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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?

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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.

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    $\begingroup$ So this is a case where the sequence of esimators can covnerge in distribution, while the sequence of their moments will not converge to the moments of the limiting distriburtion (depending on the value of $p$)? $\endgroup$ – Alecos Papadopoulos Dec 21 '16 at 16:11
  • $\begingroup$ @Alecos. Yes. Obviously it's artificial, but such artificial examples help us understand better the conditions needed to construct good estimators and assure appropriate asymptotic properties. $\endgroup$ – whuber Dec 21 '16 at 16:14
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    $\begingroup$ Thanks @whuber. I was hoping that since we have at least convergence to Normal, and we keep on talking about asymptotic normality (everyone tries to get to that), at least we should expect some convergence of moments, no? Your counterexample is very nice. $\endgroup$ – Landon Carter Dec 21 '16 at 16:15
  • $\begingroup$ @LandonCarter See this thread for the conditions needed to have convergence of moments also, stats.stackexchange.com/a/88519/28746. In whuber's example, obviously if $p>1$, $\mu_n$ diverges, and if $p>1/2$ ,$\sigma^2_n$ diverges $\endgroup$ – Alecos Papadopoulos Dec 21 '16 at 16:20

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