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If the formula is used for forecasting purposes, then it looks to me like the second one is the only usable one (and also looks more natural somehow). Or am I missing something? (I'm very, very rusty on this now!). Hope someone can clear this up.--
A bit iffy11:18, 3 March 2007 (UTC)reply
The choice is a matter of convention (for strictly periodic data), but I think that the first one (which we use here) is more natural, since people would compute the smoothed value as soon as possible and would naturally want to label it with the time that they computed it.
JRSpriggs11:27, 4 March 2007 (UTC)reply
Agreed to this, when there is a large amount of terms, is does not matter whether x_t or x_{t-1} is used. For few terms the first one should be used.
137.56.63.144 (
talk)
15:33, 21 October 2022 (UTC)reply
The textbooks that I use for teaching time series use the second one (with the lagged value of the series). See Moore, McCabe, Duckworth and Sclove,
The Practice of Business Statistics. In Minitab, on the other hand, the smoothed value at time t is defined using the first formula, while the fitted value at time t is the smoothed value at time (t-1).
EconProf8616:20, 28 May 2007 (UTC)reply
In my view it should be X_t and not X_{t-1}. The simple exponential filter is analageous to a single pole low pass IIR filter. In both cases you are saying that the current out put is formed from the sum of the new input plus a fraction of the old output — Preceding
unsigned comment added by
212.77.61.18 (
talk)
08:09, 9 September 2011 (UTC)reply
Both are correct. As said above, X_t is used when the focus is smoothing a series and X_{t-1} is used when prediction is the focus. As the article mostly follows the
NIST handbook we use X_{t-1}.--
Muhandes (
talk)
16:55, 10 September 2011 (UTC)reply
Another related point: the following paragraph states "In the limiting case with α = 1 the output series is just the same as the original series". This will only be the case with X_t. Using X_{t-1} means adding a lag in the time series.
Eric thiebaut (
talk)
17:28, 24 December 2011 (UTC)reply
Since there is only one term in the equation, the results will, in all cases, be the same except for one time step delay.
Gah4 (
talk)
08:45, 11 April 2018 (UTC)reply
Now the formula is x(t-1), but the text says: "the smoothed statistic s(t) is a simple weighted average of the current observation x(t) and the previous smoothed statistic s(t−1)" --
85.212.105.213 (
talk) —Preceding
undated comment added
19:20, 29 September 2019 (UTC)reply
Corrected an Error
I removed an innacuracy that stated that simple exponential smoothing was the same as Brown exponential smoothing. This is not the case; Brown's method is double exponential smoothing. JLT 1451, 16 Dec 1009 (CST) —Preceding
unsigned comment added by
131.10.254.62 (
talk)
Division by 0 in triple exponential smoothing
How does the method cope with scenarios in which either c(t) is 0 or s(t) is 0? The problem occurs in the next time iteration:
Initial value of b_0 in double exponential smoothing
Should the initial value of b[0] be β(x[1] - x[0]) instead of x[1] - x[0] in double exponential smoothing?
I implemented the model in a script and had some issues with the initial value of b[0] being too dominating compared to what it ultimately ended up to be. This caused me to observe a large transient in the beginning of smoothed data. — Preceding
unsigned comment added by
85.76.48.199 (
talk)
09:55, 7 January 2022 (UTC)reply