The problem goes thus:

${\{X_n\}}$ is an $iid$ sequence of random variables with mean 0 and variance $\sigma^2$. If the third moment is finite, show that $$\lim_{n \to \infty} \mathbb{E} \left(\left(\frac{S_n}{\sigma\sqrt n} \right)^3\right)=0$$ where $S_n=X_1+X_2+\cdots +X_n$

Seeing the $\frac{S_n}{\sigma\sqrt n}$ immediately reminded me of the central limit theorem which says that:

$$\lim_{n \to \infty} \mathbb{P}\left( \frac{S_n}{\sigma\sqrt n} \leq x \right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{\frac{-t^2}{2}}dt$$

But I don't really see a connection between the third moment and and the probability given here (distribution function to be more precise)

I did recall the proof of law of large numbers that went along similar lines by considering fourth order moments but we never had any $\sigma\sqrt{n}$ to be seen there and this isn't just a constant that I can tackle easily. It depends on $n$.


1 Answer 1


You don't use CLT to get this result. What is needed is a direct evaluation of the term $E[S_n^3]$.

To begin with, note that for $n \geq 3$: \begin{align*} S_n^3 = (X_1 + X_2 + \cdots + X_n)^3 = \sum_{i = 1}^nX_i^3 + \sum_{i \neq j} X_i^2X_j + \sum_{i, j, k \text{ distinct}}X_iX_jX_k. \end{align*} There are $3n(n - 1)$ terms in the second summation and $n(n - 1)(n - 2)$ terms in the third summation, but for the reason you will see shortly, counting them is not really needed.

Now by the i.i.d. and zero-mean assumption, when $i \neq j$, $E[X_i^2X_j] = E[X_i^2]E[X_j] = 0$; when $i, j, k$ are all distinct, $E[X_iX_jX_k] = E[X_i]E[X_j]E[X_k] = 0$, whence (say $E[X_1^3] = \xi$) \begin{align*} E[S_n^3] = \sum_{i = 1}^nE[X_i^3] = nE[X_1^3] = n\xi. \end{align*} It then follows that \begin{align*} E\left[\left(\frac{S_n}{\sigma\sqrt{n}}\right)^3\right] = \frac{1}{n^{3/2}\sigma^3}E[S_n^3] = \frac{\xi}{\sqrt{n}\sigma^{3}} \to 0 \end{align*} as $n \to \infty$.


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