3
$\begingroup$

In statistics, the moving average is usually defined for discrete data sets. Is there a moving average concept for continuous randomly fast-oscillating functions? I am seeking for the moving average determined in terms of integrals rather than sums.

$\endgroup$

1 Answer 1

4
$\begingroup$

You could simply use the mean value of the function in each window.

A "backward-looking" moving average with window size $w$ is $$ \mbox{MA}_w^\text{b}(t) = \frac{1}{w} \int_{t-w}^t f(x)dx , $$ and a "centered" moving average is $$ \mbox{MA}_w^\text{c}(t) = \frac{1}{w} \int_{t-w/2}^{t+w/2} f(x)dx . $$

$\endgroup$
2
  • $\begingroup$ Seems nice! Do you have a reference to a textbook or paper where this definition is utilized? $\endgroup$
    – freude
    Apr 27, 2014 at 19:38
  • $\begingroup$ I don't have a reference where the suggested function is used to compute the moving average; I just customized the first mean value theorem to your needs. $\endgroup$
    – QuantIbex
    Apr 28, 2014 at 5:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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