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$X_1,\ldots,X_n$ 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?

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

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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.

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  • $\begingroup$ The constants $a_i$ are all bounded. $\endgroup$
    – Gourab Mukherjee
    Commented Jun 3, 2017 at 1:36

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