# Are the mean and variance of a set equals the sum of its means and variaces

I'm not so fit in statistics and I found some controverse answers on the internet so I'm asking here.

I have a set $$A$$ with 10439 samples. The set is not of unique values and many of the sample values occur more than once. Now I want to calculate the mean and the variance of the set, but there is a limitation. I can't calculate for the whole set at once so I had to split it into $$N$$ batches. Now lets say I have $$N=73$$ batches $$A_{1}, ... ,A_{73}$$ created form the big set $$A$$. Do those hold?

1. $$mean(A) = \frac{mean(A_{1}) + ... + mean(A_{73})}{N}$$
2. $$var(A) = \frac{var(A_{1}) + ... + var(A_{73})}{N}$$

If not then how can I achieve my goal using the batches?

## 1 Answer

Statement 1 and Statement 2 are both untrue.

Statement 1 is not true because the means in the average are not weighted by the size of the set. For example if $$A_1 = \{1\}$$ and $$A_2 = \{2,2,2,2\}$$ then $$\operatorname{mean}(A_1 \cup A_2) = 1.8$$ but $$\tfrac{\operatorname{mean}(A_1) + \operatorname{mean}(A_2)}{2} = \frac{1+2}{2} = 1.5$$. The correct formula is $$\operatorname{mean}(A) = \frac{|A_1|\operatorname{mean}(A_1) + |A_2|\operatorname{mean}(A_2) + \cdots + |A_N|\operatorname{mean}(A_N)}{|A_1|+|A_2|+\cdots+|A_N|}$$ where $$|A_i|$$ gives the number of elements in the set $$A$$. To see this, expand $$\operatorname{mean}(A) = \frac{1}{|A|}\sum_{a\in A}a = \frac{1}{|A|}\biggl(\frac{|A_1|}{|A_1|}\sum_{a\in A_1}a + \frac{|A_2|}{|A_2|}\sum_{a\in A_2}a + \cdots + \frac{|A_2|}{|A_2|}\sum_{a\in A_N}a\biggr)$$ and observe that $$\tfrac{1}{|A_i|}\sum_{a \in A_i} a$$ is exactly $$\operatorname{mean}(A_i)$$.

Statement 2 is untrue for effectively the same reason. One way to compute the correct $$\operatorname{var}(A)$$ in batches is to compute $$s_i = \sum_{a\in A_i}a^2$$ for every batch $$i=1,2,\ldots,N$$. At this point $$\operatorname{var}(A) = \frac{s_1+s_2+\cdots+s_N}{|A_1| + |A_2| + \cdots + |A_N|} - \operatorname{mean}(A)^2,$$ where $$\operatorname{mean}(A)$$ was computed as above.

• Thank you very much I've calculated the first Statement correctly. I've had to calculate the variance of each batch separately and sequentially for each of the 73 batches with $$var(A_i) = \frac{1}{|A_i|}\cdot\sum_{i=1}^{|A_i|} (a_i - \mu_{A_i})^2$$ Is there a way to add my sequential results to get the whole variance of (A)? – TalG Mar 8 at 23:06
• It turns out that $\sigma_{A_i}^2 = \tfrac{s_i}{|A_i|} - \mu_{A_i}^2$ as well, so with $\sigma_{A_i}^2$ already computed you can find $s_i = |A_i|(\sigma_{A_i}^2 + \mu_{A_i}^2)$ and use the formula above. – cdipaolo Mar 8 at 23:09
• Thanks again! The calculation of the second Statement worked with $$var(A)=\frac{|A_1|(\sigma^2_{A_1} + \mu^2_{A_1}) + |A_2|(\sigma^2_{A_2} + \mu^2_{A_2}) + \cdots + |A_N|(\sigma^2_{A_N} + \mu^2_{A_N})}{|A_1| + |A_2| + \cdots + |A_N|}-mean(A)^2$$ – TalG Mar 9 at 0:03