# Variance bound of a functional

$X_1,\ldots,X_n$ are i.i.d standard normal random variables.

$a_1,\ldots, a_n$ are constants with $a_i \in [\kappa_1, \kappa_2]$ for all $i$ and $\kappa_1>0$.

$\hat c_n$ is given as the solution to the equation:

$$\sum_{i=1}^n \frac{c -X_i^2}{(c+a_i)^2}=0.$$

Can we prove that $$\operatorname{Variance} (\hat c_n)=O(n^{-1}) \text{ as } n \to \infty.$$

Note that the above can be proved easily when all $a_i$' s are equal. What will happen in the general case?

• Are you claiming that this equation always has a unique solution? That doesn't look obvious. Commented Jun 3, 2017 at 0:33
• The equation has a unique solution as it arrives from minimizing the convex criterion $$\sum_i \log(c a_i^{-1}+1)+(1+X_i^2)/(ca_i^{-1}+1)$$ Commented Jun 3, 2017 at 3:14
• "standard" = "standard Gaussian" or something else? Commented Jun 4, 2017 at 1:36
• Also $t\mapsto \log t+\frac Kt$ is not convex Commented Jun 4, 2017 at 1:41
• Oh sorry...i was wrong on this...will need to modify the problem accordingly..thanks for pointing out this issue... Commented Jun 4, 2017 at 17:29

## 1 Answer

This is not rigorous, but if one constant $a_1$ is small but the others $a_i$ are large, say $a_i\gg n$ for $i>1$ then it seems the $i>1$ terms are negligible and $c\approx X_i^2$, so $\mathrm{Var}(\hat c)$ does not go to 0.

• The constants $a_i$ are all bounded. Commented Jun 3, 2017 at 1:36