# $a_n$ consistency and other consistency

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!

• What on Earth is $n$ indexing here? – AdamO Jun 17 '13 at 22:41
• @AdamO: $n$ is sample size. – Tim Jun 17 '13 at 22:46
• How is $\{a_n\}$ a sequence of constants? – AdamO Jun 17 '13 at 22:51
• it is assumed to be. – Tim Jun 17 '13 at 23:01
• 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. – AdamO Jun 18 '13 at 0:49

It's easy to show $a_n[T_n(X_n) - \theta] = Op(1)$ implies $\frac{a_n}{b_n}[T_n(X_n) - \theta] = op(1)$ for any $b_n \to \infty$. Taking $b_n = a_n$, we get that $a_n$ consistency implies weak consistency.