# Showing that asymptotic normality implies consistency.

In a statistics book i'm reading, it is postulated that asymptotic normality of an estimator implies consistency. That is $$\hat{\theta}_n \stackrel{as}{\sim} \mathcal{N}(\theta_0, \frac{1}{n}\sigma(\theta_0)) \Rightarrow P_{\theta_0}(|\hat{\theta}_n - \theta_0|>\epsilon ) \to 0$$ when $n\to\infty$, for all $\theta_0 \in \Theta$ and $\epsilon>0$.

I am trying to prove this, but i can't seem to get a breakthrough.

If anyone could shed some light on how this is proven, i would be very grateful.

If the distribution of $X_n$ is $\mathcal N(x,\sigma_n^2)$ and $\sigma_n^2\to0$, then $X_n\to x$ in probability, that is, $P[|X_n-x|\geqslant\epsilon]\to0$ for every positive $\epsilon$.

To prove this, an easy route is to use Bienaymé-Chebyhev inequality, to the effect that $E[X_n]=x$ and $\mathrm{var}(X_n)=\sigma_n^2$ hence $$P[|X_n-x|\geqslant\epsilon]=P[|X_n-E[X_n]|\geqslant\epsilon]\leqslant\frac{\mathrm{var}(X_n)}{\epsilon^2}=\frac{\sigma_n^2}{\epsilon^2}.$$

• Thanks for the quick response. The only thing that i cant comprehend is that in order to use Chebyshev's inequality i need to know that the variance of $X_n$ is in fact $\sigma^2_n$, and i'm not really sure what argument i need to make in order to use it on something that has an asymptotic variance.
– John
Jun 2, 2013 at 17:09
• Isn't this an assumption in your question?
– Did
Jun 2, 2013 at 17:27
• Correct me if i am wrong, but as i have understood it when $$X_n \stackrel{as}{\sim} N (x , \sigma^2_n)$$ Then $Var(X_n)\neq \sigma^2_n$ (equality only holds in the limit.) And as i see it i need to use Chebyshev's inequality before taking the limit in n. Correct me if im wrong, but really sure about this.
– John
Jun 2, 2013 at 17:43
– John
Jun 2, 2013 at 17:49
• Which limit? Since $\sigma_n^2\to0$, in the limit $n\to\infty$, $X_n\sim$... what exactly? It seems you need to seriously explain what you mean by this hypothesis if not what I considered in my post.
– Did
Jun 2, 2013 at 19:04

A more concise answer follows using the $O_p$ and $o_p$ notation for stochastic orders. Your premise is that

$$\sqrt{n}(\theta_n-\theta_0)\overset{d}{\rightarrow}Z$$

where $Z$ is normally distributed with mean $0$ and variance $\sigma(\theta_0)$. Therefore

$$\theta_n-\theta_0=O_p\left(\frac{1}{\sqrt{n}}\right)\Rightarrow \theta_n-\theta_0= o_p(1)$$

hence $\theta_n-\theta_0\overset{p}{\rightarrow}0$, which finishes the proof.

• Can you give more intuition about the last part? I mean, why does $\hat{\theta}_{n} - \theta_{0}$ being $o_{p}(1)$ implies convergence in probability to zero? Feb 27, 2020 at 0:01
• It is the definition of $o_p(1)$, so it is an "if and only if" statement. Am I understanding your question correctly, @S.Cow ? Mar 5, 2020 at 5:03