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Let's say I have three groups of values where each group has the same number of values. However, it is unknown how many values there are per group (the values are not available anymore). For each group I do have available the mean and the variance. How can I calculate the mean and the variance from the total population? For the mean that should be easy: It is simply the mean of the means. But how about the variance?

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  • $\begingroup$ The answer depends on what formula you are referring to by "variance." Could you clarify? Would it perhaps be in the same sense as the question at stats.stackexchange.com/questions/10441 (as found by Alex Nikiforov)? $\endgroup$ – whuber Aug 29 '17 at 17:26
  • $\begingroup$ There's an old rule of thumb in statistics that the variance of the sums is equal to the sum of the variances. It may apply in your case. $\endgroup$ – DJohnson Aug 29 '17 at 17:58
  • $\begingroup$ When you're combining non-overlapping subgroups the variation between the means comes in as well (many, many posts on site deal with that issue). The wrinkle here is doing it when $n$ is unknown. $\endgroup$ – Glen_b Aug 30 '17 at 2:45
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First of all, the mean is not exactly the mean of the means. But, considering $N=n_1+n_2+n_3$ the population (that in this sense is the union of the the three groups) average is $\mu=\frac{n_1\mu_1+n_2\mu_2+n_3\mu_3}{n_1+n_2+n_3}$. Thus, you have a set of averages in a simplex generated by the constraints $N=n_1+n_2+n_3$ and $n_1,n_2,n_3>0$. For the variance ($\sigma^2$), you can use a similar approach. The population variance is the sum of the Between Group Variance and the Within Group Variance as follows: $$N\cdot \sigma^2=\sum\limits_{g=1}^3 n_g(\mu_g-\mu)^2+\sum\limits_{g=1}^3 n_g \sigma^2_g$$ Also in this case, considering that $$\sum\limits_{g=1}^3 n_g(\mu_g-\mu)^2=\sum\limits_{g=1}^3 n_g\mu_g^2-N\cdot\mu^2$$ your solution is one of the possible inside the simplex. But remember that $\mu$ and $\sigma$ depends both on the choice of $n_1$,$n_2$, and $n_3$. In your case, $n_1=n_2=n_3$ the total variance is $$\sigma^2=\frac{1}{3}\sum\limits_{g=1}^3 \left[(\mu_g-\mu)^2+\sigma^2_g\right]$$

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  • $\begingroup$ The mean is the mean of means if $n_1=n_2=n_3$, right? Which is the case in my case. $\endgroup$ – Make42 Feb 3 at 16:04
  • $\begingroup$ @antonioirpino Please add your comment into the answer where the subpops are all the same size. $\endgroup$ – CMCDragonkai Feb 11 at 6:52
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Well, variance estimation can be obtained for two groups (for simplicity) as follows:

${\hat{\sigma}^2} = \frac{1}{2N}\sum_{i=1}^{2N}{(X_i-\mu)^2 = \frac{1}{2N}\sum_{i=1}^{N}{(X_i-\mu)^2} + \frac{1}{2N}\sum_{i=N+1}^{2N}{(X_i-\mu)^2}=\frac{1}{2}(\hat{\sigma}^2_1 + \hat{\sigma}^2_2}) = \hat{\sigma}^2$

where ${X_i}$ - is a random variable (values in your case, which are not available anymore) and ${\mu}$ - is a mean.

so, variance of the total population is average of variances for every group ${\frac{1}{2}(\hat{\sigma}^2_1 + \hat{\sigma}^2_2)}$ where ${\hat{\sigma}^2_{1}}$ is a variance of group 1, the same for group 2.

quick test on Octave, where ${x, y}$ - are two groups:

octave:1> x = 3*randn(1000, 1);
octave:2> y = 3*randn(1000, 1);
octave:3> var(x)
ans =  9.0051
octave:4> var(y)
ans =  8.8170
octave:5> 0.5*(var(x) + var(y))
ans =  8.9111
octave:6>

${\hat{\sigma}^2_{1} = 9.0051}$, ${\hat{\sigma}^2_{2} = 9.0051}$, ${\hat{\sigma}^2 = 8.9111}$

Think of your estimation as a random variable, it has it's own mean and variance.

[EDIT] there is a better answer (and more correct).

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    $\begingroup$ There are multiple errors in the first line that need to be corrected, such as the wrong denominator in the first formula and the disappearance of the $X_i$ for $i\gt N$ in the second formula. When you do that, could you explain what your symbols are intended to refer to and what assumptions you are making about them? For many possible interpretations your results are incorrect, so if you would like them to be understood as you intended, including such explanation is essential. $\endgroup$ – whuber Aug 29 '17 at 18:21
  • $\begingroup$ Hi, could you please suggest what else I should add? Formulas have been fixed, thanks! $\endgroup$ – Alex Nikiforov Aug 29 '17 at 18:27
  • $\begingroup$ I can only repeat myself: explain your notation and tell us how you are interpreting this (inherently ambiguous) question. BTW, the equalities are still incorrect: your first sum references $X_1, \ldots, X_{2N}$ whereas the sums after the equality reference only $X_1, \ldots, X_N$. What happened to $X_{N+1}, \ldots, X_{2N}$? What's the reason for putting a hat on "$\sigma$"? What are $\hat\sigma_1$ and $\hat\sigma_2$? What do you mean by "average"? What do the $X_i$ represent in the original question? How does this post address the question about three variances? $\endgroup$ – whuber Aug 29 '17 at 18:43
  • $\begingroup$ (Continued) Exactly what is "the mean" $\mu$? How is it related to the $X_i$? How is it related to the three means mentioned in the question? $\endgroup$ – whuber Aug 29 '17 at 18:44
  • $\begingroup$ What do you mean by "more correct"? I see that you are not using the notorious $-1$, while the other answer seems to. But the other answer also uses the Variance of the means, which confuses me. I am not sure how this relates to your formula. Can you explain? $\endgroup$ – Make42 Aug 30 '17 at 8:10

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