Is there a way to incrementally calculate (or estimate) the average of a vector (a set of numbers) without knowing their count in advance?
For example you have a = [4 6 3 9 4 12 4 18] and you want to get an estimate of the average but you don't have all the values at hand so you want to have a running average without keeping all the previous values available.
4 Answers
You need to keep at least two pieces of information: the number of terms so far and the running mean (or something equivalent to it).
Let's suppose the $n$th component of the vector is $a_n$ and the running mean up to this is $m_n$ so $$m_n= \frac{1}{n}\sum_{i=1}^n a_i.$$
Starting with $m_0=0$, you can use the obvious single pass $$m_n = \frac{(n-1)m_{n-1}+a_n}{n}$$ but precision errors are likely to be smaller if you use the equivalent $$m_n = m_{n-1} + \frac{a_{n}-m_{n-1}}{n}.$$
Intuition
It's an easy question and @Henry basically answered it. However, I think it would be nice to add some intuition on the second equation:
$$\mu_k=\mu_{k-1}+\frac{x_k-\mu_{k-1}}{k}$$
The idea is to represent the new value $x_k$ value by a part that is equal to the previous mean $\mu_{k-1}$ plus the remaining part $x_k-\mu_{k-1}$. The new mean $\mu_k$ we then contains $k$ times the previous mean and one time the remaining part of the new value, divided by the total count. Okay, so how to get there?
