Completeness and Unbiased Estimation

Why Completeness Matters

A statistic can summarize the sample without being sufficient to determine an unbiased estimator uniquely. The key question is whether some nonzero function of that statistic can have expectation zero for every parameter value. If such a function exists, it can be added to an unbiased estimator without changing its expectation. Completeness rules out exactly this kind of ambiguity.

Completeness is therefore important when identifying a unique unbiased estimator based on a statistic. It does not, by itself, guarantee that an unbiased estimator exists, and uniqueness here means uniqueness among unbiased estimators that are functions of the specified statistic.

Definition: Let $T$ have a distribution indexed by $\theta\in\Theta$. The statistic $T$ is complete if, for every measurable function $g$ such that $E_\theta[|g(T)|]<\infty$ for every $\theta\in\Theta$, $$ E_\theta[g(T)]=0\text{ for every }\theta\in\Theta \quad\Longrightarrow\quad P_\theta(g(T)=0)=1\text{ for every }\theta\in\Theta. $$

In words, the only integrable function of a complete statistic whose expectation is zero throughout the parameter space is a function that is zero almost surely under every parameter value. The qualification “almost surely” matters: the function need not equal zero at sample points that have probability zero under the model.

Definition: The statistic $T$ is boundedly complete if the same implication holds for every bounded measurable function $g$. Completeness implies bounded completeness, because every bounded function is integrable under every probability distribution. Bounded completeness tests a smaller class of functions; it does not, in general, establish completeness.

The parameter space is part of the definition. A zero-mean identity must hold for every parameter value in the model, not merely at one selected value or at the true value as guessed from the data. Likewise, the conclusion must hold under every parameter value. A proof that silently restricts the parameter space may establish a different property.

Completeness as a Zero-Mean Test

To show that a statistic is not complete, it is enough to find one integrable function $g$ that has expectation zero for all parameter values but is nonzero with positive probability for at least one of them. To prove completeness, the task is broader: show that every function satisfying the zero-mean identity must be zero almost surely.

Write the model for the statistic.
Identify the distribution of $T$ and its parameter space $\Theta$.

Set an arbitrary function's expectation to zero.
Write $E_\theta[g(T)]=0$ as an equation that holds for every $\theta\in\Theta$.

Use the parameter variation.
Compare coefficients, differentiate an integral identity when justified, or otherwise use the fact that the equation holds throughout the parameter space.

Reach the almost-sure conclusion.
Show that $g(T)=0$ with probability one under every parameter value. If a nonzero function satisfies the zero-mean identity instead, the statistic is incomplete.

The following finite-support example illustrates how a family of expectation equations can force every value of $g$ to vanish. The probabilities are chosen to make the calculation transparent; they describe a hypothetical statistical model, not observed data.

Worked Example: Testing Completeness in a Finite Model

Suppose $T$ takes values $0,1,2$, with parameter space $-0.1<\theta<0.1$, and probabilities $$ P_\theta(T=0)=\frac13+\frac{\theta}{10},\qquad P_\theta(T=1)=\frac13+\frac{\theta^2}{10},\qquad P_\theta(T=2)=\frac13-\frac{\theta+\theta^2}{10}. $$ Determine whether $T$ is complete. Then find an unbiased estimator of $\theta$ that is a function of $T$.

Step 1: Check the probabilities. They sum to $$ \left(\frac13+\frac{\theta}{10}\right) +\left(\frac13+\frac{\theta^2}{10}\right) +\left(\frac13-\frac{\theta+\theta^2}{10}\right)=1. $$ For $-0.1<\theta<0.1$, each is positive: the smallest possible probability is greater than $0.32$. Thus these are valid probabilities on the stated parameter space.

Step 2: Test completeness. Let $g_j=g(j)$ for $j=0,1,2$. Since the support is finite, every such function is bounded and integrable. If $E_\theta[g(T)]=0$ for every $\theta$, then $$ g_0\left(\frac13+\frac{\theta}{10}\right) +g_1\left(\frac13+\frac{\theta^2}{10}\right) +g_2\left(\frac13-\frac{\theta+\theta^2}{10}\right)=0. $$ Collecting the constant, $\theta$, and $\theta^2$ terms gives $$ \frac{g_0+g_1+g_2}{3} +\frac{g_0-g_2}{10}\theta +\frac{g_1-g_2}{10}\theta^2=0 \quad\text{for every }\theta\in(-0.1,0.1). $$ A polynomial that is zero throughout an interval has every coefficient equal to zero. Therefore $g_0+g_1+g_2=0$, $g_0-g_2=0$, and $g_1-g_2=0$. The last two equations give $g_0=g_1=g_2$; substituting into the first gives $3g_2=0$. Hence $g_0=g_1=g_2=0$, proving that $T$ is complete.

