# $a_n$ 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? Jun 17, 2013 at 22:41
• @AdamO: $n$ is sample size.
– Tim
Jun 17, 2013 at 22:46
• How is $\{a_n\}$ a sequence of constants? Jun 17, 2013 at 22:51
• it is assumed to be.
– Tim
Jun 17, 2013 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. Jun 18, 2013 at 0:49

#### $$a_n$$-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 $$\theta$$.

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.