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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]

The independent and identically distributed hypothesis factors the joint probability mass function:Symbols outside the support of a probability distribution $p$ never occur, almost surely. On its finite positive probability support, $\log p(X_i)$ is bounded, so the strong law of large numbers applies:This is the asymptotic equipartition property with the sign appropriate to a logarithm of a probability mass.

For fixed $\epsilon >0$, the asymptotic equipartition property gives $\mathrm{Pr}(X^n\notin A^{n,\epsilon })\to 0$. The strong law of large numbers applied to $X_i$, which are bounded because their numerical alphabet is finite, gives $\mathrm{Pr}(X^n\notin B^{n,\epsilon })\to 0$. The union bound therefore yieldsSince $\delta >0$, eventually $\mathrm{Pr}(X^n\in S_n)\ge 2^{-\delta }$. Each word in $S_n$ has probability mass at most $2^{-n(H(p)-\epsilon )}$, soRearranging provesfor sufficiently large $n$. The key is that the factor $2^{-\delta }$ is a fixed number below one, however small $\delta$ is. This is an instance of typical sets with an additional high-probability constraint.

Use base-two logarithms and put $H=H(U_1)=-{\sum }_i{p}_i{\log }_2{p}_i$, with $0{\log }_{2}0=0$. Zero-probability letters can be omitted. The random variables $I_j=-{\log }_2{p}_{U_j}$ are self-information values, are IID, and have finite mean $H$ because the alphabet is finite. Independence givesFor an arbitrary $\delta >0$, define the typical setThe weak law of large numbers implies $P(T_{n,\delta })\to 1$. Every string in this set satisfiesThis is the asymptotic equipartition property for the given finite IID source, here obtained directly from the weak law of large numbers.

For the upper bound, summing the lower bound on typical-string probabilities yields $|T_{n,\delta }|\le 2^{n(H+\delta )}$. For sufficiently large $n$, its probability is at least $1-\epsilon$, so it is an admissible set in the definition of $M(n,\epsilon )$. HenceFor the lower bound, let $A$ be any admissible set and write ${\eta }_n=P(T_{n,\delta }^c)\to 0$. ThenEach member of this intersection has probability at most $2^{-n(H-\delta )}$. Therefore, for all sufficiently large $n$,This applies in particular to a minimum-size admissible set, giving a liminf at least $H-\delta$. Letting $\delta \downarrow 0$ proves both existence and the value of the minimal high-probability source-set exponent:For a deterministic source, $H=0$ and $M(n,\epsilon )=1$, consistent with the formula.

For coding, label the strings in an admissible set with distinct binary indices of length $\lceil {\log }_{2}M(n,\epsilon )\rceil$. A decoder reconstructs those strings exactly and may fail outside the set. Thus asymptotically $H$ bits per source symbol describe any fixed fraction $1-\epsilon$ of the probability mass. More strongly, every rate $R>H$ permits block codes with error tending to zero: choose $\delta <R-H$, encode the typical strings individually and reserve one index for all other strings. The required number of indices is at most $2^{n(H+\delta )}+1\le 2^{nR}$ for large $n$.

Conversely, a block encoder with at most $2^{nR}$ outputs can reconstruct at most that many distinct input strings correctly. If $R<H$, choose $0<\delta <H-R$. Its success probability is at mostThis gives the strong converse for fixed-rate source coding. It is the operational significance of the information entropy limit for reliable source encoding at a rate. The allowance of a small reconstruction error matters: a fixed-length code that must encode every possible string without error generally needs the larger rate ${\log }_2|\{i:p_i>0\}|$.

Let $Y_i=-{\log }_2{P}_Z(Z_i)$. Since the alphabet is finite, these independent and identically distributed random variables are finite almost surely, have mean $H(Z)$, and have finite variance $v$. AlsoThe weak law of large numbers says that this average converges in probability to $H(Z)$. ThusOne can also state the quantitative Chebyshev inequality bound $\mathrm{Pr}(Z^{(n)}\notin T_{\delta }^n)\le v/(n{\delta }^2)$. If $v=0$, every word of positive probability is in the typical set. This is the finite-alphabet asymptotic equipartition property.

Fix $0\le ϵ<1$ and $\delta >0$. By the asymptotic equipartition property, all sufficiently large $n$ satisfy $\mathrm{Pr}(Z^{(n)}\notin T_{\delta }^n)\le (1-ϵ)/2$. For any epsilon-sufficient set $S_n$,Each word in this intersection has probability at most $2^{-n(H(Z)-\delta )}$. Hencewhich gives the minimum size of a sufficient set for an IID source:The Chebyshev inequality used above supplies the sufficient quantitative condition $n\ge 2v/[{\delta }^2(1-ϵ)]$. No assumption that $S_n$ itself is a typical set is needed; its intersection with one provides the bound.

LetHere $S_n$ has the binomial distribution with parameters $(n,\lambda /n)$. Part a, now with independent coordinates, givesBy Pinsker's inequality, the probability mass functions therefore converge in total variation, and in particular $S_n$ converges in distribution to $Z\sim \mathrm{Poisson}(\lambda )$.

The joint probability of the observed row depends only on $S_n$:Taking logarithms in any fixed base and choosing $c_n=\log n$ giveswhereThe convergence lemma supplied in the question now yieldsThis sparse triangular array therefore has a random limiting normalized self-information rather than the constant limit in the usual asymptotic equipartition property.

The asymptotic equipartition property is $-n^{-1}{\log }_{2}p(X_1,\dots ,X_n)\to H$ in probability. For an i.i.d. source this is the weak law applied to $-{\log }_{2}p(X_i)$, and independence gives $H(X_1,\dots ,X_n)=nH(X)$. Thus its information rate and AEP constant are both $H(X)$.

For a typical set of probability at least $1-\delta$,The upper bound sums the lower probability bound for each word; the lower bound sums the upper probability bound and uses the set's total probability. These bounds turn the asymptotic equipartition property into exponential estimates of source-coding dimension.

Let $A^{n,\epsilon }$ be the weak typical set of a finite alphabet independent and identically distributed source with Shannon entropy $H$, and let $B_n$ be any sets with $\mathrm{Pr}(X^n\in B_n)\to 1$. The asymptotic equipartition property and union bound give $\mathrm{Pr}(X^n\in A^{n,\epsilon }\cap B_n)\to 1$. Every word in the intersection has probability mass between $2^{-n(H+\epsilon )}$ and $2^{-n(H-\epsilon )}$. Summing these bounds givesFor every fixed $\delta >0$, the first probability eventually exceeds $2^{-\delta }$, proving the title bound. A sample mean within a fixed positive tolerance of the expected value supplies $B_n$ by the strong law of large numbers.