Step 3: Find an unbiased estimator of $\theta$. Seek values $a_0,a_1,a_2$ such that $E_\theta[a(T)]=\theta$. Matching coefficients in the expectation gives the equations $$ a_0+a_1+a_2=0,\qquad a_0-a_2=10,\qquad a_1-a_2=0. $$ The third equation gives $a_1=a_2$. Then $a_0+2a_2=0$, while $a_0-a_2=10$, so $-3a_2=10$. Thus $a_2=a_1=-10/3$ and $a_0=20/3$.

The estimator is $a(T)=20/3$ when $T=0$, and $a(T)=-10/3$ when $T=1$ or $T=2$. Its expectation is $$ \frac{20}{3}\left(\frac13+\frac{\theta}{10}\right) -\frac{10}{3}\left[\left(\frac13+\frac{\theta^2}{10}\right) +\left(\frac13-\frac{\theta+\theta^2}{10}\right)\right]=\theta. $$ The constants cancel, the $\theta^2$ terms cancel, and the remaining term is $\theta$. The calculation verifies unbiasedness. Completeness also means that no other unbiased estimator that is a function of this $T$ can differ from $a(T)$ with positive probability.

A Continuous Example and Its Justification

A useful completeness proof turns the zero-mean condition into an integral identity indexed by the parameter. When that identity holds for every endpoint, differentiating it can reveal the function itself. The differentiation is justified when the relevant integrals exist and the integral as a function of its endpoint is absolutely continuous.

Worked Example: Completeness of a Sample Maximum

Let $X_1,X_2$ be independent observations from a uniform distribution on $(0,\theta)$, where $\theta>0$, and let $M=\max(X_1,X_2)$. Show that $M$ is complete and use it to construct an unbiased estimator of $\theta$.

Step 1: Find the density of $M$. For $0<m<\theta$, $$ P_\theta(M\le m)=P_\theta(X_1\le m,\ X_2\le m) =\left(\frac{m}{\theta}\right)^2. $$ Differentiating with respect to $m$ gives $f_M(m)=2m/\theta^2$ on $(0,\theta)$.

Step 2: Prove completeness. Suppose $g$ is integrable under each distribution of $M$ and $E_\theta[g(M)]=0$ for all $\theta>0$. The density gives $$ 0=E_\theta[g(M)]=\frac{2}{\theta^2}\int_0^\theta t g(t)\,dt, \qquad \theta>0. $$ Since $2/\theta^2$ is positive, $H(\theta)=\int_0^\theta t g(t)\,dt=0$ for every $\theta>0$. The integral $H$ is absolutely continuous on each bounded interval on which the integrand is integrable, and its derivative is $H'(\theta)=\theta g(\theta)$ for almost every $\theta>0$. Because $H$ is identically zero, $H'(\theta)=0$ almost everywhere. Since $\theta>0$, this implies $g(\theta)=0$ for Lebesgue-almost every $\theta>0$. The density of $M$ is positive throughout $(0,\theta)$, so $g(M)=0$ almost surely for every $\theta$. This proves completeness.

Step 3: Estimate $\theta$ without bias. Using the density, $$ E_\theta[M]=\int_0^\theta m\frac{2m}{\theta^2}\,dm =\frac{2}{\theta^2}\left[\frac{m^3}{3}\right]_0^\theta =\frac{2\theta}{3}. $$ Therefore $\widehat{\theta}=3M/2$ is unbiased because $E_\theta[3M/2]=(3/2)(2\theta/3)=\theta$. Completeness ensures uniqueness among integrable unbiased estimators that are functions of $M$.

This proof also illustrates why the full parameter range matters. The zero-mean equation is available for every upper endpoint $\theta>0$. That gives an integral identity at every point, which can then be differentiated. A single expectation equation at one fixed parameter value would not determine $g$ throughout the support.

How Incompleteness Creates Nonuniqueness

When a statistic is incomplete, a nonzero zero-mean function can produce a second unbiased estimator. The following example shows the mechanism directly. It also distinguishes “zero expectation” from “equal to zero”: a random variable may have mean zero while taking nonzero values with positive probability.

Worked Example: Two Unbiased Estimators from an Incomplete Statistic

Let $X$ have a uniform distribution on $(-\theta,\theta)$, with $\theta>0$, and use $T=X$. Show that $T$ is incomplete and construct two different unbiased estimators of $\theta$ that are functions of $T$.

Step 1: Exhibit a zero-mean function. Take $g(T)=T$. By symmetry, $$ E_\theta[T]=\int_{-\theta}^{\theta}x\frac{1}{2\theta}\,dx =\frac{1}{2\theta}\left[\frac{x^2}{2}\right]_{-\theta}^{\theta}=0 $$ for every $\theta>0$. But $P_\theta(T=0)=0$, so $g(T)$ is not zero almost surely. Thus $T$ is incomplete.

