 Three-point quadratic reproduction forces the identity filter

A symmetric three-point smoother with weights $(a,b,a)$ reproduces constants only if $2a+b=1$. Reproducing quadratic trends also requires $2a=0$, so $a=0$ and $b=1$. Thus the identity estimator $X_t$ is unbiased, but there is no nontrivial three-point linear smoother reproducing every quadratic. An assertion excluding even the identity estimator is false. This clarifies why the five-point symmetric quadratic trend smoother is the first symmetric nontrivial example.