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The Efron–Stein inequality bounds the variance of a complicated output by the effect of replacing one input at a time. Let $X_1,\dots ,X_n$ be mutually independent random variables, not necessarily identically distributed, and $Z=f(X_1,\dots ,X_n)$ for a measurable function $f$, with $\mathbb{E}{Z}^2<\infty$. Take $X_1^{\prime },\dots ,X_n^{\prime }$ with $X_i^{\prime }$ having the same probability distribution as $X_i$, all $2n$ inputs mutually independent. DefineUse the same function $f$, keeping every other input fixed. ThenThe expected value averages over both the original inputs and their replacements. An infinite right side gives a trivial bound; under the stated square-integrable condition every term is finite.^[ref]

For a concrete four-state example, take independent Bernoulli random variables with $\mathbb{P}(X_1=1)=0.3$ and $\mathbb{P}(X_2=1)=0.6$, and set $Z=\max \{X_1,X_2\}$. This output records whether either input is one. The states $(0,0),(1,0),(0,1),(1,1)$ have respective probabilities $0.28,0.12,0.42,0.18$, soReplacing $X_1$ changes $Z$ only when $X_2=0$ and $X_1^{\prime }\ne X_1$. A fresh Bernoulli draw can repeat the old value: the replacement is not a forced flip. By independence, the expected value of the squared change isSimilarly,Thus Efron–Stein inequality gives $0.2016\le (0.168+0.336)/2=0.252$. The larger second contribution reflects both its replacement probability and how often the other input leaves the output sensitive to it.

For the same inputs, consider instead $W=X_1+2X_2$. The variance of a sum of independent random variables givesHere $\mathbb{E}[(W-W_1^{\prime }{)}^2]=0.42$ and $\mathbb{E}[(W-W_2^{\prime }{)}^2]=1.92$, so the Efron–Stein inequality is exact: $(0.42+1.92)/2=1.17$. More generally, for $W=c+{\sum }_i{g}_i(X_i)$ with square-integrable summands, each contribution $\frac{1}{2}\mathbb{E}[(W-W_i^{\prime }{)}^2]$ equals $\mathrm{Var}(g_i(X_i))$ and these variances add exactly.

Independence is essential. Let $X_1=X_2=B$ for a single Bernoulli random variable with parameter $1/2$, and take $W=X_1+X_2=2B$. Its variance is one. If each coordinate is nevertheless replaced by an independent draw from its marginal Bernoulli distribution, each squared-change expected value is $1/2$, giving the false proposed upper bound $1/2$. Replacement now creates off-diagonal states $(0,1)$ and $(1,0)$ that the original input never visits. The original coordinates fail the hypothesis of Efron–Stein inequality.

The attraction is local calculation: a function may involve many inputs, yet changing one can have a simple effect. Each half squared-change expected value supplies a coordinate's contribution to an upper budget for the total variance. Applying Chebyshev's inequality then bounds a tail probability using that budget. These contributions need not partition the actual variance, as the maximum example shows.

The name commemorates Bradley Efron and Charles M. Stein. Their 1981 paper The Jackknife Estimate of Variance appeared in The Annals of Statistics, volume 9, issue 3, pp. 586–596.^[ref]

By Bradley Efron and Charles M. Stein, printed as B. Efron and C. Stein. Published in The Annals of Statistics 9(3), 586–596, 1981. The Institute of Mathematical Statistics is the publisher; Project Euclid hosts the article record.

Article record: projecteuclid.org/journals/annals-of-statistics/volume-9/issue-3/The-Jackknife-Estimate-of-Variance/10.1214/aos/1176345462.full

Bibliographic verification: Publications. The journal publication is distinct from the October 1978 Stanford technical report. Full text was unavailable during inspection; only the title and publication metadata are used.