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Suppose we have an experiment involving $N$ independent samples of single variable functions, $y_1(t),...,y_N(t)$ where $$y_k(t) = \dfrac{1}{M}\sum_{j = 1}^{M} x_j(t); \ \ k = 1,...,N.$$ I am interested in the average over independent samples, $$\bar{y}(t) = \dfrac{1}{N} \sum_{k = 1}^N y_k(t)$$

How should I combine the uncertainty in the measurements $x_j(t)$, $y_k(t)$ and the final average $\bar{y}(t)$? For instance, if I produce a plot of each $y_i(t)$, I could draw error bars at each time point equal to the standard deviation $$\tilde{y}_i(t) = y_i(t) \pm \sqrt{\text{Var}[y(t)]}$$ to give some indication of uncertainty. The same could be done for $\bar{y}$, $$\hat{y}_i(t) = \bar{y}(t) \pm \sqrt{\text{Var}[\bar{y}(t)]}.$$ Clearly if for all $t$, $$ \sqrt{\text{Var}[y(t)]} \le \sqrt{\text{Var}[\bar{y}(t)]},$$ we could just consider the error in the final average; under what conditions would that be true? If the $x_j(t)$'s are i.i.d with finite variance and the CLT applies, then $$\sqrt{\text{Var}[y(t)]} \sim \mathcal{O}(M^{-1/2})(?)$$ Then $N \ll M$ would justify us to indicate only the width (SD) of $\bar{y}(t)$ as a simple error estimate. I imagine this situation is very common, any standard texts or references discussing it? This post is similar, but I am hoping for a conceptual explanation.

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If $X$ has standard deviation of $\sigma$, the distribution of the means of $n_1$ samples has standard deviation $\dfrac{\sigma}{\sqrt{n_1}}$.

Now you can apply the same idea. The means of $n_2$ samples of the distribution of the mean of $n_1$ samples will have a standard deviation of $\dfrac{\sigma}{\sqrt{n_1} \cdot \sqrt{n_2}}$.

This works as long as you can apply Central Limit Theorem.

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  • $\begingroup$ Thanks! This combines the sample sizes in the final result of SD in $\bar{y}$ but should there not be a $\sigma_1$ and $\sigma_2$ for each averaged quantity? To clarify my question, I would like to have some measure of the error of the $\bar{y}$ object which accounts for errors in the determination of the $y$'s. In your answer, the relationship between SD's of each object $\bar{y}$ and $y$ is independent of $\sigma$. $\endgroup$
    – algae
    Apr 29, 2020 at 9:06
  • $\begingroup$ I am sorry, I am not sure of what you want. I hope someone can give some information. $\endgroup$ Apr 30, 2020 at 12:54
  • $\begingroup$ My fault, you are in the right direction. The error of the means is defined as you say: en.wikipedia.org/wiki/Standard_error. It's just about combining them. $\endgroup$
    – algae
    May 10, 2020 at 2:28

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