Why Completeness Is a Family Property
An estimator can have expectation zero for every parameter value only if it is zero with probability one under every distribution in a complete family. This condition is stronger than a statement about a single distribution: it depends on both the random variable and the full range of parameter values being considered.
For Bernoulli and binomial models, the condition can be checked using elementary polynomial identities. The key is that a polynomial which is zero for every probability in an interval must have all its coefficients equal to zero. This gives a direct proof of completeness, without requiring a general theorem about exponential families.
Completeness is about a whole collection of distributions indexed by a parameter. If an identity holds only at one parameter value, it does not meet the definition. Likewise, the parameter range matters: an identity that holds at a few selected probabilities need not hold throughout an interval.
The Bernoulli Family
Let \(X\sim\operatorname{Bernoulli}(p)\), where \(p\in(0,1)\). Thus \(X\) takes values 0 and 1, with \(P_p(X=1)=p\) and \(P_p(X=0)=1-p\). The family allows every success probability strictly between 0 and 1.
For any function \(g\) on \(\{0,1\}\), its expectation is a linear polynomial in \(p\):
If this expectation is zero for every \(p\in(0,1)\), then the linear polynomial is zero throughout an interval. Its coefficients must therefore be zero. In particular, the coefficient of the constant term in the expanded expression is \(g(0)\), and the coefficient of \(p\) is \(g(1)-g(0)\). Hence \(g(0)=0\) and \(g(1)=0\). Both outcomes have positive probability for every \(p\in(0,1)\), so \(g(X)=0\) with probability one.
Worked Example: Testing a Zero-Expectation Identity for a Bernoulli Observation
Let \(X\sim\operatorname{Bernoulli}(p)\), with \(0<p<1\), and suppose a function \(g\) satisfies \(E_p[g(X)]=0\) for every such \(p\). Use the expectation equation to determine both values of \(g\).
Method. Write the expectation using the two possible values of \(X\), then compare polynomial coefficients.
Since \(X\) can be 0 or 1, $$ E_p[g(X)]=(1-p)g(0)+pg(1) =g(0)+p\bigl(g(1)-g(0)\bigr). $$
For this expression to equal zero for every \(p\in(0,1)\), its constant coefficient and its coefficient of \(p\) must both be zero: $$ g(0)=0,\qquad g(1)-g(0)=0. $$
The first equation gives \(g(0)=0\). Substituting that result into the second gives \(g(1)=0\). Thus \(g(X)=0\) for either possible outcome, so the Bernoulli family is complete.
Check. Substituting the results into the expectation gives \((1-p)0+p0=0\) for every allowed \(p\). The common mistake is to check the equation at just one value of \(p\); that gives one equation for two unknown function values and cannot establish completeness.
Completeness of a Fixed-Size Binomial Count
Now let \(K\sim\operatorname{Binomial}(n,p)\), where \(n\) is a fixed positive integer and \(p\in(0,1)\). The support is \(\{0,1,\ldots,n\}\), and $$ P_p(K=k)=\binom{n}{k}p^k(1-p)^{n-k}. $$ The assumption that \(n\geq 1\) is fixed is part of the model. The case \(n=1\) is the Bernoulli family.
Suppose \(E_p[g(K)]=0\) for every \(p\in(0,1)\). Directly, this expectation is a sum involving powers of \(p\) and \(1-p\). The odds substitution \(t=p/(1-p)\) turns it into an ordinary polynomial with one coefficient for each possible value \(k\).
For a function \(g\) on the binomial support, \(E_p[g(K)]=\sum_{k=0}^{n}g(k)\binom{n}{k}p^k(1-p)^{n-k}\).
Set \(t=p/(1-p)\). Since \(0<p<1\), \(t>0\), \(p=t/(1+t)\), and \(1-p=1/(1+t)\). The expectation equals \((1-p)^n\sum_{k=0}^{n}\binom{n}{k}g(k)t^k\).
The factor \((1-p)^n\) is positive. A zero expectation for every \(p\in(0,1)\) therefore makes the polynomial \(\sum_{k=0}^{n}\binom{n}{k}g(k)t^k\) zero for every \(t>0\).
A polynomial that is zero at infinitely many values must be the zero polynomial. Its coefficient of \(t^k\) is \(\binom{n}{k}g(k)\). Since \(\binom{n}{k}>0\), \(g(k)=0\) for every \(k=0,\ldots,n\).
This proves the theorem. It also explains precisely why the open interval matters: as \(p\) ranges over \((0,1)\), the odds \(t\) range over all positive real numbers, giving infinitely many roots of the polynomial. The endpoints need not be included for this argument.
Worked Example: Expanding the Binomial Identity When \(n=3\)
Let \(K\sim\operatorname{Binomial}(3,p)\), with \(0<p<1\). Suppose \(E_p[g(K)]=0\) for every allowed \(p\). Show explicitly that all four values of \(g\) must be zero.
Method. Expand the expectation as a polynomial in \(p\), then equate coefficients. This makes the zero-coefficient conclusion visible without relying only on the general theorem.
