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<DIV><FONT color=#000000 size=2>Hi Laurent,</FONT></DIV>
<DIV><FONT color=#000000 size=2>try 1-800-option1.Would you care to discuss your
option pricing model privately?</FONT></DIV>
<DIV><FONT color=#000000 size=2>Best regards,</FONT></DIV>
<DIV><FONT color=#000000 size=2>gordon</FONT></DIV></BODY></HTML>
</x-html>From ???@??? Tue Mar 16 08:34:40 1999
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From: "Walter Lake" <wlake@xxxxxxxxx>
To: "Metastock bulletin board" <metastock@xxxxxxxxxxxxx>
Subject: moving averages in excel
Date: Tue, 16 Mar 1999 10:32:01 -0500
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Hi Cab
I haven't "cracked" this puppy open yet.
http://home.earthlink.net/~jfritch/TA.html
Was waiting until the data download stuff was finished and Dale might have a
chance to explain whether or not we had to use Quotes Plus to look at the
indicators. Probably not, but just being careful (more like intimidated).
check out the following from Mr. Fritch's web site
"... Procedures for calculation of technical indicator values and trading
signals are found in the Indicators module. "
and
"... Exponential Moving Average. The EMA of "full cycle period" n is
mathematically equivalent to Wilder's modified moving average of "half-cycle
period" N, where n = 2N-1, or, equivalently, N = (n+1)/2. Therefore this EMA
procedure is used instead of Wilder's modified moving average in calculation
of the RSI and DMI indicator values. This is the reason why TA.xls requires
RSI and DMI indicator periods to be input as "full cycle" periods (typically
27 days) instead of the traditional "half-cycle" periods (typically 14
days). For a full-cycle EMA period of n, the percentage of the EMA value
attributable to prices before half-cycle period (n+1)/2 is
100%*[(n-1)/(n+1)]^[(n+1)/2], which has values of 29.6, 31.6, 32.7, 33.5,
36.4,and 36.8% for n = 5, 7, 9, 11, 100, and infinity. The percentage of the
EMA value attributable to prices before full-cycle period n is
100%*[(n-1)/(n+1)]^n, which has values of 13.2-13.5% for n = 5 to infinity.
The conventional procedure of setting the first day's EMA equal to the first
day's value and then using the same recursive EMA equation on all subsequent
days results in the ratio of the first and second days' contributions being
(n-1)/2 instead of the correct value of (n-1)/(n+1). Thus the first day's
contribution is a factor of (n+1)/2 higher than it should be, which can skew
EMA values significantly, even beyond the EMA period n.
To overcome this problem, the EMA procedure here applies an adjustment
factor during the first days up to and including completion of the first EMA
period. With this adjustment factor, the weights for the variable values
always total to exactly one and always are in appropriate proportion, even
for NumDay = 1 to n (the EMA period). This adjustment factor is applied as a
temporary fix to EMA for NumDay up to EMA period n (for StartDay = 1), when
it is applied as a permanent and final fix to Sx."
Best regards
Walter
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