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Let $\hat{\theta}_n$ be an estimator of the parameter $\theta$ from the sample $\Omega_n$ of $n$ observations, satisfying that $\sqrt{n} (\hat{\theta}_n-\theta) \overset{d}{\longrightarrow} \mathcal{N} (0,\sigma^2)$. Let $\hat{\theta}_{1,n}$ and $\hat{\theta}_{2,n}$ be the analogues of $\hat{\theta}_{n}$ but computed over the first half and the other half of the sample $\Omega_n$. Prove that:

$\hat{\theta}_n - 0.5 (\hat{\theta}_{1,n} + \hat{\theta}_{2,n}) = o_p(n^{-1/2})$.

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  • $\begingroup$ What have you tried yet? Where were you stuck in the proof? $\endgroup$ – Easymode44 Dec 12 '18 at 8:27
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This is not a well-posed exercise, because when we consider $n\to \infty$, the concepts of "first half of the sample" and "second half of the sample" are not really defined. What is $\infty/2$?

Informally of course, we understand that since both the "first half" and the "second half" will contain an infinite number of observations, the related estimators will become asymptotically equivalent with the full-sample estimator and so the required $o_p$ relation does hold.

But again, one should use a different way to formulate the problem. For example, instead of "first half" and "second half" one could define two sub-samples by $S_1 = \{\text{observation index is odd}\}$ and $S_2 = \{\text{observation index is even}\}$ and define the estimators over them.

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