The binomial probabilities give $$ E_p[g(K)] =g(0)(1-p)^3+3g(1)p(1-p)^2+3g(2)p^2(1-p)+g(3)p^3. $$
Expanding each term and collecting equal powers of \(p\) gives $$ E_p[g(K)] =g(0)+p\bigl(-3g(0)+3g(1)\bigr) +p^2\bigl(3g(0)-6g(1)+3g(2)\bigr) +p^3\bigl(-g(0)+3g(1)-3g(2)+g(3)\bigr). $$
The polynomial is zero for all \(p\in(0,1)\), so its four coefficients are zero. Starting with the constant term, $$ g(0)=0. $$ The coefficient of \(p\) then gives \(3g(1)=0\), so \(g(1)=0\). The coefficient of \(p^2\) becomes \(3g(2)=0\), so \(g(2)=0\). Finally, the coefficient of \(p^3\) becomes \(g(3)=0\).
Check. All four values are zero, and therefore the expectation is zero for every \(p\). The common mistake is to set the expectation equal to zero at one chosen probability and conclude all four values are zero; a single equation cannot determine four values.
Why the Full Parameter Range Matters
The theorem does not say that every binomial model is complete under every possible parameter set. Completeness must be checked for the particular family being used. When \(p\) is allowed to vary over the entire open interval, the expectation polynomial is constrained at infinitely many points. If only a few parameter values are allowed, that reasoning no longer forces every coefficient to vanish.
Worked Example: A Restricted Binomial Family Is Not Complete
Suppose \(K\sim\operatorname{Binomial}(2,p)\), but the parameter space is restricted to \(p\in\{1/3,2/3\}\). Define \(g(0)=1\), \(g(1)=-5/4\), and \(g(2)=1\). Check whether this nonzero function has expectation zero throughout the restricted family.
Method. Substitute the three function values into the binomial expectation and evaluate it at each allowed parameter value.
For general \(p\), $$ E_p[g(K)] =(1-p)^2-\frac{5}{2}p(1-p)+p^2. $$
At \(p=1/3\), the three terms are \(4/9\), \(-5/9\), and \(1/9\), respectively. Their sum is $$ \frac{4}{9}-\frac{5}{9}+\frac{1}{9}=0. $$ At \(p=2/3\), the terms are \(1/9\), \(-5/9\), and \(4/9\), so $$ \frac{1}{9}-\frac{5}{9}+\frac{4}{9}=0. $$
Thus \(E_p[g(K)]=0\) for both allowed parameter values, although \(g\) is not identically zero. Each of the three outcomes has positive probability at both parameter values, so \(g(K)\) is not zero with probability one. The restricted family is not complete.
Check. At \(p=1/2\), which is not in the restricted parameter set, the expectation would be \(1/4-5/8+1/4=-1/8\), not zero. That does not contradict the calculation: the restricted family requires the identity only at \(1/3\) and \(2/3\). The common mistake is to use the full-interval completeness result after changing the parameter space.
Exam Checks and Common Traps
When applying these results, make the model and parameter space explicit before using a completeness argument. In particular, distinguish a single Bernoulli observation from a binomial count based on a fixed positive number of trials, and state whether \(p\) ranges over the full open interval.
- Confusing one parameter value with a family. Completeness requires a zero-expectation identity for every parameter value in the family, not just at a selected value.
- Forgetting the support. For a binomial count, the conclusion is \(g(k)=0\) for each \(k=0,\ldots,n\). These are all positive-probability outcomes for \(0<p<1\).
- Dropping a binomial coefficient. The polynomial coefficient is \(\binom{n}{k}g(k)\). It implies \(g(k)=0\) because \(\binom{n}{k}\) is nonzero.
- Using the result after restricting \(p\). A polynomial can vanish at finitely many points without all its coefficients being zero.
The proof technique is reusable: express an expectation as a polynomial, use the assumed range of the parameter to get enough roots, and then read off what the coefficients imply. Here that argument establishes completeness for Bernoulli and fixed-\(n\) binomial families with \(0<p<1\), aligned with the SOA VEE Mathematical Statistics syllabus.
Check Your Understanding
For each problem, use the support and parameter range given; do not assume a parameter varies unless the question says so.
- Let \(X\sim\operatorname{Bernoulli}(p)\), \(0<p<1\), and let \(g(0)=a\), \(g(1)=b\). If \(E_p[g(X)]=0\) for every \(p\), find \(a\) and \(b\). (Answer: \(a=0,\ b=0\).)
- Let \(K\sim\operatorname{Binomial}(2,p)\), \(0<p<1\). Apply the odds substitution to \(E_p[g(K)]=0\) and state the three resulting coefficient equations. (Answer: \(g(0)=0,\ 2g(1)=0,\ g(2)=0\), so all three function values are zero.)
- For \(K\sim\operatorname{Binomial}(4,p)\), compute \(E_p[K(K-1)]\). Does this function have expectation zero for every \(p\in(0,1)\)? (Answer: \(12p^2\); no, it is positive for every \(p\in(0,1)\).)
- For \(K\sim\operatorname{Binomial}(2,p)\) with \(p\in\{1/3,2/3\}\), verify that \(g=(1,-5/4,1)\), listed in support order \(0,1,2\), has zero expectation at both parameter values. What does this imply? (Answer: The expectations are \(0\) at both values, but \(g\) is nonzero, so the restricted family is not complete.)
- Let \(K\sim\operatorname{Binomial}(3,p)\), \(0<p<1\), and suppose \(E_p[g(K)]=0\) throughout the parameter range. Use the constant and linear coefficients of the expanded expectation to find \(g(0)\) and \(g(1)\), then state the values of \(g(2)\) and \(g(3)\). (Answer: \(g(0)=g(1)=0\); the quadratic and cubic coefficients then give \(g(2)=g(3)=0\).)