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Say we take a sample every time unit, we have a calibrated value which needs tracking for drift, however every sample has a given variance causing jumping if we were to use it directly as the new calibrated value, therefore we use $(1-a)*\textrm{current} + a*\textrm{sample}$ with $a$ some suitable value between $0$ and $1$. What is this kind of method called?

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$(1-a)x + a y$ is called a convex combination of $x$ and $y$ (assuming $a \in [0,1]$)... –  copper.hat Oct 29 '12 at 2:53
    
@copper.hat thanks, but I believe there is some name in the context of sampling/data-series, I am just utterly unable to find it... –  wich Oct 29 '12 at 2:57
    
Filtering of some sort? –  copper.hat Oct 29 '12 at 2:58
    
I think it is called exponential smoothing. If sample is constant, then the value converges to this value exponentially. –  copper.hat Oct 29 '12 at 3:00
    
@copper.hat Looks like you are correct, but I don't understand why the wikipedia entry for exponential smoothing gives $s_{t} = \alpha x_{t-1} + (1-\alpha)s_{t-1}$ as definition instead of $s_{t} = \alpha x_{t} + (1-\alpha)s_{t-1}$. By the way, please make an answer, then I can accept it. –  wich Oct 29 '12 at 3:08
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It is called exponential smoothing. If sample is constant, then the value converges to this value exponentially.

If $u_n$ represents the 'raw' data, then $x_n$ is the 'exponentially smoothed' version in the following (with $\alpha \in (0,1]$):

$$x_{n+1} = (1-\alpha) x_n + \alpha u_n$$

The explicit solution is given by $$ x_n = (1-\alpha)^n x_0 + \sum_{k=0}^{n-1} (1-\alpha)^k \alpha u_{n-1-k}$$ If $u_n = \hat{u}$ for all $n$, then this gives $$x_n = (1-\alpha)^n x_0 + \hat{u} (1 - (1-\alpha)^n)$$ from which we see that $x_n \to \hat{u}$ with rate $(1-\alpha)$.

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