Direct Domination Tests for Binomial Estimators

From an Estimator to a Risk Comparison

For a binomial observation, a proposed improvement is not established by checking a few likely values of the success probability. Domination is a statement over the entire parameter space, including its boundaries. The useful feature of the binomial model is that the risk of any estimator based on the count is a polynomial in the success probability. That makes a direct comparison possible: derive the risk difference, then determine its sign on the whole interval.

Throughout this tutorial, $X\sim\operatorname{Bin}(n,\theta)$, where $n$ is a fixed positive integer and $0\leq\theta\leq1$. The action is an estimate of $\theta$, and the loss is squared error, $L(\theta,a)=(a-\theta)^2$. An estimator is a deterministic function $\delta$ of the observed count, with values $\delta(0),\ldots,\delta(n)$. Its risk is the expected loss at parameter value $\theta$.

Definition: The squared-error risk of an estimator $\delta$ in the binomial model is $$ R(\theta,\delta)=E_{\theta}\big[(\delta(X)-\theta)^2\big] =\sum_{x=0}^{n}{n\choose x}\theta^x(1-\theta)^{n-x}\big(\delta(x)-\theta\big)^2. $$ An estimator $\delta_1$ dominates $\delta_0$ on $[0,1]$ if $R(\theta,\delta_1)\leq R(\theta,\delta_0)$ for every $\theta\in[0,1]$, with strict inequality for at least one value of $\theta$.

At $\theta=0$, the binomial count is $X=0$ with probability one; at $\theta=1$, it is $X=n$ with probability one. Thus the risk formula at the boundaries is interpreted directly from the distribution, or equivalently by evaluating its polynomial extension there. Boundary checks can rule out a proposed domination immediately, but passing those checks alone does not establish domination in the interior.

Risk of an Affine Estimator

A useful starting point is an estimator of the form $\delta(X)=aX/n+b$, where $a$ and $b$ are fixed real numbers. The sample proportion $X/n$ is the special case $a=1,b=0$. The next derivation separates the risk into its variance and squared bias, making parameter dependence visible.

Result: For $X\sim\operatorname{Bin}(n,\theta)$ and $\delta(X)=aX/n+b$, $$ R(\theta,\delta)=\frac{a^2\theta(1-\theta)}{n} +\big((a-1)\theta+b\big)^2. $$

To derive this expression, use $E_{\theta}[X]=n\theta$ and $\operatorname{Var}_{\theta}(X)=n\theta(1-\theta)$. Then $E_{\theta}[\delta(X)]=a\theta+b$, so the bias is $(a-1)\theta+b$, and $\operatorname{Var}_{\theta}(\delta(X))=a^2\theta(1-\theta)/n$. For any random variable with finite second moment,

E_{\theta}\big[(\delta(X)-\theta)^2\big] =\operatorname{Var}_{\theta}(\delta(X)) +\big(E_{\theta}[\delta(X)]-\theta\big)^2.

Substituting the variance and bias gives the stated risk. This decomposition is exact; it is not an approximation or a large-sample result. It also applies at the boundaries, where the variance term is zero.

Worked Example: Risk of a Linear Adjustment to the Sample Proportion

Let $n=4$ and consider $\delta(X)=\tfrac{1}{2}(X/4)+\tfrac{1}{4}$, an estimator that averages the sample proportion with $1/2$. Find its squared-error risk.

Here $a=1/2$ and $b=1/4$. The variance term is $\tfrac{1}{4}\theta(1-\theta)/4=\theta(1-\theta)/16$. Its expectation is $\theta/2+1/4$, so its bias is $1/4-\theta/2=(1-2\theta)/4$. Therefore,

R(\theta,\delta) =\frac{\theta(1-\theta)}{16} +\frac{(1-2\theta)^2}{16} =\frac{1-3\theta+3\theta^2}{16}.

For a check, at $\theta=0$ the observed count is zero and the estimate is $1/4$, giving risk $1/16$, which agrees with the formula. At $\theta=1$, the estimate is $3/4$, again giving squared error $1/16$. These boundary checks confirm the bias contribution has been retained.

Reduce Domination to a Polynomial Inequality

For any fixed estimator $\delta$, the risk formula is a finite sum of products of powers of $\theta$ and $1-\theta$, multiplied by squared errors that are quadratic in $\theta$. Consequently, $R(\theta,\delta)$ is a polynomial in $\theta$, of degree at most $n+2$. A difference of two risks is therefore also a polynomial. This observation turns the definition of domination into a global sign-checking problem.

