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Classification in statistical learning with two possible labels. Given the conditional probability $\eta (x)=\mathbb{P}(Y=1\mid X=x)$, a Bayes classifier under zero-one loss predicts one exactly when $\eta (x)\ge 1/2$, with arbitrary decisions at ties.

Fano's inequality turns remaining uncertainty into a constraint on decision errors. If an observation leaves many plausible answers, a decoder cannot usually recover the answer reliably. Its elegant idea is to separate two uncertainties: whether the guess was wrong, and which alternative was correct when it was wrong.

Let $X,Y$ be discrete random variables, with $X$ taking values in a finite alphabet $\mathcal{X}$ of cardinality $M\ge 2$. Choose any deterministic decoder $\hat{X}=f(Y)$ whose output lies in $\mathcal{X}$, and define its average error probability $p_e=\mathbb{P}\{\hat{X}\ne X\}$. With base-two logarithms, Fano's inequality statesHere $H(X\mid Y)$ is conditional entropy and $h_2$ is binary entropy, with $0{\log }_{2}0=0$ at the endpoints. Both sides are measured in bits; using natural logarithms throughout gives the same inequality in nats. No uniform source, independence of $X,Y$, or optimality of the decoder is assumed. Requiring its output to belong to the same alphabet is what leaves at most $M-1$ alternatives after an error.^[ref]

For an exact toy example, give $X$ the discrete uniform distribution on $\{A,B,C,D\}$. A channel reports $Y=X$ with probability $1-q$ and reports each of the other three symbols with probability $q/3$, where $0\le q\le 1$. Use the decoder $f(y)=y$. The output is uniform, and the conditional probabilities after any report $y$ areThus $p_e=q$ and direct calculation of the conditional entropy givesThis is equality in Fano's inequality for every $q\in [0,1]$. At $q=0.3$, each report leaves probabilities $(0.7,0.1,0.1,0.1)$, up to their order, and $H(X\mid Y)\approx 1.3568$ bits. The two contributions are $h_2(0.3)\approx 0.8813$ bits for the error flag and $0.3{\log }_{2}3\approx 0.4755$ bits for identifying the correct alternative after an error. These are theoretical probabilities, not measured data.

The identity decoder in this example minimizes error only for $0\le q\le 3/4$: its guessed symbol has the largest posterior probability exactly when $1-q\ge q/3$. At $q=3/4$, $X,Y$ are independent random variables and all four guesses tie. For $q>3/4$, guessing any one of the three other symbols has error probability $1-q/3<q$. Equality in Fano's inequality therefore does not certify an optimal decoder.

For usefulness, combine $h_2(p_e)\le 1$ with ${\log }_2(M-1)\le {\log }_{2}M$ to obtain the simpler consequenceFor a uniform $X$, $H(X\mid Y)={\log }_{2}M-I(X;Y)$, so this becomes $p_e\ge \max \{0,1-(I(X;Y)+1)/{\log }_{2}M\}$. This connects mutual information to unavoidable decision error in channel coding or classification in statistical learning; it can rule out reliable recovery regardless of the chosen decoder.^[ref]

The sharper right-hand side $F_M(p)=h_2(p)+p{\log }_2(M-1)$ increases from zero to ${\log }_{2}M$ on $0\le p\le 1-1/M$, then decreases: in the interior, $F_M^{\prime }(p)={\log }_2((M-1)(1-p)/p)$. When interpreting the inequality by inverting a curve, use its increasing branch to obtain a lower error threshold. A decoder that chooses a largest posterior probability, as in maximum a posteriori decoding, always has $p_e\le 1-1/M$, since the largest of $M$ conditional probabilities is at least $1/M$. Arbitrary decoders can have larger error. In particular, low conditional entropy does not force every decoder to work well: with a fair binary $X$ and $Y=X$, the rule $f(Y)=1-Y$ has $p_e=1$ even though $H(X\mid Y)=0$. Fano's inequality supplies a constraint, not an exact error prediction for each rule.

The misclassification rate is the expected zero-one loss of a classification in statistical learning rule. Its empirical version is the fraction of incorrectly labeled observations. Evaluation on the fitting sample gives a training error; evaluation on an independent test set estimates prediction error. For a majority-class leaf of a classification tree, the training error count is $n_t-{\max }_k{n}_{tk}$, unlike classification-tree deviance, which also uses the fitted class probabilities.

For binary classification, write $\eta (x)=\mathbb{P}(Y=1\mid X=x)$ and assume feature-measurable nearest-neighbour tie-breaking. If $\mathbb{E}|\eta (X_{(1)}(X))-\eta (X)|\to 0$, then the expected conditional misclassification risk of the one-nearest-neighbour classifier tends to $\mathbb{E}[2\eta (X)(1-\eta (X))]$. Couple each label and an oracle label using one uniform variable. Their disagreement has the preceding expected absolute difference. The oracle and the test label are conditionally independent Bernoulli random variables of parameter $\eta (X)$, so their disagreement probability is exactly $2\eta (X)(1-\eta (X))$. The difference of classification in statistical learning error probabilities is at most the coupling disagreement.

Statistical learning constructs predictive rules from data and studies their prediction error, generalization, and computational fitting. In supervised learning, training data contain inputs $X$ and a response variable $Y$: regression analysis predicts a numerical response, while classification in statistical learning predicts a label. Unsupervised learning instead seeks structure without supplied responses, as in cluster analysis and principal component analysis. These tasks connect learning to probability and mathematical optimization. ^[ref]

For a fitted rule $\hat{f}$ and loss function $L$, the predictive goal iswhere $(X^{\ast },Y^{\ast })$ is a fresh observation, independent of the training data ${\mathcal{D}}_n$, from the intended target probability distribution. This prediction error fixes the fitted rule and averages over future observations; averaging again over training data gives the learning procedure's average performance. Its practical beauty is that a rule compresses past observations into useful future predictions. Its central difficulty is that excellent training error need not imply good generalization: cross-validation supports model selection, and a reserved test set assesses the resulting choice.

Supervised learning fits a prediction rule from training data containing both inputs and associated target responses. Regression analysis predicts numerical responses; classification in statistical learning predicts class labels. A loss function specifies which prediction errors matter, and generalization concerns performance beyond the observations used to fit the rule.