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Example

The variance of the error of a sample mean can be expressed in terms of the variance of the distribution of the population that is sampled.

$$E[(\bar{X}-\mu)^2] = \frac{1}{n} \cdot E[(X-\mu)^2] $$

In this case, the expression on the right-hand side can be partitioned in a product of a function that is only dependent of $n$ and a function that is an expectation of some function of the variable.

Generalisations?

Which are other functions $g(x) \neq x^2$ for which we can write the expectation $E[g(\bar{X}-\mu)]$ as a product like below? What can we say about these functions?

$$E[g(\bar{X}-\mu)] = h(n) \cdot E[f(X-\mu)]$$

So in the example case we have $g(x) =x^2$ in which case $h(n) = 1/n$ and $f(x) = x^2$.

(to be clear: $f(x)$ is not a function that can change depending on $n$.)

Background: In this answer I use an expression of the variance of the error of the sample mean in terms of an expectation of the population distribution that is computed with an integral of the quantile function $E[f(X)] = \int_{\Omega}f(X(\omega)) dp(\Omega)$. The question remains open for which other cost functions we could use a similar construction.

Sextus Empiricus
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  • For the 3rd central moment we can also partition the expression. Because we have [the additive property](https://en.wikipedia.org/wiki/Central_moment#Properties). $$\mu_3(X+Y) = \mu_3(X) + \mu_3(Y)$$ from which follows $$\mu_3(\bar{X}) =\frac{1}{n} \mu_3(X)$$. – Sextus Empiricus Jan 10 '22 at 08:43
  • I made a mistake in the previous comment. We have $$\mu_3(\bar{X}) = \mu_3( \sum_{i=1}^n \frac{1}{n} {X_i}) = n \mu_3(\frac{1}{n} X) = n \left( \frac{1}{n}\right)^3 \mu_3( X) = \frac{1}{n^2} \mu_3(X)$$ – Sextus Empiricus Jan 10 '22 at 09:27

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