Unbiased estimation and Rao–Blackwell improvement
Average away irrelevant randomness and prove the resulting variance improvement with conditional expectation.
On this page
The question for today
What you’ll understand
- Distinguish unbiasedness from small mean squared error.
- Derive Rao–Blackwell improvement using total variance.
- Use completeness and sufficiency to establish a unique minimum-variance unbiased estimator.
Suppose you estimate a common success probability using only the first observation while ignoring the rest. The estimate is unbiased, yet wastes information and fluctuates unnecessarily. Conditional expectation provides a systematic repair: average that estimator over all records compatible with a sufficient summary. This procedure is not averaging the unknown parameter or inventing extra samples. It replaces the estimator's avoidable variation by a deterministic function of the summary, while preserving its expectation under each candidate law. The proof reveals exactly which performance criterion improves.
Unbiasedness is one constraint, not a complete ranking
An estimator of is unbiased when its expectation equals the target for every parameter value. Under squared-error loss, mean squared error decomposes into variance plus squared bias whenever the second moment exists. Unbiasedness therefore makes risk equal variance, but does not guarantee the smallest risk among all estimators. A mildly biased estimator may reduce variance enough to improve mean squared error for some or all parameter values. Always name both the target and the loss before describing an estimator as better.
For iid Bernoulli observations, the first observation and the sample mean are both unbiased for the common probability. The first observation has variance , while the sample mean has variance . Independence makes the cross-covariance terms vanish in the latter calculation. Repeated copies of one observation would not gain this reduction. Finite-sample variance formulas here are exact under the iid model; no normal approximation or large-sample argument is being used.
Condition on a sufficient statistic
Let be sufficient and let have finite second moment under each model law. Define using a parameter-free conditional law supplied by sufficiency. This common version is an actual statistic, computable without knowing the parameter. Iterated expectation preserves the mean, so an unbiased produces an unbiased . Without sufficiency, a conditional expectation may depend on the unknown parameter and fail to define an available estimator even though the formal variance identity remains true for each fixed law.
Total variance splits the original variance into the variance of the conditional mean plus the average conditional variance. The second term is nonnegative, proving the improvement. Equality at a parameter value occurs precisely when the original estimator already equals its conditional expectation almost surely under that law. More generally conditional Jensen gives no increase in risk for convex loss as a function of the reported action, when the relevant expectations and conditional mean exist. This is a finite-sample theorem, not an asymptotic efficiency slogan.
Completeness removes remaining unbiased ambiguity
If the sufficient statistic is also complete, any two unbiased integrable functions of it estimating the same target agree almost surely under every model law. Subtract them: the difference is a function of the statistic with zero expectation everywhere, so completeness forces that difference to vanish. Combining this uniqueness with Rao–Blackwell gives the Lehmann–Scheffé conclusion: an unbiased function of a complete sufficient statistic with finite variance is uniformly minimum variance among all unbiased finite-variance estimators of that target.
Uniformly means the variance comparison holds at each parameter value, not merely on average over a prior. Minimum variance is restricted to unbiased estimators; it does not rule out a biased estimator with a better chosen risk tradeoff. Existence also matters: some targets have no unbiased estimator, or no one with finite variance. In practice, identify a useful starting estimator, find a sufficient statistic, compute the conditional mean, and then check completeness if you want a uniqueness and optimality claim. Each step supplies a distinct part of the argument.
Work it through
Improve the first binary observation
For iid Bernoulli observations, Rao–Blackwellize using the total . Compute the conditional estimate when and compare exact variances at .
Given a total of three, the four compatible records differ only in the location of the one failure and are equally likely. The first entry is a success in three of these four records.
Thus . The same symmetry gives for every possible total, so the improved estimator is the sample mean.
At , the first-observation variance is . The improved variance is . Total variance attributes the removed to average conditional variation.
The reported estimate is 0.75 when the total is three. Variance falls from 0.25 to 0.0625 at p = 0.5, while unbiasedness holds for every p.
The living notebook
Explore the model
Interactive lab · sampling
See the distribution of the improved estimator
Reset and set both ; predict sample-mean variance . Reset and set both ; predict . Inspect the exact mean-distribution bars and readouts. This compares the distributions of the first observation and four-observation mean under iid sampling; it does not animate the conditional averaging operation itself.
- 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 forAveraging independent information reduces variance without shifting the estimator's mean.
Pause & reason
Check your understanding
Your turn
Try it yourself
For four iid Bernoulli observations, is unbiased for . Find its Rao–Blackwell version given total , and evaluate it when .
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Conditional on the total, choose two distinct ordered positions. Count how many ordered pairs have both entries equal to one.
Compare with the solution
There are ordered pairs of distinct positions, of which contain two successes. Hence the conditional expectation is , giving at total three. It is unbiased for and has no larger variance than . Because the binomial count is complete sufficient, this is also the unique minimum-variance unbiased estimator up to model-null events.
Keep the idea
Take it with you
- Unbiasedness preserves the target on average but does not rank all risks.
- Rao–Blackwell removes conditional variation using a sufficient summary.
- Completeness upgrades an unbiased summary-based estimator to a unique UMVU estimator.
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