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In Lehmann-Casella (Theory of Point Estimation) they state without proof that if $T \sim Bin(n,p)$, then $g(p)$ is estimable only if it is a polynomial in $p$ of degree $\leq n$. How does one go about proving this?

Here is their somewhat justification:

In fact, it follows from Equation (1.2) that a function $g(p)$ can be U-estimable only if it is a polynomial of degree $\leq n$. Let $\delta(k)$ be the estimator. Then $g(p)$ has to satisfy Eqn 1.2: $$(1.2) ~~~~~~~\sum_{k=0}^n \delta(k) {n \choose k} p^k (1-p)^{n-k} = g(p), ~\forall 0 < p < 1.$$

Ben
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Dhamma K
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1 Answers1

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The left-hand-side of $(1.2)$ is a weighted sum of polynomials with degree $n$, which is itself a polynomial of degree (no greater than) $n$. If you want to see this more clearly then you can use the binomial theorem to write it in its explicit form:

$$\begin{align} LHS &\equiv \sum_{k=0}^n \delta(k) {n \choose k} p^k (1-p)^{n-k} \\[12pt] &= \sum_{k=0}^n \delta(k) {n \choose k} p^k \sum_{i=0}^{n-k} {n-k \choose i} (-p)^i \\[12pt] &= \sum_{k=0}^n \sum_{i=0}^{n-k} {n-k \choose i} {n \choose k} (-1)^i \delta(k) p^{k+i} \\[12pt] &= \sum_{k=0}^n \sum_{r=k}^n {n-k \choose r-k} {n \choose k} (-1)^{r-k} \delta(k) p^r \\[12pt] &= \sum_{r=k}^n \Bigg[ \sum_{k=0}^n {n-k \choose r-k} {n \choose k} (-1)^{r-k} \delta(k) \Bigg] p^r. \\[12pt] \end{align}$$

Ben
  • 124,856