# Averages in series

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$[2]$)?

-

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.

-

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.

-