Say I have an exponential smoothing for certain $\Delta t$, $t_{i+1} = t_i + \Delta t$. In this sampling, I choose a particular $\alpha$ to filter signal $z_i$ like $$ v_1 = z_0 \\ v_{i+1} = \alpha\:z_i + (1-\alpha)\:v_i $$

and we have some sort of agreement as to what $\alpha$ to use.

Now, someone comes along and tells me that I am able to get, say, five times more samples in the same $\Delta t$ period. I need to modify the smoothing factor $\alpha$. It seems to me, intuitively, that the new factor, say $\alpha'$, that has the same discount effect for previous values but with five times more iterations should be $$ (1-\alpha')^5 = 1-\alpha $$

Is this correct? Is there any principled approaches to justify this?

Note: The formula above seems sensible in the sense that the original asymptotic variance (for a constant model) $$ \mathbb{V}ar[v] \to \frac{\alpha}{2-\alpha} \sigma^2 $$ becomes $$ \mathbb{V}ar[v'] \to \frac{\alpha'}{2-\alpha'} \sigma^2 $$ which, under the equation above, is actually smaller (I checked this). This makes sense as more data would give more precision for $v_t$ as estimator of a constant $z_t = k + \epsilon$, where $\epsilon \sim \mathcal{N}(0,\sigma)$.


1 Answer 1


I think it's instructive to take a look at the impulse responses, taking absolute time into account. I'm assuming the following recursion:

$$y_n=\alpha x_n+(1-\alpha)y_{n-1}\tag{1}$$

The impulse response corresponding to the recursive system described by (1) is


Assume now that we have a discrete-time system with sample interval $T$, and another one with sample interval $T^{\prime}=T/L$. Let $\beta=1-\alpha$, and let $\alpha^{\prime}$ and $\beta^{\prime}$ be the constants of the system with sample interval $T^{\prime}$. In order for the two systems to have the same time constant and the same scaling we require (considering a point in time $t_n=nT=nLT^{\prime}$)


From (2) we see that

$$\alpha=\alpha^{\prime},\quad \beta=\beta^{\prime L}\quad \text{(i.e. $\beta^{\prime}\neq 1-\alpha^{\prime}$)}$$

In this case the impulse response of the system with the higher sampling rate is simply an interpolated version of the other system's impulse response. Note that the condition on $\beta^{\prime}$ is the one that you came up with intuitively, and it is also the more important of the two because it affects the time constant of the system. The choice $\alpha=\alpha^{\prime}$ only affects the scaling and in this sense is secondary. Choosing $\alpha^{\prime}=1-\beta^{\prime}$ is of course also possible and just changes the scaling.

  • $\begingroup$ View at the impulse response, great idea! $\endgroup$
    – carlosayam
    May 29, 2014 at 0:59
  • $\begingroup$ What does it mean to say that setting $\alpha^{\prime}$ to $1 - \beta^{\prime}$ "only affects the scaling"? How will the system behave if the $\alpha^{\prime}$ value is changed to no longer equal $\alpha$? $\endgroup$
    – mstahl
    Oct 24, 2022 at 16:05
  • $\begingroup$ Intuitively, it seems to me that I must use $\alpha^{\prime} = 1 - \beta^{\prime}$. Exponential smoothing computes the current smoothed value as the weighted sum of the current data value and the previous smoothed value. If $\alpha^{\prime}$ is anything other than $1 - \beta^{\prime}$,then the sum of the weights would not be equal to $1$, and it wouldn't be "exponential smoothing". $\endgroup$
    – mstahl
    Oct 25, 2022 at 13:04

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.