Step 2: Construct an unbiased estimator. The density is $1/(2\theta)$ on $(-\theta,\theta)$, so $$ E_\theta[|T|] =\frac{1}{2\theta}\left(\int_{-\theta}^0(-x)\,dx+\int_0^\theta x\,dx\right) =\frac{1}{2\theta}\left(\frac{\theta^2}{2}+\frac{\theta^2}{2}\right) =\frac{\theta}{2}. $$ Hence $\delta_1(T)=2|T|$ is unbiased for $\theta$.

Step 3: Add the zero-mean function. Define $\delta_2(T)=2|T|+T$. Its expectation is $2(\theta/2)+0=\theta$, so it is also unbiased. The two estimators differ by $T$, which is nonzero with probability one under this continuous model. They are therefore genuinely different. The method is to add a nonzero zero-mean function; the trap is to assume that zero expectation means the added function is zero.

Uniqueness of Unbiased Estimation

The central uniqueness result follows immediately from the definition. Suppose $T$ is complete and $\delta_1(T)$ and $\delta_2(T)$ are integrable unbiased estimators of the same target $\tau(\theta)$. Their difference has expectation zero for every parameter value: $$ E_\theta[\delta_1(T)-\delta_2(T)] =\tau(\theta)-\tau(\theta)=0. $$ Completeness, applied to $g(T)=\delta_1(T)-\delta_2(T)$, gives $\delta_1(T)=\delta_2(T)$ almost surely under every $\theta$. This proves uniqueness within the stated class.

Theorem: If $T$ is complete and two integrable functions of $T$ are unbiased for the same parameter function $\tau(\theta)$ for every $\theta$, then the two estimators are equal almost surely for every $\theta$. Completeness establishes uniqueness, not existence; an unbiased estimator must first be found.

If $T$ is only boundedly complete, the same argument applies when the difference of the two estimators is bounded. For unbounded differences, bounded completeness alone does not provide the needed implication. In any application, check both the class of functions in the definition and the integrability conditions required to take expectations.

Common Mistakes and Exam Traps

Checking only one parameter value. Showing $E_{\theta_0}[g(T)]=0$ for one $\theta_0$ says nothing about completeness. The identity must hold for every $\theta\in\Theta$.
Confusing zero mean with zero almost surely. In the symmetric uniform example, $E_\theta[T]=0$ while $T$ is almost surely nonzero. The former does not imply the latter unless completeness applies to the zero-mean identity.
Claiming uniqueness beyond the stated class. Completeness of $T$ proves uniqueness for unbiased estimators that are functions of $T$. By itself it does not compare every possible estimator based on the full sample.
Forgetting the boundedness condition. Bounded completeness only tests bounded $g$. Do not use it for an unbounded difference unless a separate argument justifies the conclusion.
Treating completeness as an existence theorem. Completeness says there cannot be two distinct unbiased functions of $T$ for the same target. It does not guarantee that even one such estimator exists.

Key takeaway: To test completeness, ask whether $E_\theta[g(T)]=0$ for every parameter value forces $g(T)=0$ almost surely. If it does, any unbiased estimator that is a function of $T$, when one exists, is unique within that class.

Check Your Understanding

For each problem, use the zero-mean definition carefully and state whether the conclusion concerns completeness or bounded completeness.

Let $T$ have probabilities $P_\theta(T=0)=1/2$ and $P_\theta(T=1)=1/2$ for every $\theta$. Is $T$ complete? (Answer: No; $g(0)=1,\ g(1)=-1$ has expectation zero but is not zero almost surely.)
For $X\sim\operatorname{Uniform}(-\theta,\theta)$, $\theta>0$, define $g(X)=X^2-\theta^2/3$. Is this a valid function of $T=X$ that demonstrates incompleteness? (Answer: No; it depends explicitly on the unknown $\theta$, so it is not a fixed function of $T$.)
For two independent observations from $\operatorname{Uniform}(0,\theta)$, let $M$ be their maximum. Given $E_\theta[M]=2\theta/3$, what multiple of $M$ is unbiased for $3\theta$, and what does completeness imply about other unbiased functions of $M$? (Answer: $9M/2$; it is the unique integrable unbiased estimator of $3\theta$ that is a function of $M$, up to almost-sure equality.)
Suppose $T$ is complete, and $a(T)$ and $b(T)$ are integrable with $E_\theta[a(T)]=E_\theta[b(T)]=\theta^2$ for every $\theta$. What follows? (Answer: $a(T)=b(T)$ almost surely for every $\theta$.)
Suppose a statistic is known only to be boundedly complete, and two unbiased functions of it differ by an unbounded random variable. Can bounded completeness alone establish that the estimators are equal almost surely? (Answer: No; the zero-mean difference is outside the bounded-function condition.)

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