_{\max }$ for every possible outcome. With $m$ such outcomes, normalization forces $p_{\max }=1/m$. If Shannon entropy is infinite, the inequality still holds and is strict, because min-entropy of a countable probability mass function is finite. No subtraction of two infinite quantities is needed.

An analytical family makes the gap visible. Give $A$ probability $q$ and each of $B,C,D,E$ probability $(1-q)/4$, with $1/5\le q\le 1$. Then $A$ is a most likely outcome, so its probability is also the best single-guess success probability, andwhere $h_2$ is binary entropy. The entropy bound with one prescribed probability explains the second expression: it separates whether $A$ occurred from which of four equally likely alternatives occurred. At $q=1/5$, all five outcomes are uniform and both quantities are ${\log }_{2}5\approx 2.3219$ bits. At $q=1/2$, Shannon entropy is $2$ bits while min-entropy is $1$ bit, reproducing the biased example in min-entropy. At $q=1$, both are zero; the four impossible alternatives contribute nothing. For every $1/5<q<1$, the probability mass function is nonuniform on its possible outcomes and the gap is strictly positive.

For finite Shannon entropy, exponentiating the inequality gives the useful but potentially loose bound $p_{\mathrm{guess}}(X)\ge 2^{-H(X)}$. Equality holds exactly for a discrete uniform distribution on the possible outcomes. Thus Shannon entropy alone supplies a lower bound on optimal success, not the upper bound needed to promise that every guess is unlikely. Shannon's source coding theorem concerns average compression under its hypotheses, while min-entropy answers the single-guess question directly. The two quantities serve different tasks even when both are expressed in bits.

For a discrete random variable $X$ with a finite or countable alphabet and probability mass function $p_x=\mathbb{P}(X=x)$, define $p_{\max }={\max }_x{p}_x$. Its min-entropy, using base-two logarithms, isThe unit is the bit. This is the smallest information content of any possible outcome: it records the easiest outcome to predict, whereas Shannon entropy averages information content over all outcomes. A normalized countable probability mass function has an attained positive maximum, so $H_{\infty }(X)$ is finite even when Shannon entropy is infinite. Outcomes of zero probability cannot minimize the surprise and are omitted from the minimum; including extra impossible symbols changes neither quantity.^[ref]

Suppose someone knows the probability mass function but receives no observation correlated with $X$, and must name exactly one outcome. Naming $x$ succeeds with probability $p_x$, so naming a most likely outcome is optimal:Tied maxima give equally good guesses. Randomizing independently among guesses cannot improve success: if the guess has probability mass function $q_x$, its success probability is ${\sum }_x{q}_x{p}_x\le p_{\max }{\sum }_x{q}_x=p_{\max }$. This is the simplest instance of maximum a posteriori decoding. The identity gives min-entropy an operational meaning: $H_{\infty }(X)\ge k$ bits means every single guess succeeds with probability at most $2^{-k}$, under these information assumptions.^[ref]

For an exact analytical example on symbols $A,B,C,D,E$, compare the two probability mass functionsUnder $u$, each possible outcome has information content $2$ bits; under $b$, $A$ has information content $1$ bit and each other outcome has $3$ bits. Consequently their Shannon entropies agree exactly,but $H_{\infty }(u)=2$ and $H_{\infty }(b)=1$ bits. A best single guess therefore succeeds with probability $1/4$ under $u$ and $1/2$ under $b$. The same average surprise can hide a factor of two in immediate predictability. The zero entry in $u$ contributes zero to Shannon entropy, using $0{\log }_{2}0=0$.

For the discrete uniform distribution on $m$ possible outcomes, $H_{\infty }={\log }_{2}m$ bits and the best single guess succeeds with probability $1/m$. A certain outcome has $H_{\infty }=0$ and success probability one. Thus min-entropy is useful when a common outcome could defeat an attempt to make outputs unpredictable, even if rare alternatives raise Shannon entropy.

For cryptography, this unconditional calculation alone does not certify security: an observer might know $X$ already. If a report $Y=X$ reveals a uniform four-outcome source, its unconditional min-entropy remains $2$ bits but guessing after the report succeeds with probability one. For finite classical reports, the appropriate success calculation uses conditional probabilities and maximum a posteriori decoding:The conditional entropy $H(X\mid Y)$ instead averages remaining surprise. A security argument must specify the observer's information and the protocol's required guarantee; neither unconditional min-entropy nor Shannon entropy supplies those assumptions.

The unit for information content or information entropy computed with natural logarithms. Converting to bits divides the numerical value by $\log 2$, so $1\mathrm{nat}=(1/\log 2)\mathrm{bits}$.

Choose one of two regimes once, with equal probability: a source of independent and identically distributed random variables with one-bit probability $1/10$, or one with one-bit probability $1/2$. Keep the chosen regime forever. The resulting source is a strictly stationary process, because shifting a block does not change either component law. It is not an ergodic stationary process: its limiting empirical frequency identifies the regime.

The source's information rate isConditioning on the single regime bit changes any block's information entropy by at most one bit, so dividing by the block length proves this rate. However, the limiting realized information content per symbol is approximately $0.468996$ in the first regime and one in the second. The alternative component's likelihood ratio decays exponentially along a typical path from the chosen component, at a rate given by the positive relative entropy between the two Bernoulli distributions. The mixture therefore has two possible pathwise limits, not one limit equal to the averaged information rate.

The count $K_n$ has binomial distribution with mean $n/4$ and variance $3n/16$. The information content per symbol is an affine function of $K_n/n$, so its variance isChebyshev's inequality therefore gives the finite-sample lower boundThe bound is often loose, but it proves that every fixed positive tolerance eventually captures nearly all the probability. Direct summation of the binomial distribution's probability masses gives $P(T_{0.1}^{(640)})\approx 0.999779$. A finite sample does not need to have a nearly full typical set: the 40-symbol example above captures only about $0.639$.