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From Jun Shao's Mathematical Statistics

Definition 2.10 (Consistency of point estimators). Let $X = (X_1 , ..., X_n)$ be a sample from $P ∈ \mathcal P$ and $T_n(X)$ be a point estimator of $θ$ for every $n$.

Let $\{a_n\}$ be a sequence of positive constants diverging to $∞$. $T_n(X)$ is called $a_n$-consistent for $θ$ if $a_n[T_n(X) − θ] = O_p(1)$ w.r.t. any $P∈\mathcal P$.

$O_p(\cdot)$ is defined here. When $a_n := \sqrt{n}$, the consistency is called root $n$ consistency.

Do one of $a_n$ consistency and weak consistency imply the other? if not, how about $a_n := \sqrt{n}$?

Do one of $a_n$ consistency and strong consistency imply the other? if not, how about $a_n := \sqrt{n}$?

Do one of $a_n$ consistency and $L^2$ consistency imply the other? if not, how about $a_n := \sqrt{n}$?

Thanks and regards!

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  • $\begingroup$ What on Earth is $n$ indexing here? $\endgroup$
    – AdamO
    Jun 17, 2013 at 22:41
  • $\begingroup$ @AdamO: $n$ is sample size. $\endgroup$
    – Tim
    Jun 17, 2013 at 22:46
  • $\begingroup$ How is $\{a_n\}$ a sequence of constants? $\endgroup$
    – AdamO
    Jun 17, 2013 at 22:51
  • $\begingroup$ it is assumed to be. $\endgroup$
    – Tim
    Jun 17, 2013 at 23:01
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    $\begingroup$ I thought the consistency of an estimator is independent of the sample size. For instance, the sample mean of a regular probability distribution is $\sqrt{n}$ consistent by the central limit theorem. $\endgroup$
    – AdamO
    Jun 18, 2013 at 0:49

2 Answers 2

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$a_n$-consistency implies weak consistency

This is only a partial answer to your full query, but I hope it gets you started by showing how you can connect the consistency property in question with the standard consistency property. Since $a_n |T_n - \theta| = O_p(1)$ there exists a function $c: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ such that:

$$\sup_{n \in \mathbb{N}} \mathbb{P}(a_n |T_n - \theta| \geqslant c(\varepsilon)) < \varepsilon \quad \quad \quad \text{for all } \varepsilon > 0.$$

Consequently, for all $\varepsilon>0$ we have:

$$\sup_{n \geqslant N} \mathbb{P} \Big( |T_n - \theta| \geqslant \frac{c(\varepsilon)}{a_n} \Big) \leqslant \sup_{n \in \mathbb{N}} \mathbb{P} \Big( |T_n - \theta| \geqslant \frac{c(\varepsilon)}{a_n} \Big) < \varepsilon.$$

Now, since $\{ a_n \}$ diverges to infinity, for any $\phi>0$ and $\epsilon>0$ you can choose $N$ sufficiently large to ensure that $a_n \geqslant c(\varepsilon)/\phi$ for all $n \geqslant N$. Consequently, for any $\phi>0$ and $\epsilon>0$ you have:

$$\begin{align} \lim_{n \rightarrow \infty} \mathbb{P}(|T_n - \theta| > \phi) &\leqslant \lim_{N \rightarrow \infty} \sup_{n \geqslant N} \mathbb{P}(|T_n - \theta| > \phi) \\[6pt] &= \lim_{N \rightarrow \infty} \sup_{n \geqslant N} \mathbb{P} \Big( |T_n - \theta| > \frac{\phi a_n}{a_n} \Big) \\[6pt] &\leqslant \lim_{N \rightarrow \infty} \sup_{n \geqslant N} \mathbb{P} \Big( |T_n - \theta| \geqslant \frac{c(\varepsilon)}{a_n} \Big) \\[8pt] &< \lim_{N \rightarrow \infty} \varepsilon \\[8pt] &= \varepsilon. \\[6pt] \end{align}$$

Since this inequality holds for all $\varepsilon>0$, we can say that for all $\phi>0$ we have:

$$\lim_{n \rightarrow \infty} \mathbb{P}(|T_n - \theta| > \phi) = 0,$$

which demonstrates weak consistency of $T_n$ in estimating $\theta$.

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We can see that if $a_n[T_n(X_n) - \theta] = O_p(1)$, for any $b_n \to \infty$:

$$\begin{align*}\frac{a_n}{b_n}[T_n(X_n) - \theta] &= \frac{1}{b_n} O_p(1) \\ &= o_p(1)O_p(1) \\ &= o_p(1)\end{align*}$$

This shows $\frac{a_n}{b_n}[T_n(X_n) - \theta] = o_p(1)$

If $a_n \to \infty$, then we can take $b_n = a_n$, and get that

$$T_n(X_n) - \theta = o_p(1)$$

Which proves that $a_n$ consistency implies weak consistency.

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