Re: Averages, weighted, rolling, attenuated?

From: Rick Miller (rmiller_at_yahoo.com)
Date: 10/25/04 

Date: Mon, 25 Oct 2004 18:16:47 +0000 (UTC)

I don't understand all the confusion here.

If you want a 4-period moving average with 10% discount per time
period, then the weights are

x(t-1): 1.000/A
x(t-2): 0.900/A
x(t-3): 0.810/A
x(t-4): 0.729/A

where A is a normalizing factor, which in this case is A = 1.000 +
0.900 + 0.810 + 0.729 = 3.439. The generalization seems obvious to
me.

If your data are x(t-1)=20, x(t-2)=10, x(t-3)=8, x(t-4)=7, then the
exponentially smoothed average is
11.801 = 20(1.000/3.439) + 10(0.900/3.439) + 8(0.810/3.439) +
7(0.729/3.439).

On Fri, 22 Oct 2004 22:52:59 -0700, Top Spin wrote:
>On Fri, 22 Oct 2004 18:55:04 +0000 (UTC), rmiller@yahoo.com (Rick
>Miller) wrote:
>
>>On Fri, 22 Oct 2004 09:23:11 -0700, Top Spin wrote:
>>>The descriptions for exponential smoothing look to me like exactly
>>>what I was calling "attenuated averaging" ... but what seems to be
>>lost is any intuitive sense of the meaning of alpha. If alpha=.9, I
am
>>discounting history by 10%. Does this alpha have some other
intuitive
>>meaning?
>>
>>The formula is x_smooth = {sum of (alpha^k)*x(t-k)} / {sum of
>>(alpha^k)}. Yes, you're discounting history by by 10% (or 1-alpha
>>percent).
>
>I don't think so.
>
>Let's take a simple case where I have just 2 data points:
>
> X1 = 10
> X2 = 20
>
>Clearly the simple mean is (10+20)/2 = 15
>
>Let's say we want to discount history by 50% (.5). That should give
us
>a smoothed average closer to 20 than the simple mean of 15.
>
>Using my formula, I would set F=.5 and would get
>
> A2 = (F * A1 + X2) / (F+1)
> = (.5*10 + 20) / (1.5)
> = 25 / 1.5
> = 16.66666667
>
>I think that is what we would expect. History (10) was discounted by
>50% (yielding 5) and then averaged with the new value (20) yielding
an
>average greater than the simple mean (15).
>
>Now if I repeat this with the exponential smoothing function where
>alpha = .5, I get
>
> S2 = alpha * X2 + (1-alpha) * S1
> = .5 * 20 + .5 * 10
> = 10 + 5
> = 15
>
>This is exactly the simple mean, which, I believe, means that history
>was not discounted at all.
>
>As it turns out, the two formulas are equivalent if we replace alpha
>with (1/(F+1)) or F with (1-alpha)/alpha).
>
>Let's convert the alpha of .5 into F:
>
> F = (1-alpha)/alpha
> = .5/.5
> = 1
>
>So we see that an alpha value of .5 is equivalent to an F value of 1
>which is a zero discounting of history which is just the simple mean.
>
>This is what was puzzling me. Unless there is something that I do not
>understand, always a good bet, my formula is equivalent to the
>standard Simple Exponential Smoothing (SES) formula but it is in
terms
>of F, which is the discounting factor, whereas the SES formula is in
>terms of alpha, which I have no idea what it is.
>
>My fundamental question, then, is why is the standard SES formula in
>terms of alpha (rather than F) and what the heck is alpha?
>
>I can't believe that it is just some arbitrary factor.
>
>
>>No, there isn't any deeper meaning to it. Like Freud once
>>said, sometimes a cigar is just a cigar.
>
>That would be helpful if I were having dreams with phallic symbols,
>but I am only trying to figure out how these infernal smoothing
>functions work.
>
>
>--
>For email, use Usenet-20031220 at spamex.com

Relevant Pages

  • Re: Averages, weighted, rolling, attenuated?
    ... >>The descriptions for exponential smoothing look to me like exactly ... >lost is any intuitive sense of the meaning of alpha. ... Let's say we want to discount history by 50%. ... which is a zero discounting of history which is just the simple mean. ...
    (sci.stat.math)
  • Re: Averages, weighted, rolling, attenuated?
    ... >>discounting history by 10%. ... The SES formula would go like this using an F ... I started by converting equation ...
    (sci.stat.math)