I have a set of measurements which is partitioned into M partitions. However, I only have the partition sizes $N_i$ and the means $\bar{x}_i$ from each partition. Because all measurements are assumed to be from the same distribution, I believe I can estimate the mean of the population, $\bar{y}$, and standard deviation of the mean, $\sigma_{mean}$: $$ N=\sum_{i=1}^M N_i $$ $$ \bar{y} = \frac{1}{N}\sum_{i=1}^MN_i\bar{x}_i $$

$$ \sigma_{mean}=\sqrt{\frac{1}{N}\sum_i N_i(\bar{x}_i-\bar{y})^2} $$

My questions:

  1. Am I right in my assumptions, that the mean $\bar{y}$ can be computed as above?
  2. How can I find the standard deviation for the population, given only the means? I read that the standard deviation of the population and standard deviation of the mean is related with $$ \sigma_{mean}=\frac{\sigma}{\sqrt{n}} \mbox{[1]} $$ where $n$ is the number of samples used in the computation of $\bar{x}_i$. So is it actually as simple as just multiplying $\sigma_{mean}$ with $\sqrt{n}$ if $n$ for all means are the same?
  3. If it's that simple, what do I do if each $\bar{x}_i$ is computed using a different number of samples?

[1] Wikipedia:Standard Deviation

  • $\begingroup$ Do you know for a fact that all of these subsets were taken from the same population? Do you know how the partitioning was done (eg, 1st 10, 2nd 15, next 7, etc)? How did these measurements come to you & what meta-information came with them? $\endgroup$ Mar 20, 2012 at 14:30
  • $\begingroup$ Yes; these are energy measurements of a CPU for an experiment I am running, and are run on the same machine under identical conditions every time. I only have the means because the granularity is too coarse to get readings on each sample, so I have to run the experiment for a number of times to get reasonable accuracy. The partitions is like this: The N1 first measurements is in partition 1, the N2 next measurements is in partition 2, and so on. Basically, what I try to get is the standard deviation for each separate e $\endgroup$
    – Hallgeir
    Mar 20, 2012 at 14:44
  • $\begingroup$ I have added the anova tag because its techniques will readily answer these questions: (1) is standard but (2) is incorrect because it neglects the within-group variation. $\endgroup$
    – whuber
    Mar 20, 2012 at 15:54
  • $\begingroup$ Thank you whuber! Pardon my ignorance, but would not variation within groups be reflected in their means collectively? I.e. wouldn't a large in-group variation typically result in larger variation in the means, than if the in-group variation was small, if the subsets are assumed to be independent and from the same distribution? I will also take a look at the topic of anova as well; thanks for that! $\endgroup$
    – Hallgeir
    Mar 20, 2012 at 17:17
  • $\begingroup$ @whuber, if we knew that they all came from the same population, they would all have to have the same variance. Wouldn't that mean that we could ignore within-group variation so long as we weighted the group means by group n's as in the (unlabeled) 3rd equation? $\endgroup$ Mar 20, 2012 at 21:04

1 Answer 1


Let $X_i$ be the mean of $N_i$ independent draws from some unknown distribution $F$ having mean $\mu$ and standard deviation $\sigma$. Altogether these values represent $N=N_1+N_2+\cdots+N_k$ draws. It follows from these assumptions that each $X_i$ has expectation $\mu$ and variance $\sigma^2/N_i$.

Part of the question proposes estimating $\mu$ from these data as

$$\hat{\mu} = \frac{1}{N}\sum_{i=1}^k N_i X_i.$$

We can verify that this is a good estimate. First, it is unbiased:

$$E[\hat{\mu}] = E\left[\frac{1}{N}\sum_{i=1}^k N_i X_i\right] = \frac{1}{N}\sum_{i=1}^k N_i \mu = \mu.$$

Second, its estimation variance is low. To compute this we find the second moment:

$$\begin{align} E\left[\hat{\mu}^2\right] &= E\left[\frac{1}{N^2}\sum_{i,j}N_i N_j X_i X_j\right]\\ &= \mu^2 + \sigma^2/N. \end{align}$$

Subtracting the square of the first moment shows that the sampling variance of $\hat{\mu}$ equals $\sigma^2/N$. This is as low as an unbiased linear estimator can possibly get, because it equals the sampling variance of the mean of the $N$ (unknown) values from which the $X_i$ were formed; that sampling variance is known to be minimum among all unbiased linear estimators; and any linear combination of the $X_i$ is a fortiori a linear combination of the $N$ underlying values.

