Hi Ritchie,

On Sat, Apr 26, 2014 at 8:25 PM, Ritchie Lee <[email protected]> wrote:

> Hi friends of NuPIC,
>
> ......However, looking closely why does the anomaly score only spike at
> the end of the first and second anomalies?  Why not spike at the beginning
> (at least) and throughout the anomaly?
>

Just to check, your anomalies are introduced at {6000, 6500, 7000, 7500)
for about 100 steps, right?
Looks like (un)lucky coincidence to me, but it seems at times 6000, 6500
you introduce the anomaly "at the middle of the wave", where sin(x)=0, so
the first first anomaly step is actually not an anomaly, the for 99 steps
same value is not an anomaly either,
only the jump at the end (from 0 to -1) triggers an anomaly. (#1)

At 7000, 7500 thhe introduced anomaly might have cought the predicted sine
at a different phase, triggering anomaly right off (and multiple times?)
...if this is true, Nupic already outsmarted us! :)


> Also, why is the prediction so good, and residual so small at the
> beginning?
>

Answer to this can be found in Nick Mitri's email: [nupic-discuss]
Confusion about shifted predictions
(#2) At the beginning, almost all predictions fail, and CLA returns
last-seen value. In your example you have 1000 steps per 5 "full sines",
thus the resolution is pretty high, change is small -> residual is actually
better at the beginning
than after learning :)


It would be interesting to see a rerun of your experiment with much smaller
resolution (100, 10 steps per period?).

My question:

is the "perfect predictor for unlearned" on high resolutions actually a
feature, or (misleading) bug? (#3)

Also, anomaly is "actual(T)-predicted(T-1)/..."; now, what is fed as
input(i) when learning=OFF? actual(i), or predicted(i-1)? I think it's
actual(i) but no weight changes are stored. But for our example, it should
be predicted(i-1).
Example: sine; introduce anomaly as a line at level 0, at the phase where
sine and 0 meet. Then as in #1 the 1st point is not anomaly, on step 2, the
context is unknown (new), so prediction would give 0 according to #2
(actual(i-1)) which leads
to 0 anomaly score! (as "bug" in #3).


Cheers, Mark
-- 
Marek Otahal :o)
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