Suppose that $X_i$ are i.i.d. following a Bernoulli distribution with $\mathbb{P}(X_i = 1) = \theta$. Let $\bar X_n$ be the sample mean of $X_1, ..., X_n$. Let $h(t) = t^3 - t$. Find $a_n$ and $b_n$ such that $a_n\left(h(\bar X_n) - h(\theta) - b_n\right)$ converges in distribution (to a non-degenerate limit).

Notice that $h'(t) = 3t^2 - 1$ and has zeros at $\pm \frac{1}{\sqrt{3}}$, so if $\theta = \frac{1}{\sqrt{3}}$ (which is plausible), $h'(\theta) = 0$.

Assume that is not the case. Then $\bar X_n$ satisfies the delta theorem and a simple application of the theorem yields:

$$\sqrt{n}\left[h(\bar X_n) - h(\theta)\right] \to N\left(0, \theta(1 - \theta)\left(h'(\theta)\right)^2\right)$$

Thus pick $a_n = \sqrt{n}$ and $b_n = 0$ and the problem is solved.

But if $\theta = \frac{1}{\sqrt{3}}$, the delta theorem cannot be applied.

Thus my question is: do I need to find a new $a_n$ and $b_n$ for the case $\theta = \frac{1}{\sqrt{3}}$?


1 Answer 1


The so-called delta-method stems from two basic results:

  1. By the usual CLT, $\sqrt{n}(\bar X_n-\theta)\to\sigma_\theta Z$ in distribution, where $Z$ is standard normal and $\sigma_\theta^2=\mathrm{var}(X_1)$.
  2. If $h'(\theta)\ne0$, then $h(x)-h(\theta)\sim h'(\theta)(x-\theta)$ when $x\to\theta$.

Putting 1. and 2. together, one sees that, if $h'(\theta)\ne0$, then $h(\bar X_n)-h(\theta)\sim h'(\theta)(\bar X_n-\theta)$, which implies:

$\sqrt{n}(h(\bar X_n)-h(\theta))\to a_\theta Z$ in distribution with $a_\theta=h'(\theta)\sigma_\theta$.

This is the result you recall since $a_\theta Z$ is centered normal with variance $a_\theta^2=h'(\theta)^2\theta(1-\theta)$.

Likewise, if $h'(\theta)=0$, our point 2. above becomes:

  1. If $h'(\theta)=0$ and $h''(\theta)\ne0$, then $h(x)-h(\theta)\sim \frac12h''(\theta)(x-\theta)^2$ when $x\to\theta$.

Thus, if $h'(\theta)=0$ and $h''(\theta)\ne0$, then $h(\bar X_n)-h(\theta)\sim \frac12h''(\theta)(\bar X_n-\theta)^2$, which implies:

If $h'(\theta)=0$, then $n(h(\bar X_n)-h(\theta))\to b_\theta Z^2$ in distribution with $b_\theta=\frac12 h''(\theta)\sigma_\theta^2$.

It remains to identify the distribution of $b_\theta Z^2$ if $Z$ is standard normal. Can you do that?

  • $\begingroup$ $Z^2 \sim \chi^2(1)$ so $b_\theta Z^2$ is a scaling of a $\chi^2$-random variable. I don't know off the top of my head what $b_\theta$ is but finding it does not look hard. Thank you! $\endgroup$
    – cgmil
    Jul 30, 2017 at 22:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .