Asymptotic normality: do the following convergences hold?

Question

Suppose we have a sequence $Z_n$ of estimators with mean $\mu_n$ and variance $\sigma_n^2$. Suppose we know that there exist constants $a_n$ and $b_n(>0)$ such that $\dfrac{Z_n-a_n}{b_n}\to\mathcal N(0,1)$ in distribution.

Then, is it true that $\dfrac{\sigma_n^2}{b_n^2}\to1$ and $\dfrac{\mu_n-a_n}{b_n}\to0$?

Can you please point me to the relevant literature?

Answer 1

This can be solved from basic principles. At the end I'll explain the underlying idea.

Let $X_n$ be a sequence of iid standard Normal variables and, independently of it, let $Y_n$ also be such a sequence. Independently of both of them let $U_n$ be a sequence of independent Bernoulli variables with parameter $1/n$: that is, $U_n$ has a probability $1/n$ of equalling $1$ and otherwise is $0$. Pick a number $p$ (to be determined below) and define

$$Z_n = U_n(Y_n + n^p) + (1-U_n) X_n.$$

Each $Z_n$ is a mixture of a standard Normal (namely $X_n$) and a standard Normal shifted to $n^p$ (namely $Y_n+n^p$). Compute the mean and variance of $Z_n$:

$$\mu_n = n^{p-1};\quad \sigma^2_n = 1 + (n-1)n^{2(p-1)}.$$

The distribution of $Z_n$ approaches a standard Normal distribution $\Phi$ because its distribution function is

$$F_{Z_n}(z) = \frac{n-1}{n}\Phi(z) + \frac{1}{n}\Phi(z-n^p) \to \Phi(z).$$

Consequently $a_n=0$ and $b_n=1$ will work, since $(Z_n-a_n)/b_n=Z_n$. Nevertheless,

$$\frac{\sigma_n^2}{b_n^2} = \frac{1 + (n-1)n^{2(p-1)}}{1} \approx n^{2p-1}$$

diverges for $p \gt 1/2$ and

$$\frac{\mu_n-a_n}{b_n} = \frac{n^{p-1}}{1}$$ diverges for $p \gt 1$.

What has happened is that moving a vanishing bit of the total probability ($1/n$ of it) doesn't change the limiting distribution, but spreading the two components far enough to counterbalance that small probability (by selecting a sufficiently large $p$) allows us to control the mean and variance and even make them both diverge.