In all derivations I've seen of the standard error formula $\sigma/\sqrt{n}$, it is assumed all the samples in the sampling distribution have the same variance ($\sigma^2$). Why is it assumed they all have the same variances, and why is each sample's variance the population variance?

For example, in this answer: https://stats.stackexchange.com/a/89159/419909, it is assumed each $X_i$ has the same $\sigma^2$

New contributor
statataka is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • 3
    $\begingroup$ Welcome to Cross Validated! These samples are all drawn from the same distribution in these derivations, right? $\endgroup$
    – Dave
    Commented Jul 11 at 3:04
  • $\begingroup$ Thank you, and yes. This is what I'm referencing: stats.stackexchange.com/a/89159/419909 the answer assumes that all the samples have the same distribution and variance, but isn't that not always the case? $\endgroup$
    – statataka
    Commented Jul 11 at 3:11
  • 1
    $\begingroup$ Hi: Under standard sampling assumptions, it is assumed ( unless otherwise stated ) that the samples are being taken from the same population. So, assuming the variance is the same for each sample is not unreasonable if each sample is taken from the same population. $\endgroup$
    – mlofton
    Commented Jul 11 at 3:37
  • 3
    $\begingroup$ Yes, the variances can differ. The rules of variance for uncorrelated random variables imply the variance of the average $(X_1+X_2+\cdots+X_n)/n$ is $1/n^2$ times the sum of their variances, so there's no problem. The issue is this: if you don't know those $n$ variances, how do you propose to estimate them if you only have this sample of $n$ values?? $\endgroup$
    – whuber
    Commented Jul 11 at 13:16