# What is a good aproximation in asymptotic normality?

I have a conceptual doubt. For example, suppose I have $$X_i \stackrel{iid}{\sim} N(\theta^*,1)$$ and I know that (I have the information) $$\theta^{*}\geq 0$$. So I have the Constrained Maximum Likelihood Estimation:

$$\hat{\theta}_{CMLE} = \max \{\bar{X}_n ,0 \}$$

I know, that if $$\theta > 0$$, then

$$\sqrt{n}(\hat{\theta}_{CMLE} - \theta^{*}) \to^{d} N(0,1)$$

I have heard that if $$\theta^{*}$$ is very close to zero, then I will not have a good approximation. For example, if $$\theta^{*}= 0.00000000000000001$$. Why should not I have a good approximation? If I know that although this $$\theta^{*}$$ is very small it will always be positive and my convergence above will be valid. In what sense would this not be a case of good approximation?

• It's a good topic for questions. Please note, however, that the sense in which the left hand side converges (in distribution) to a standard Normal law is that the estimator $\hat\theta$ is the random variable, not $\theta^{*}.$ Because you truncate the sample mean at zero, what does that do to the distribution of the estimator? If you're not sure, run a quick simulation. The insight this easy exercise provides will carry over to much more complex situations.
– whuber
Commented Mar 16, 2019 at 15:00
• If I did not have the restriction, then the estimator would be $\bar{X}_n$. And this estimator has a distribution in finite samples given by $N (0,1)$. Let me see if I understand your question. Do you want to know the distribution of the constrained estimator? I can deduce what the distribution of $\sqrt{n}(\hat{\theta}_{CMLE} - \theta^{*})$ will be, independent of the true parameter will be zer or not. Is that what you want?
– Fam
Commented Mar 16, 2019 at 15:12
• Be a little careful. The distribution of ${\bar X}_n$ is not standard Normal: it is Normal with mean $\theta^{*}$ and variance $1/n.$ Thus, the definition of $\hat\theta$ censors this distribution by replacing all negative values with $0.$ It can help to draw the PDF of ${\bar X}_n$ and to sketch the area corresponding to the censored values. How large must $n$ become before that area becomes inconsequentially small?
– whuber
Commented Mar 16, 2019 at 16:18
• You're right. I got confused. Actually, I meant that $\sqrt{n}(\hat{\theta}_{CMLE} - \theta^*)$ has a $N(0,1)$ distribution. I'm a little confused. When you say that $\hat{\theta}$ censors the distribution, you refer to $\bar {X} _n$ or $\hat{\theta}_{CMLE}$?
– Fam
Commented Mar 16, 2019 at 17:12
• Look at your definition of $\hat \theta:$ when ${\bar X}_n$ is less than $0,$ you replace it by $0.$ That's what (left) censoring is.
– whuber
Commented Mar 16, 2019 at 18:35

If you run maximum likelihood estimation while imposing the constraint $$\theta^* >0$$, and if $$\theta^* \sim 0$$, then you have the situation where the true value of the parameter lies virtually on the boundary of the parameter space.
In such a case, $$\hat{\theta}_{CMLE}$$ will approach $$\theta^{*}$$ predominantly from above. It will then take a very-very large number of samples to obtain the asymptotic normality result (and this is why you were told that "you will not have a good approximation"), while for sample sizes that usually suffice for a good approximation, in this case they will not be enough, because then the values of the MLE will not be symmetrically distributed around the true value.