To address the other parts of the question, let us seek an unbiased estimator of the variance $\sigma^2$ in the form of a weighted sample variance. Write the weights as $\omega_i$. Computing in a similar vein we obtain

$$\begin{align} E\left[\widehat{\sigma^2}\right] &= E\left[\sum_i \omega_i(X_i-\hat{\mu})^2\right] \\ &= \sum_i \omega_i E\left[X_i^2 - \frac{2}{N}\sum_j N_j X_i X_j + (\hat{\mu})^2\right]\\ &= \sum_i \omega_i \left((\mu^2 + \sigma^2 / N_i)\left(1 - 2\frac{N_i}{N}\right) - \frac{2}{N}\sum_{j\ne i} N_j \mu^2 + (\mu^2 + \sigma^2/N)\right)\\ &= \sigma^2 \sum_i \omega_i\left(\frac{1}{N_i} - \frac{1}{N}\right). \end{align}$$

A natural choice (inspired by ANOVA calculations) is $$\omega_i = \frac{N_i}{k-1}.\tag{*}$$ For indeed,

$$E\left[\widehat{\sigma^2}\right] = \sigma^2 \sum_i^k \frac{N_i}{k-1}\left(\frac{1}{N_i} - \frac{1}{N}\right) = \sigma^2 \frac{1}{k-1}\sum_i^k \left(1 - \frac{N_i}{N}\right) = \sigma^2\frac{k-\frac{N}{N}}{k-1} = \sigma^2.$$

This at least makes $\widehat{\sigma^2}$ unbiased. With more than $k=2$ groups, there are many other choices of weights that give unbiased estimators. When the group sizes are equal, it's easy to show that this choice gives a minimum-variance unbiased estimator. In general, though, it appears that the MVUE depends on the first four moments of $F$. (I may have done the algebra wrong, but I'm getting some complicated results for the general case.) Regardless, it appears that the weights provided here will not be far from optimal.

As a concrete example, suppose that each of $X_1$, $X_2$, and $X_3$ is the average of $N_i=4$ draws. Then $N=12$, $k=3$, and the weights as given in formula $(*)$ are all given by $\omega_i = \frac{4}{3-1}=2$. Consequently we should estimate $$\widehat{\sigma^2} = 2((X_1-\hat{\mu})^2 + (X_2-\hat{\mu})^2 + (X_3-\hat{\mu})^2)$$ and, of course, $$\hat{\mu} = \frac{1}{12}(4X_1 + 4X_2 + 4X_3) = (X_1+X_2+X_3)/3.$$

  • $\begingroup$ Thank you so much for a very thorough and in-depth answer; this completely answers my question! :) $\endgroup$
    – Hallgeir
    Mar 23, 2012 at 8:04
  • 1
    $\begingroup$ I've run some simulations on this to verify the answer and I encountered a few problems. It appears that choosing $w_i=N_iN/(N-1)$ does not give the correct result (it is scaled with about $k$). Plugging this weight into your formula for $E(\hat{\sigma^2})$, gives $E[\hat{\sigma^2}]=\sigma^2(\frac{kN}{N-1}-1)$. Choosing $w_i=\frac{N_iN}{N-N_i}\frac{1}{k}$ solves this, but the resulting estimator underestimates $\sigma^2$ when the number of groups is small. I wonder if the expression for $E[\hat{\sigma}^2]$ is correct; I've tried the calculations but I have a hard time following them. $\endgroup$
    – Hallgeir
    Mar 24, 2012 at 15:23
  • 1
    $\begingroup$ Thanks for catching that. I had erroneously summed $k$ $1$'s and arrived at $N$ :-). That has now been fixed, both in the formula and the example at the end. I ran simulations with $N$ and $k$ varying between $2$ and $17$ (both with constant and varying $N_i$ and including cases with groups of one element each) with accurate results. $\endgroup$
    – whuber
    Mar 24, 2012 at 18:55
  • $\begingroup$ Thanks for clearing that up! :) The new weights you propose seems to work well. $\endgroup$
    – Hallgeir
    Mar 25, 2012 at 9:54

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