Timeline for What type of time series model would be good?
Current License: CC BY-SA 3.0
27 events
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| Jul 23, 2012 at 23:48 | comment | added | user603 | No: but i think you are confused about basic time series stuff. Try to look for "moving window estimation". Otherwise ask a separate questions. | |
| Jul 23, 2012 at 23:46 | comment | added | Damien | @user602: So we are using 36 windows of size 12 to get the forecasts. To get the projected values for the next 10 time periods, we would need to use 46 windows? | |
| Jul 23, 2012 at 23:22 | comment | added | user603 | @Damien: i'm not even sure what the question means. The important property of the recursive median in this context is that it bulges very little --minimally among all estimator of central tendency, in fact--, when up to width/2-1 of the last width observations have been replaced by arbitrarily data. | |
| Jul 23, 2012 at 23:14 | comment | added | Damien | @user603: Would you say median filters are good to detect any violations of apparent trends in a dataset? | |
| Jul 23, 2012 at 22:32 | comment | added | user603 | yes. But again, you are encouraged to go read the references quoted in the manual | |
| Jul 23, 2012 at 22:22 | comment | added | Damien | @user603: The model uses 1-12 to get 13, 2-13 to get 14, etc...? | |
| Jul 23, 2012 at 22:05 | comment | added | user603 | @Damien: as with all models that allow level shifts, there is no simple linear expression for $y_{t+k}|y_t$: you have to recursively fit the forecast in the model to get a new forecast. After 12 periods the forecast will be a constant.because the model only uses the last 12 observations to build a forecast. This is the "width" parameter. But again, this is pretty simple to do in R from the code i posted (it's just a loop). | |
| Jul 23, 2012 at 22:01 | comment | added | Damien | @user603: How would we get the next predicted observation after the last observed one? What about the next one after that....etc...? 14.5 is just the predicted value at t=48. But we already have observed this value. What about the predicted value at t=49? | |
| Jul 23, 2012 at 21:59 | comment | added | user603 | @Damien: no: extrapolate=TRUE only concerns the data for which we don't have a model. Since online is TRUE, the extrapolation only affects the first 11 observations for which we don't have a model --and which would otherwise be coded as NA--. | |
| Jul 23, 2012 at 21:57 | vote | accept | Damien | ||
| Jul 23, 2012 at 21:55 | comment | added | Damien | @user602: It seems that all the value in mod4a$level[,1] are the square root of the forecasted values up to the last data point we have. But if we wanted to extrapolate, we could just change extrapolate = TRUE to get the next prediction? | |
| Jul 23, 2012 at 21:51 | comment | added | user603 | it had 48, and the next period forecast would be 14.5 :). But you can do the whole analysis for yourself: R is open source, free and so is the robfilter library and furthermore the methodology is fully explained in the peer reviewed papers referenced in the package's documentations. In a word, it's not a black box and your are encouraged to play with it. | |
| Jul 23, 2012 at 21:49 | history | edited | user603 | CC BY-SA 3.0 |
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| Jul 23, 2012 at 21:47 | comment | added | Damien | @user603: This has 47 and I just used your example for it. Thanks: 2 1 4 5 4 8 7 11 4 4 11 7 10 7 0 19 13 13 11 9 8 16 10 12 9 7 21 9 10 6 7 19 18 9 19 15 14 17 9 10 10 13 15 20 15 12 15 16 | |
| Jul 23, 2012 at 21:47 | comment | added | Damien | @user603: Sorry I realized that you were working off of the old data which had more data. | |
| Jul 23, 2012 at 21:44 | comment | added | Damien | @user603: Also I don't see how i have >100 data points. I only had 37 data points | |
| Jul 23, 2012 at 21:43 | history | edited | user603 | CC BY-SA 3.0 |
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| Jul 23, 2012 at 21:37 | comment | added | user603 | @Damien: it's the last entry of mod4a$level[,1] raised to the power 2 -- since this is based on a model for the square root of your data--. | |
| Jul 23, 2012 at 21:34 | comment | added | Damien | @user603: Where is the final forecast for the next period? | |
| Jul 23, 2012 at 21:16 | comment | added | IrishStat | @Damien The AUTOBOX forecast is 10.8572 ( the robust/outlier adjusted mean of the last 44 values ) | |
| Jul 23, 2012 at 21:00 | comment | added | user603 | @Damien: yes, as long as you use "online=TRUE" this approach can be used for forecasting (we only use the past). The final forcast for the next period is 11...not very different from IrishStats's forcast. | |
| Jul 23, 2012 at 20:55 | history | edited | user603 | CC BY-SA 3.0 |
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| Jul 23, 2012 at 20:41 | comment | added | whuber♦ | Good point about the potential online nature (+1). Another mild improvement can be achieved by analyzing the square roots of the data (because these evidently are counts). Alternatively--for sophisticated analysts--a Poisson GLM with splines or changepoints would do a fine job. | |
| Jul 23, 2012 at 20:41 | comment | added | Damien | @user602: I want to predict data | |
| Jul 23, 2012 at 20:38 | comment | added | user603 | @whuber: --this is a one sided filter: as far as i understood the option "online" makes sure it doesn't use data from $t+i$, $i>0$ at time $t$. More generally, I agree with you: I also tough of asking the OP what was the end purpose (is he, for example, interested in the value of an ar coefficient for a given lag)? | |
| Jul 23, 2012 at 20:33 | comment | added | whuber♦ | This is the right idea, because (a) there is a trend but it's not easily characterized and (b) there are no significant serial correlations at any lag. However, loess will do a much better job than a median filter at characterizing these data. All this begs the question of why the OP is fitting the data: median filters or loess will do little for predicting future values, for instance. | |
| Jul 23, 2012 at 19:11 | history | answered | user603 | CC BY-SA 3.0 |