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While I think it is reasonable, I cannot show this result.

Suppose $\hat{\theta}$ is an estimator of $\theta$ and $P(\hat{\theta}\neq \theta) \rightarrow 0$ as the sample size increases, that is, $P(\hat{\theta}\neq \theta)=o(1)$. Then I can write $\hat{\theta}=\theta+o_p(1)$.

How can I show this?

*Note the difference between, $o(1)$ and $o_p(1)$. While "little oh" deal with sequences of numbers, the "little oh in probability" is related with convergence in probability (to zero).

*The only similar result that I know is $\hat{\theta}=E(\hat{\theta})+(Var(\hat{\theta}))^{1/2} O_p(1)$. In my case, $\hat{\theta}=\min_{\theta_1}\{ RSS(\theta_1)+\lambda \theta_1\}$

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    $\begingroup$ I'm actually not aware of the difference between $o(1)$ and $o_p(1)$. Does it have to do with strict boundedness vs. boundedness in probability? $\endgroup$
    – AdamO
    Jun 3, 2019 at 15:50
  • $\begingroup$ @AdamO ok. I Updated my question! $\endgroup$ Jun 3, 2019 at 15:57

1 Answer 1

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That $\hat{\theta} = \theta + o_{p}(1)$ is a restatement of $\hat{\theta} \to_{p} \theta$. (Aware that saying a sequence $(x_{n})$ converges to some $x$ is equivalent to saying $x_{n} = x + o(1)$.) The latter statement holds iff $\mathbb{P}(|\hat{\theta} - \theta| \geq \epsilon) \to 0$ for every $\epsilon > 0$, by definition. Since $|\hat{\theta} - \theta| \geq \epsilon$ implies $\hat{\theta} \neq \theta$, we have $$ \mathbb{P}(|\hat{\theta} - \theta| \geq \epsilon) \leq \mathbb{P}(\hat{\theta} \neq \theta) \to 0 $$ by assumption.

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  • $\begingroup$ Nice. I believe the key point is the last inequality. For a given $\epsilon$ , in fact, the inverse image $\{\hat{\theta} \neq \theta\}$ contains $\{\mid \hat{\theta}-\theta\mid\}$. Using the monotonicity of the measure, the result follows. But I also need an argument about the equivalence that you claimed in the first sentence. $\endgroup$ Jun 3, 2019 at 16:36
  • $\begingroup$ Ok, I got it now. hehe $\endgroup$ Jun 3, 2019 at 16:42
  • $\begingroup$ @Danmat, The first claim is by convention. The point seems to be the conclusion of a convergence in probability. Unless there is another meaning assigned to the statement pertaining to $o_{p}(1)$, the first claim is not much crucial. $\endgroup$
    – Yes
    Jun 3, 2019 at 16:59
  • $\begingroup$ @GaryMoore Yes, I thought about the first calim, and concluded that the equivalence follows directly by the definition of convergence in probability and the definition of $o_p$. Thanks for the answer! $\endgroup$ Jun 3, 2019 at 17:08
  • $\begingroup$ @Danmat, you are welcome. I guess it was perhaps the casual "so" that confused you. $\endgroup$
    – Yes
    Jun 3, 2019 at 17:42

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