Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a series of numbers: $1,2,3$

I'm calculating a simple average between, them $(2)$.

Now, I'm deleting the series and have no information regarding how many elements there were in the series or what the series item values are, but I keep the average value $(2)$.

Now, I have another series: $4,5,6$. I calculate the average $(5)$.

Is it correct to say that if I take the previous series average $(2)$ and do a new average between it and the new series average $(5)$, I will always get an accurate result for the average of the combination of the two series (series$[1]$ and series$[2]$)?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Yes, if the series contain the same amount of numbers. No, otherwise.

  • If the series contain the same amount of numbers, you have $a_1,\dots, a_n$ and $b_1,\dots,b_n$. Now the average of the first is $a=\frac{a_1+\cdots+a_n}{n}$ and the second $b=\frac{b_1+\cdots b_n}{n}$. The average of $a$ and $b$ is $$\frac{a+b}{2} = \frac{\frac{a_1+\cdots+a_n}{n}+\frac{b_1+\cdots+b_n}{n}}{2}=\frac{a_1+\cdots+a_n + b_1+\cdots+b_n}{2n}$$ which is the average of $\{a_1,a_2,\dots,a_n,b_1,b_2,\dots,b_n\}$.

  • If the series are unbalanced, the general answe is no. If one series is $\{0\}$ and the second series is $\{1,1,1,1,1,1,1,1,\dots,1\}$ ( a set of $k$ ones), then the average of averages is always $\frac12$, while the real average of the combination is $\frac{k}{k+1}$. As $k$ becomes large, this value approaches $1$.

In general, the average of the combination of the two series is a convex combination of the individual averages. This can be seen by manipulating the formulas for the averages. If the series are $a_1,\dots, a_m$ and $b_1,\dots, b_n$, then the average is $$\frac{a_1+\cdots+a_m+b_1+\cdots b_n}{m+n} = \frac{a_1+\cdots +a_m}{m+n}+\frac{b_1+\cdots +b_n}{m+n} =\\= \frac{m}{m+n}\frac{a_1+\cdots +a_m}{m}+\frac{n}{m+n}\frac{b_1+\cdots +b_n}{n} = \alpha a + \beta b$$ where $alpha = \frac{m}{m+n}$ and $\beta = \frac{n}{m+n}$ and $a,b$ are the averages of the individual series.

share|improve this answer

No it isn't.

If you know the number of terms in each series and their averages then you can re-construct the totals and get the correct averages, but otherwise if the number of terms differ you will not get the correct result.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.