| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I have been told about the exponential average of n datas
Something like
EA(n) = a Y(t) + (1-a) EA(t-1)
With EA : expon. aver. of the last n datas.
Y(t) data value at time t
Y(t-1) data value the time before (t-1)
a = 2 / (n+1)
Does somebody known the formula which gives EA(t) without EA(t-1) ?
PAscal;
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1783.1 | FORTY2::PALKA | Tue Aug 24 1993 13:05 | 14 | ||
Something like.
i=t
__
\ (t-i)
EA(t) = / a (1-a) Y(i)
--
i=0
Note that EA depends on all the previous Y(t), not just the previous n.
However the effect of the older ones is very small.
Andrew
| |||||
| 1783.3 | <thanks> | EVOAI2::$TRICHET | $ Spawn/nowait dir /output=nl: | Tue Aug 24 1993 14:46 | 7 |
Thanks for quick answer
It is going to be used for financial purpose
as MACD theory (Moving Average Convergence Divergence)
Pascal
| |||||