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Let $X_1,X_2,\dots,X_n$ be a sample of $n$ independent and identically distributed observations of a continuous population random variable $X$. Define $Z_n$ to be the inverse of the sample variance: $$ Z_n = \bigg(\frac{1}{n}\sum_{i=1}^n X_i^2 - \overline{X}_n^2\bigg)^{-1}. $$

Assuming that $Z_n$ is integrable for a fixed $n$ seems to be a relatively weak assumption based on comments in my previous question.

Now I am wondering what are the implications of assuming the set $\{Z_n\}_{n \ge 3}$ to be uniformly integrable. Note that we start the indexing at $n = 3$ as the expectation does not exist for $n=1,2$.

Is uniformly integrability of $\{Z_n\}_{n \ge 3}$ a weak assumption or a strong assumption? If we assume that this set is uniformly integrable what kind of population random variables $X$'s are we allowing/excluding?

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    $\begingroup$ You had better start indexing at $n=3,$ because $Z_2$ never has an expectation! $\endgroup$
    – whuber ♦
    Commented Mar 18, 2021 at 14:25
  • $\begingroup$ My curiosity as a non native speaker: is 'inverse' the correct term? The reciprocal variance is usually called precision. I can't help with the question of uniform intregrability, but would be interested, why this is of interest. $\endgroup$
    – cherub
    Commented Mar 18, 2021 at 17:09

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