core31 min studyLesson 250 in Mathematics

Decision theory and minimax risk

Compare full risk functions and prove a finite-sample minimax result using a constant-risk Bayes estimator.

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What you’ll understand

  • Define loss, risk, Bayes risk, and minimax risk.
  • Calculate the bias-variance tradeoff of binomial shrinkage.
  • Prove minimaxity by matching an attainable worst-case risk with a Bayes lower bound.

Two estimators can exchange advantages across a parameter space: one works better near the center, another near the extremes. Decision theory makes the comparison explicit by specifying the cost of an error and averaging that cost over possible datasets at each parameter. Minimax analysis then asks for the smallest achievable worst-case risk. It is a criterion for choosing a procedure, not a guarantee that its realized error is always small. A small binomial example lets us prove a complete finite-sample minimax claim instead of relying on a plot or an appeal to intuition.

A procedure has a whole risk function

A decision rule δ(X)\delta(X) maps observations to an action. A loss L(θ,a)L(\theta,a) measures the cost of action aa if the parameter is θ\theta, and risk averages that loss under the sampling law at the fixed parameter. Squared-error risk combines bias and variance; asymmetric costs or classification decisions lead to other losses. Changing the loss changes the optimization problem, so an optimal estimator must always be described together with its model, target, action space, and loss.

An estimator dominates another if its risk is no larger everywhere and smaller somewhere. When risk curves cross, dominance cannot rank them. Bayes analysis averages risk against a specified proper prior and minimizes that average. Minimax analysis minimizes the supremum of risk over the parameter space. The prior can be used as a mathematical device to obtain a lower bound on worst-case performance even if no subjective interpretation of it is intended. Neither criterion excuses checking whether the model and loss represent the actual application.

R(θ,δ)=EθL(θ,δ(X)),r∗=inf⁡δsup⁡θ∈ΘR(θ,δ),rπ=inf⁡δ∫R(θ,δ) π(dθ)≤r∗R(\theta,\delta)=E_\theta L(\theta,\delta(X)),\qquad r^*=\inf_\delta\sup_{\theta\in\Theta}R(\theta,\delta),\qquad r_\pi=\inf_\delta\int R(\theta,\delta)\,\pi(d\theta)\le r^*
For a proper prior, an average cannot exceed the worst case for a fixed rule; taking infima yields the Bayes lower bound on minimax risk.↔ Scroll wide equations horizontally.

Shrinkage trades bias for a flatter risk

Let K∼Binomial⁡(n,p)K\sim\operatorname{Binomial}(n,p) with p∈[0,1]p\in[0,1], and estimate pp under squared error. The sample proportion has risk p(1−p)/np(1-p)/n, maximized at the center. Consider instead δa=(K+a)/(n+2a)\delta_a=(K+a)/(n+2a) for a>0a>0. Its mean is drawn toward one half. Its bias is a(1−2p)/(n+2a)a(1-2p)/(n+2a), and its variance is np(1−p)/(n+2a)2np(1-p)/(n+2a)^2. Adding their squared-bias and variance contributions gives a risk polynomial that can be inspected exactly.

Set a=n/2a=\sqrt n/2. Since (1−2p)2=1−4p(1−p)(1-2p)^2=1-4p(1-p), the coefficient of p(1−p)p(1-p) in the numerator cancels, leaving risk constant in pp. This rule sacrifices zero risk at the endpoints to reduce the largest risk near the center. Constant risk alone does not prove minimaxity: there could still exist a different rule with a lower constant or lower worst case. A lower bound matching this achieved constant is the missing ingredient.

R(p,δa)=np(1−p)+a2(1−2p)2(n+2a)2,a=n2⇒R(p,δa)=14(n+1)2R(p,\delta_a)=\frac{np(1-p)+a^2(1-2p)^2}{(n+2a)^2},\qquad a=\frac{\sqrt n}{2}\Rightarrow R(p,\delta_a)=\frac1{4(\sqrt n+1)^2}
This exact finite-sample calculation shows how a particular shrinkage amount flattens the squared-error risk over the entire closed probability interval.↔ Scroll wide equations horizontally.