Theorem: Let $\delta_0,\delta_1$ be two deterministic estimators in this binomial model, and define $$ D(\theta)=R(\theta,\delta_1)-R(\theta,\delta_0). $$ Then $\delta_1$ dominates $\delta_0$ on $[0,1]$ exactly when $D(\theta)\leq0$ for every $\theta\in[0,1]$ and $D(\theta)<0$ for at least one $\theta$. To verify the weak inequality over the interval, it is sufficient and necessary to check $D$ at $\theta=0$, $\theta=1$, and every interior point where $D'(\theta)=0$.

Proof. The first statement follows directly from the definition of domination and the definition of $D$. For the finite check, $D$ is a polynomial, hence continuous on the compact interval $[0,1]$. It attains a maximum there. If a maximum occurs at an endpoint, its value is among the endpoint checks. If a maximum occurs in the interior, the derivative is zero there by the usual necessary condition for an interior extremum. Thus $D\leq0$ at the endpoints and at all interior stationary points implies its maximum is at most zero, so $D\leq0$ throughout the interval. Conversely, if $D\leq0$ throughout, it is nonpositive at every listed point. Strict inequality at at least one parameter value is still required for domination rather than risk equality everywhere. If $D$ is constant, the endpoint check already determines its sign. $\square$

This result does not say that checking only the endpoints is enough. An interior stationary point can be where the risk difference is largest, and a polynomial can change sign between the boundaries and that point. Finding the stationary points may require solving $D'(\theta)=0$; when that is inconvenient, factoring $D$, bounding its terms, or identifying its roots can provide another full-interval sign argument.

Worked Example: The Sample Proportion and Uniform Smoothing

Compare the sample proportion $\delta_0(X)=X/n$ with the smoothed estimator $\delta_1(X)=(X+1)/(n+2)$ under squared-error loss. Does either dominate the other?

For the sample proportion, the affine formula gives $R(\theta,\delta_0)=\theta(1-\theta)/n$. The smoothed estimator has expectation $(n\theta+1)/(n+2)$, bias $(1-2\theta)/(n+2)$, and variance $n\theta(1-\theta)/(n+2)^2$. Its risk is therefore

R(\theta,\delta_1) =\frac{n\theta(1-\theta)+(1-2\theta)^2}{(n+2)^2}.

Subtract the sample-proportion risk and write $q=\theta(1-\theta)$:

D(\theta)=R(\theta,\delta_1)-R(\theta,\delta_0) =\frac{n-(8n+4)q}{n(n+2)^2}.

At $\theta=0$, $q=0$, so $D(0)=1/(n+2)^2>0$. Thus the smoothed estimator cannot dominate the sample proportion. At $\theta=1/2$, $q=1/4$, giving

D(1/2)=\frac{n-(8n+4)/4}{n(n+2)^2} =-\frac{n+1}{n(n+2)^2}<0.

The risk difference has opposite signs at these two parameter values. Hence neither estimator dominates the other: the smoothed estimator has lower risk at $1/2$, but higher risk at $0$. In particular, a reduction in risk near the center of the parameter space does not establish a uniform improvement.

Direct Comparisons: Crossing and Domination

A direct test should display the risk difference and the reasoning that establishes its sign. The next comparison shows a simple crossing, while the following one shows how a global inequality can certify domination. They also illustrate why the parameter interval and the action space must be stated when interpreting a result.

Worked Example: Two Constant Estimators Have Crossing Risks

Consider $\delta_0(X)=1$ and $\delta_1(X)=1/2$, both under squared-error loss for $0\leq\theta\leq1$. Since both estimators are constant, their risks do not depend on the binomial count:

R(\theta,\delta_0)=(1-\theta)^2,\qquad R(\theta,\delta_1)=(1/2-\theta)^2.

The difference, using $\delta_1$ minus $\delta_0$, is

D(\theta)=(1/2-\theta)^2-(1-\theta)^2 =\theta-\frac{3}{4}.

