Identifiability and estimability

Question

I am somewhat confused about this identifiability and estimability concept with application to binomial example in David Freedman's book (statistical models: theory and practice Page 125-P126).

let $g(x)$ be some estimator. Then,

$E_p(g(x))=(1-p)g(0)+pg(1)$

This is a linear function in p. However, $\sqrt{p}$ is not a linear function of p. So, $E_p(g(x)) \ne \sqrt{p}$. And $\sqrt{p}$ is not estimable.

I still can not understand why $\sqrt{p}$ is not estimable. If I can get M.L.E (sample proportion) of $\hat{p}=sum(1(x=1))/N$. Can't I use $\sqrt{\hat{p}}$ as an estimator?

or because this $\sqrt{\hat{p}}$ is a biased (even though a consistent) estimator of $\sqrt{p}$?

Any help with clarification of this concept will be appreciated!

Answer 1

Yes, $E[\hat{p}] = p$, but you cannot use $\sqrt{\hat{p}}$ because of Jensen's inequality. In particular:

$$ E[\sqrt{\hat{p}}] \neq \sqrt{E[\hat{p}]}. $$

To see why no such function $g(\cdot)$ works, do a proof by contradiction. Assume to the contrary that there exists some $g(\cdot)$ such that $E[g(X)] = \sqrt{p}$. Then $(1-p)g(0)+pg(1) = \sqrt{p}$. After squaring both sides you'll see this is quadratic in $p$. This polynomial can only be equal to $0$ if $g(1) = g(0)$. But then again, any function that does that maps both $0$ and $1$ to the same thing has a constant expectation. This constant can not be equal to $p$ for every $p$.