Match the lower bound to prove minimaxity

Use the proper beta prior with equal positive shape parameters aa. Multiplying its density by the binomial likelihood gives a beta posterior with shapes K+aK+a and n−K+an-K+a. Under squared-error loss the posterior mean minimizes posterior expected loss, because completing the square separates posterior variance from squared distance to that mean. Hence δa\delta_a is Bayes for this prior. At the selected value of aa, its constant risk is also its integrated Bayes risk.

Every competing rule has worst-case risk at least its prior-averaged risk, which is at least the Bayes optimum. The candidate attains exactly that value at every parameter, so the lower and upper bounds coincide and prove minimaxity. This proof covers arbitrary measurable rules with well-defined risks, including randomized rules: conditional averaging cannot increase squared-error loss. It establishes the criterion for this binomial experiment and squared-error target only. A different parameter restriction or loss can select a different procedure, and minimaxity does not mean uniform dominance over the unbiased sample proportion.

A complete four-trial minimax calculation

For K∼Binomial⁡(4,p)K\sim\operatorname{Binomial}(4,p) and squared-error loss for estimating pp, find the constant-risk shrinkage rule, prove its minimax risk, and report its estimate when K=3K=3.

  1. Set a=4/2=1a=\sqrt4/2=1, giving δ=(K+1)/6\delta=(K+1)/6. Its risk numerator is 4p(1−p)+(1−2p)2=14p(1-p)+(1-2p)^2=1, so risk is 1/361/36 for every pp.

  2. A Beta(1,1) prior is uniform on the probability interval. Its posterior after observing K has mean (K + 1)/6, so this rule is Bayes under squared error. Its Bayes risk equals its constant frequentist risk, 1/36.

  3. Any rule has worst-case risk at least this Bayes optimum. Our rule attains 1/361/36 everywhere, proving minimaxity. At K=3K=3 it reports 4/6=2/34/6=2/3, compared with the unbiased sample proportion 3/43/4.

The result

The exact minimax risk is 1/36. The sample proportion has worst-case risk 1/16, although it has smaller risk than shrinkage at the endpoints.

Explore the model

Interactive lab

Interactive lab · probability

Recover risk from a sampling law

Reset, set both p=0.5,n=4p=0.5,n=4, and predict sample-proportion risk 1/161/16. Weight the five squared errors of (K+1)/6(K+1)/6 by the exact bars to obtain 1/361/36. Reset and set both p=0,n=4p=0,n=4: all mass is at zero, so sample-proportion risk is zero while shrinkage risk remains (1/6)2(1/6)^2. The lab shows the count law; calculate the two risks rather than reading a nonexistent risk curve.

Recover risk from a sampling lawThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forLower worst-case risk can require accepting higher risk at some parameter values.

Check your understanding

Which argument proves that a constant-risk rule with risk r is minimax?

Try it yourself

For nine iid Bernoulli observations, use a=n/2a=\sqrt n/2 to find the minimax squared-error estimator of pp, its constant risk, and its estimate at six successes. Compare its worst-case risk with that of the sample proportion.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The square root of nine is three. Substitute the resulting a into both the estimator and constant-risk formulas before evaluating the observed count.

Compare with the solution

Here a=1.5a=1.5 and δ=(K+1.5)/12\delta=(K+1.5)/12. Its constant risk is 1/[4(3+1)2]=1/64=0.0156251/[4(3+1)^2]=1/64=0.015625. At six successes the estimate is 7.5/12=0.6257.5/12=0.625. The sample proportion has worst-case risk 1/(4⋅9)=1/36≈0.0277781/(4\cdot9)=1/36\approx0.027778. The same Beta(1.5,1.5) Bayes lower-bound argument proves minimaxity; this comparison does not assert that shrinkage is better at every probability.

Take it with you

  • Risk averages loss over repeated data at each fixed parameter.
  • Minimax risk optimizes the worst case, while Bayes risk averages against a prior.
  • A constant-risk Bayes rule proves minimaxity when its achieved risk matches the Bayes lower bound.

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