RE: vectorized moving average?

[Tim Rueth]

The last instruction with "long_ma" should have read: "avg = avg(n+1 :
end);" which effectively trims off the computed values from negative time.
But, as you say, it looks like I didn't need to do this because the history
is completely captured in avg(1) = data(1), so no need to compute a "run-in"
time.  Thanks Francesco.

Sherman had found that I can set the initial condition by specifying a 4th
parameter in filter() equal to the first data point.  I tried this, and got
very similar (but not quite exact) results when compared to the for-loop
below with no negative time values.  But this small difference dissipated
within "ndays" and isn't a big deal.  Thanks Sherman.

In summary, to calculate the exponential moving average of "data" for
"ndays", the following code:

 alpha = 2/(ndays+1);
 n = length(data);
 avg = zeros(n,1);
 avg(1) = data(1);
 for i = 2 : n
     ao = avg(i-1);
     avg(i) = ao + alpha*(data(i) - ao);
 endfor;

...is close, but not quite equal to:

 alpha = 2/(ndays+1);
 avg = filter(alpha, [1 alpha-1], data, data(1));

...for roughly the first ndays of avg.

--Tim


> -----Original Message-----
> From: Francesco Potortì [mailto:address@hidden 
> Sent: Wednesday, May 12, 2010 11:22 PM
> To: address@hidden
> Cc: 'Octave-ML'; 'James Sherman Jr.'
> Subject: Re: vectorized moving average?
> 
> >Your filter code below works just fine when compared to what 
> I had been 
> >doing, except for a number of initial days, due to what values are 
> >assumed in negative time.  I had been using the following code:
> > 
> ># "ndays" is the number of days to be used when computing the 
> >exponential moving average of "data" (data is a column vector)
> > data = [repmat(data(1), ndays, 1); data];   # repeat 
> data(1) ndays times at
> >the beginning of data for negative time values  alpha = 
> 2/(ndays+1);  n 
> >= length(data);  avg = zeros(n,1);
> > avg(1) = data(1);
> 
> The above instruction is all you need to "invent" past memory 
> for negative values.  You should do the same for the filter 
> function, but I could not say how to do it offhand.
> 
> > for i = 2 : n
> >     ao = avg(i-1);
> >     avg(i) = ao + alpha*(data(i) - ao); endfor
> > 
> ># trim off run-in period for negative time values
> >   long_ma = long_ma(lma_days+1 : end);
> 
> I don't understand the above instruction.  What is long_ma?
> 
> >For small values of ndays, the number of initial days where 
> there's a 
> >discrepancy with your filter() implementation is minimal, but for 
> >larger values of ndays, the number of initial days of 
> discrepancy grows 
> >(obviously, due to the nature of an exponential MA having a 
> long-tail 
> >memory).  Note, I add similar negative time values to the 
> front of the 
> >vector when using
> >filter() as well.  I'm just not sure what is the convention when it 
> >comes to calculating exponential moving averages for points 
> in "data" where "ndays"
> >reaches back into negative time.  Thanks again.
> 
> -- 
> Francesco Potortì (ricercatore)        Voice: +39 050 315 
> 3058 (op.2111)
> ISTI - Area della ricerca CNR          Fax:   +39 050 315 2040
> via G. Moruzzi 1, I-56124 Pisa         Email: address@hidden
> (entrance 20, 1st floor, room C71)     Web:   http://fly.isti.cnr.it/
>