3
$\begingroup$

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!

$\endgroup$
5
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

1
$\begingroup$

$a_n$-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 $\theta$.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.