An estimator for inverse of variance

Question

I want to find some function $f$ such that $$ \mathbb{E}\left[ f(X) \right] = \frac{1}{\sigma^2}, $$ where $X \sim \mathcal{N}(0,\sigma^2)$. Obviously, $f$ should not depend on $\sigma$. $1/\mathbb{E}[X^2]$ is not what I want.

Notes

For a normal random variable, $\mathbb{E}[1/X^2]$ does not exist.

Answer 1

I don't believe that you would get an estimate of $1/\sigma^2$ based on a single observation. Instead, assume that we have $n$ iid observations, $X_1, \dots, X_n.$ Define $S^2 = \sum_{i=1}^n X_i^2.$ It is known that $S^2/\sigma^2\sim \chi^2_n.$ Using the pdf of $\chi^2_n,$ you can show that $$\mathbb{E}(S^{2r}) = \sigma^{2r}\times \frac{\Gamma(n/2+r)}{2^r \Gamma(n/2)}.$$ Set $r=-1$ and get the desired estimator (works for $n>2$).