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Can we safely assume that the Standard Error of mean = $\sigma/\sqrt N$ for non-Gaussian Distribution (e.g. Chi-squared)?

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  • $\begingroup$ Standard error is respect to a statistic. Are you referring to the sample mean? $\endgroup$
    – Flowsnake
    Commented Jul 30, 2017 at 5:32

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Given a random sample of size $N$ from any distribution with standard deviation $\sigma$, the sample mean $\overline X = \frac1N\sum_{i=1}^N X_i$ has standard deviation $\sigma/\sqrt{N}$. This applies to any distribution for which a standard deviation $\sigma$ exists (i.e., any distribution with finite second moment), so in particular this includes all $\chi^2$ distributions.

Note that this quantity $\sigma/\sqrt N$ is called the standard deviation of the sample mean. Since it based on the parameter $\sigma$, usually this quantity $\sigma/\sqrt N$ is unknown, and in particular it cannot be exactly computed based on the data in the random sample. On the other hand, the standard error of the (sample) mean (SEM) refers to the related quantity $S/\sqrt N$, where $S$ is the sample standard deviation $S=\sqrt{\frac1{N-1}\sum_{i=1}^N (X_i - \overline X)^2}$. The SEM can be computed from the data and is an approximation to the standard deviation of the sample mean. This approximation is valid for all distributions (again provided only that $\sigma$ exists), in the sense that as the sample size $N$ increases, the sample standard deviation $S$ converges to $\sigma$ with probability 1 (as can be shown by the strong law of large numbers).

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