For $0\leq\theta<3/4$, $D(\theta)<0$, so the estimator $1/2$ has lower risk. At $\theta=3/4$, the risks are equal. For $3/4<\theta\leq1$, $D(\theta)>0$, so the estimator $1$ has lower risk. Neither dominates the other, because each is worse on a nonempty part of the parameter interval. The crossing point follows from solving the risk equality, not from comparing the numerical sizes of the estimates.

Worked Example: A Full-Interval Domination Certificate

For a transparent domination test, let $\delta_0(X)=1+X/n$ and $\delta_1(X)=1$. These are real-valued estimators under squared-error loss; the first can exceed one, so it would be excluded if actions were constrained to $[0,1]$. Determine whether the constant estimator dominates it on the stated unrestricted action space.

Set $Y=X/n$, so $0\leq Y\leq1$ and $E_{\theta}[Y]=\theta$. The risk difference is

\begin{aligned} D(\theta) &=R(\theta,\delta_1)-R(\theta,\delta_0)\\ &=E_{\theta}\left[(1-\theta)^2-(1+Y-\theta)^2\right]\\ &=-2(1-\theta)E_{\theta}[Y]-E_{\theta}[Y^2]\\ &=-2\theta(1-\theta)-\left(\theta^2+\frac{\theta(1-\theta)}{n}\right)\\ &=-2\theta+\theta^2-\frac{\theta(1-\theta)}{n}. \end{aligned}

Here $E_{\theta}[Y^2]=\operatorname{Var}_{\theta}(Y)+(E_{\theta}[Y])^2=\theta(1-\theta)/n+\theta^2$. Since $0\leq\theta\leq1$, both $2\theta(1-\theta)$ and $E_{\theta}[Y^2]$ are nonnegative, so the displayed expression for $D$ as their negatives is nonpositive everywhere. It is strictly negative for every $\theta>0$, because $E_{\theta}[Y^2]>0$ there; at $\theta=0$, $X=0$ with probability one and the risks are equal. Therefore $\delta_1$ dominates $\delta_0$. The calculation also shows why a domination claim must identify its rule class: with the action constraint $[0,1]$, $\delta_0$ is not an available estimator.

How to Audit a Proposed Comparison

For estimators specified by their values at each possible count, substituting into the binomial risk sum gives a direct polynomial risk function. For affine estimators, the bias-variance formula is usually shorter. Either way, the comparison concerns the entire risk function, not merely its value at the parameter estimate, a central parameter value, or a convenient subset of cases.

Write both risks.
Use the binomial probability mass function for general rules, or the bias-variance decomposition when its hypotheses apply.

Form one signed difference.
For a proposed estimator $\delta_1$ to dominate $\delta_0$, define $D=R(\theta,\delta_1)-R(\theta,\delta_0)$. The required direction is $D\leq0$.

Test the whole interval.
Check endpoints and interior stationary points, or give another argument that proves the polynomial is nonpositive for every $0\leq\theta\leq1$.

Check strictness and the rule class.
Establish strict improvement somewhere, and confirm that both estimators satisfy any stated action constraints.

A common mistake is to calculate a lower risk at one value, such as $\theta=1/2$, and call the estimator dominant. The smoothed-estimator comparison above disproves that reasoning: it improves at $1/2$ but has higher risk at $\theta=0$. The correction is to keep the risk difference as a function of $\theta$ and establish its sign over the full parameter interval. Another mistake is to check only the endpoints. The polynomial criterion shows why that is insufficient: an interior stationary point can determine the maximum risk difference.

Key takeaway: In the binomial model under squared-error loss, direct domination testing means deriving the risk difference and proving its sign on all of $[0,1]$. Endpoint checks can refute a claim quickly; a complete positive claim needs a full-interval argument, including interior behavior.

Check Your Understanding

For each question, attend to the sign of the risk difference and the entire parameter space.

For $\delta(X)=aX/n+b$, identify its bias and variance, and state how they combine to give squared-error risk.
Why does $D(0)>0$, for $D=R(\theta,\delta_1)-R(\theta,\delta_0)$, rule out $\delta_1$ dominating $\delta_0$?
In the comparison of $X/n$ with $(X+1)/(n+2)$, what does the sign of $D(1/2)$ establish, and what does $D(0)$ establish?
Why can checking only $\theta=0$ and $\theta=1$ fail to prove a polynomial risk difference is nonpositive on $[0,1]$?
What additional condition, beyond nonpositive risk difference everywhere, is required for domination as defined here?