4:35 p.m.
Hi,
using current git on Ubuntu 19.04, I obtain wrong values for the
centered moving average:
<hansl>
clear
set verbose off
nulldata 10
setobs 1 1 --time-series
series S = 0
S = S(-1) + 2
scalar y0 = 99
scalar p = 2
# Works
series mavg_p2 = movavg(S, p)
series mavg_p2_y99 = movavg(S, p, 0, y0) # y0 is ignored
# WRONG VALUES!
series cmavg_p2 = movavg(S, p, 1, 99) # y0 is ignored
series cmavg_p4 = movavg(S, 2*p, 1, y0) # y0 is ignored
# Works
series expmavg_p2 = movavg(S, 0.3, 1, y0) # y0 is considered
print S mavg_p2 mavg_p2_y99 cmavg_p2 cmavg_p4 expmavg_p2 -o
</hansl>
Thanks,
Artur
5:29 p.m.
Am 11.09.2019 um 18:35 schrieb Artur Tarassow:
scalar y0 = 99
scalar p = 2
# Works
series mavg_p2 = movavg(S, p)
series mavg_p2_y99 = movavg(S, p, 0, y0) # y0 is ignored
Yes, as it should be for p>1, right?
# WRONG VALUES!
series cmavg_p2 = movavg(S, p, 1, 99) # y0 is ignored
series cmavg_p4 = movavg(S, 2*p, 1, y0) # y0 is ignored
So is it not centered, or are the values wrong in an "unknown" way?
cheers
sven
5:33 p.m.
Am 11.09.19 um 19:29 schrieb Sven Schreiber:
Am 11.09.2019 um 18:35 schrieb Artur Tarassow:
> scalar y0 = 99
> scalar p = 2
>
> # Works
> series mavg_p2 = movavg(S, p)
> series mavg_p2_y99 = movavg(S, p, 0, y0) # y0 is ignored
Yes, as it should be for p>1, right?
Exactly, that's all fine. Just wanted
to show that the returned values
are correct for the standard moving-average case.
> # WRONG VALUES!
> series cmavg_p2 = movavg(S, p, 1, 99) # y0 is ignored
> series cmavg_p4 = movavg(S, 2*p, 1, y0) # y0 is ignored
So is it not centered, or are the values wrong in an "unknown" way?
It's just replicating the input series S but no (centered) moving
averages are computed.
Best,
Artur
5:48 p.m.
Am 11.09.2019 um 19:33 schrieb Artur Tarassow:
Am 11.09.19 um 19:29 schrieb Sven Schreiber:
> Am 11.09.2019 um 18:35 schrieb Artur Tarassow:
>> # WRONG VALUES!
>> series cmavg_p2 = movavg(S, p, 1, 99) # y0 is ignored
>> series cmavg_p4 = movavg(S, 2*p, 1, y0) # y0 is ignored
>
> So is it not centered, or are the values wrong in an "unknown" way?
It's just replicating the input series S but no (centered) moving
averages are computed.
Well, but your S is a linearly increasing sequence, the centered moving
average _will_ be the value in the middle. To me the question is how
actually a centered MA is calculated for an even number of terms -
shouldn't it be the center plus the same "tails" left and right such
that overall it will be an odd number p?
cheers
sven
5:57 p.m.
Am 11.09.19 um 19:48 schrieb Sven Schreiber:
Am 11.09.2019 um 19:33 schrieb Artur Tarassow:
> Am 11.09.19 um 19:29 schrieb Sven Schreiber:
>> Am 11.09.2019 um 18:35 schrieb Artur Tarassow:
>>> # WRONG VALUES!
>>> series cmavg_p2 = movavg(S, p, 1, 99) # y0 is ignored
>>> series cmavg_p4 = movavg(S, 2*p, 1, y0) # y0 is ignored
>>
>> So is it not centered, or are the values wrong in an "unknown" way?
>
> It's just replicating the input series S but no (centered) moving
> averages are computed.
Well, but your S is a linearly increasing sequence, the centered moving
average _will_ be the value in the middle.
Oh dear, you are totally right. I guess
I should call it a day ;-) Here
an example that the centered mov-avg works well:
<hansl>
clear
set verbose off
nulldata 10
setobs 1 1 --time-series
series S = 0
S = S(-1) + 1.5 + normal()
series cmavg_p2 = movavg(S, 2, 1)
print S cmavg_p2 -o
</hansl>
To me the question is how
actually a centered MA is calculated for an even number of terms -
shouldn't it be the center plus the same "tails" left and right such
that overall it will be an odd number p?
This was exactly the question I originally had in mind.
Best,
Artur
6:04 p.m.
Am 11.09.2019 um 19:57 schrieb Artur Tarassow:
Am 11.09.19 um 19:48 schrieb Sven Schreiber:
> To me the question is how
> actually a centered MA is calculated for an even number of terms -
> shouldn't it be the center plus the same "tails" left and right such
> that overall it will be an odd number p?
This was exactly the question I originally had in mind.
Apart from the answer "Use the source, Luke!" here is some evidence:
<hansl>
nulldata 3
series x = 0
x[2] = 2
x[3] = 1
eval (0 + 2 + 1) / 3 # 1
eval 0.25 * 0 + 0.5 * 2 + 0.25 * 1 # 1.25
eval movavg(x, 2, 1) # NA,1.25,NA
</hansl>
So I guess an appropriately weighted average is used.
cheers
sven
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