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Given a estimator $\hat \theta$ of $\theta$, I want to show that $\sqrt{n}(\hat\theta -\theta-B)\to N(0,V_\theta)$ as $n\to\infty$, given that the limit $V_\theta$ exists and $B>0$ possibly dependent of $n$.

In my case, $\hat \theta$ is not feasible, but there is a feasible estimator, say $\tilde\theta$, in the oracle case.

I plan to show this convergence from $$\sqrt{n}(\hat\theta-\theta)=\sqrt{n}(\hat\theta-\tilde\theta)+\sqrt{n}([\tilde\theta-E\tilde\theta]+[E\tilde\theta-\theta]-B)$$ by proving the following results:

  1. $\sqrt{n}(\hat\theta-\tilde\theta)=o(1)$;
  2. $(E\tilde\theta-\theta)=B+o(1).$;
  3. $\sqrt{n}(\tilde\theta-E\tilde\theta)\to N(0,V_\theta)$

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Proving 2. does not prove that $\sqrt{n}\cdot o(1) \to 0$.

So in this approach you need to prove that

$$(E\tilde\theta-\theta)=B+o(n^{-1/2})$$

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  • $\begingroup$ Yes! I noticed that. Have you ever worked with Local linear estimator? In my case, the third term (in third step above) is $\sqrt{nh}\sum_{i}^n W_i(x)\epsilon_i$ where $W_n(x)$ is the local linear weight at $x$, $h$ the bandwidth and $\epsilon$ the error process. Also, $V_\theta=Var((\frac{h}{n})^{1/2}\sum_i K((i/n-x)/h))\epsilon_i$. I'm struggling to show this convergence in distribution because there is a term that is diverging in my derivations. $\endgroup$
    – Celine Harumi
    Commented Dec 20, 2019 at 22:19

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