U-statistics and Hoeffding decomposition
Separate first-order information from a degenerate pair remainder and identify when the usual square-root-n normal approximation fails.
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The question for today
What you’ll understand
- Write the centered order-two Hoeffding decomposition.
- Calculate its exact variance components.
- Distinguish nondegenerate Gaussian limits from a concrete degenerate limit.
Many useful statistics average a function over pairs or larger subsets of observations. Their terms overlap, so treating all pairs as independent badly underestimates uncertainty. Hoeffding decomposition organizes this dependence by projecting the statistic onto contributions from individual observations, then leaving a conditionally centered remainder. When the first projection is nonzero, ordinary square-root-n normal behavior emerges. When it vanishes, the rate and limit can change. Both the kernel's moments and its relation to the underlying distribution determine which case applies.
Average over distinct subsets without pretending they are independent
For n≥2 iid observations and a symmetric measurable order-two kernel h with , define θ as its expectation and average h over all unordered distinct pairs. Every term has expectation θ, so the U-statistic is unbiased. A V-statistic instead includes diagonal terms under a different normalization, and can have different finite-sample bias. The letter kernel here means a function of observations; it is not necessarily a nonnegative density-smoothing kernel.
Two pair terms sharing one observation can be correlated even though the original observations are independent. There are quadratically many pair terms, but their overlap often leaves a variance of order one over sample size rather than one over its square. Counting terms alone is therefore misleading. Symmetrization is useful when an original pair function is not symmetric, since averaging the two argument orders preserves its target expectation under iid sampling and creates the standard unordered-pair representation.
Project out the single-observation information
The first projection has expectation zero, and the canonical remainder satisfies almost surely. Expanding each pair kernel and counting how often each observation appears gives the centered statistic as twice the average first projection plus the average remainder. The two components are orthogonal in L². Remainder terms sharing one argument also have zero covariance because of conditional centering, which gives an exact variance identity.
Let zeta1 be the variance of the first projection and zeta2 the variance of the canonical remainder. The first contribution is four zeta1 divided by n, while the second is two zeta2 divided by n times n minus one. These symbols refer to the centered canonical components; some sources use the same zeta notation for overlapping-kernel covariances instead. When zeta1 is positive, the ordinary iid central limit theorem applies to the first sum and the remainder is negligible at square-root-n scale.
Degeneracy changes what must be approximated
If the first projection vanishes almost surely, the statistic is called degenerate at this distribution. Its variance is then of order one over n squared, provided the canonical variance is positive. A square-root-n normalization converges to zero rather than to a nontrivial normal distribution. The next-order limit is often a quadratic Gaussian expression, and depends on the kernel and distribution. Degeneracy is not a claim that the original observations have zero variance or that the statistic is identically constant.
This distinction matters in testing and resampling. A nondegenerate standard-error formula can collapse under a null where the first projection disappears, and a bootstrap designed for a linear projection may need modification. Moment conditions also matter: estimating variance with a squared-difference kernel needs a finite fourth moment for the square-integrable analysis used here. An algebraic unbiasedness identity can require fewer moments than a variance formula or limit theorem, so these levels of conclusion should not be bundled into one assumption-free statement.
Work it through
Sample variance is a pair average
Use h(x,y)=(x−y)²/2 and data 1,2,4. Calculate U3 and relate the kernel projections to the population mean mu and variance sigma², assuming a finite fourth moment.
The three unordered pair values are 1/2,9/2 and 2. Their average is seven-thirds. The ordinary unbiased sample variance also equals seven-thirds, using sample mean seven-thirds.
For an independent copy Y, E[(x−Y)²/2]=[(x−mu)²+sigma²]/2. The target is sigma², so h1(x)=[(x−mu)²−sigma²]/2.
Subtracting the target and both first projections leaves h2(x,y)=−(x−mu)(y−mu). Conditional expectation given x is zero. The finite fourth moment supplies the square integrability needed for the displayed projection variance formulas.
The pair terms overlap, but their Hoeffding components organize that dependence exactly. The first projection may itself vanish for special populations.
The living notebook
Explore the model
Interactive lab · sampling
A pair statistic that equals an ordinary mean
For Bernoulli data choose the order-two kernel h(x,y)=(x+y)/2; its U-statistic equals the sample mean exactly. Pause and Reset, then set both controls to p=0.5 and n=25; predict variance 0.01. Reset and set p=0.5 and n=100; predict variance 0.0025. These exact binomial bars show a nondegenerate pair statistic, with no random resampling. They do not represent the degenerate product kernel in the practice problem.
- Mean sample proportion
- 0.5
- Sample-mean variance
- 0.0125
- Uniform CDF error bound
- ≤ 0.1062
Exact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forQuadratically many overlapping pairs can carry only the same first-order information as an ordinary sample average.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let Xi be iid standard normal and h(x,y)=xy. Find the target and first projection, express Un using the sample sum and sum of squares, and derive the limit of nUn.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Expand the square of the sample sum. Use the normal distribution of the standardized sum and the law of large numbers for squared observations.
Compare with the solution
The target is zero and h1(x)=xE[X]=0, so h2=xy with variance one. The statistic is [(sum Xi)²−sum Xi²]/[n(n−1)], with variance 2/[n(n−1)]. Thus nUn equals n/(n−1) times [(sum Xi/sqrt(n))²−(sum Xi²/n)]. The first standardized sum is standard normal and the second term converges in probability to one. Slutsky's theorem gives the nonnormal limit Z²−1. Independence between those two terms is not needed for this argument.
Keep the idea
Take it with you
- Overlapping subset averages require covariance analysis rather than pair counting.
- Hoeffding projections separate a linear information term from a canonical remainder.
- A vanishing first projection changes the rate and can produce a nonnormal limit.
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