4

Let $X_1, \ldots, X_n$ be independent real random variables, with $n > 1$. Define the sample mean by :

\begin{equation*} \overline{X} = \frac{1}{n} \sum_{i=1}^n X_i \end{equation*}

Define the sample variance by :

\begin{equation*} S^2 = \frac{1}{n-1} \sum_{i=1}^n(X_i-\overline{X})^2 \end{equation*}

If, for each $i$, $X_i$ is normal with mean $\mu_i$ and variance $\sigma_i^2$, then we know from Cochran's theorem that $\overline{X}$ and $S^2$ are independent.

What about the converse? Suppose that $\overline{X}$ and $S^2$ are independent, does it follow that each $X_i$ must be normal with some mean and variance? Does the result change if we assume furthermore integrability or boundedness on each $X_i$ (or even $X_i$'s are identically distributed)?

Amir
  • 11,124
温泽海
  • 3,335
  • This paper (https://academic.oup.com/jrsssb/article-abstract/3/2/178/7026432) seems to claim that it is true. I can unfortunately only see the first page of the paper. – PhoemueX May 05 '24 at 19:35
  • Let me spend some time to check it out. It is going to take a while. – 温泽海 May 05 '24 at 19:39

1 Answers1

2

Yes it is a characterization for the normal distribution shown first by Geary (1936) that

the sample mean and sample variance are independent if and only if the population distribution is normal.

See Apeendix C in this paper for a proof using modern notation.

Moreover, note that the sample mean and sample variance are uncorrelated for any population distribution with finite third moment and symmetric distribution about its mean as we have

$$\text{Cov}(\bar{X}, S^2) = \frac{\mathbb E ((X-\mu)^3)}n.$$

See here for a proof of the above formula and here for an elegant proof of the above observation (without using the above formula).

Amir
  • 11,124